:: Fermat's Little Theorem via Divisibility of {N}ewton's Binomial
:: by Rafa{\l} Ziobro
::
:: Received June 30, 2015
:: Copyright (c) 2015-2016 Association of Mizar Users
:: (Stowarzyszenie Uzytkownikow Mizara, Bialystok, Poland).
:: This code can be distributed under the GNU General Public Licence
:: version 3.0 or later, or the Creative Commons Attribution-ShareAlike
:: License version 3.0 or later, subject to the binding interpretation
:: detailed in file COPYING.interpretation.
:: See COPYING.GPL and COPYING.CC-BY-SA for the full text of these
:: licenses, or see http://www.gnu.org/licenses/gpl.html and
:: http://creativecommons.org/licenses/by-sa/3.0/.
environ
vocabularies ABIAN, NUMBERS, NAT_1, INT_1, FINSEQ_1, ARYTM_3, XXREAL_0,
CARD_1, ARYTM_1, INT_2, COMPLEX1, RELAT_1, REAL_1, XCMPLX_0, XREAL_0,
ORDINAL1, ORDINAL4, NEWTON, POWER, PREPOWER, FUNCT_1, CARD_3, SUBSET_1,
TARSKI, RFINSEQ, XBOOLE_0, REALSET1, PARTFUN1;
notations TARSKI, XBOOLE_0, SUBSET_1, ORDINAL1, XCMPLX_0, XREAL_0, NUMBERS,
XXREAL_0, ABIAN, POWER, INT_1, INT_2, NEWTON, ABSVALUE, PREPOWER,
RELAT_1, FUNCT_1, RECDEF_1, PARTFUN1, RFINSEQ, RVSUM_1, FINSEQ_1,
IRRAT_1;
constructors ABIAN, SQUARE_1, NAT_D, NEWTON, POWER, PREPOWER, RFINSEQ,
WSIERP_1, RECDEF_1, RELSET_1;
registrations ABIAN, ORDINAL1, XXREAL_0, XREAL_0, NAT_1, INT_1, NEWTON,
XCMPLX_0, WSIERP_1, POWER, NEWTON01, NAT_6, FUNCT_1, NUMBERS, FINSEQ_1,
XBOOLE_0, VALUED_0, CARD_1, RVSUM_1, RELAT_1, NAT_3, INT_6, LAGRA4SQ;
requirements REAL, NUMERALS, BOOLE, SUBSET, ARITHM;
equalities FINSEQ_1;
expansions ABIAN, INT_1, INT_2, FINSEQ_1;
theorems INT_2, NAT_1, ABSVALUE, INT_1, EULER_1, PREPOWER, NEWTON, XCMPLX_1,
GR_CY_3, XREAL_1, XXREAL_0, NAT_D, ABIAN, WSIERP_1, XREAL_0, COMPLEX1,
POWER, RADIX_2, INT_4, NEWTON01, NAT_5, INT_6, TAYLOR_2, NAT_6, PEPIN,
INT_5, PYTHTRIP, NAT_3, SERIES_4, NUMERAL2, SEQ_2, FINSEQ_1, FUNCT_1,
RFINSEQ, FINSEQ_2, RVSUM_1, FINSEQ_3, RELAT_1, FINSEQ_5, ORDINAL1,
IRRAT_1, GROUP_17, NAT_4, ASYMPT_1, PARTFUN1, MATRIXC1, TOPREALC;
schemes NAT_1;
begin :: Divisibility of Newton's Binomial
reserve a,b,c,d,m,x,n,j,k,l for Nat,
t,u,v,z for Integer,
f,F for FinSequence of NAT;
reserve p,q,r,s for real number;
registration
let a be Complex;
reduce 1*a|^0 to 1;
reducibility by NEWTON:4;
end;
registration
let n be non zero Nat;
reduce 0|^n to 0;
reducibility by NEWTON:84;
end;
registration
let a be Nat;
reduce |.a.| to a;
reducibility by ABSVALUE:29;
end;
registration
let a be Nat;
reduce a gcd 0 to a;
reducibility by NEWTON:52;
end;
registration
let t,z;
reduce (t mod z) mod z to t mod z;
reducibility by NUMERAL2:14;
end;
registration
let t;
reduce 0 mod t to 0;
reducibility by INT_4:12;
end;
registration
let u,z;
reduce (0 + u*z) mod z to 0;
reducibility
proof
thus (0 + u*z) mod z = 0 mod z by EULER_1:12
.= 0;
end;
end;
registration
let r be non zero Real;
let n be even natural number;
cluster r |^ n -> positive;
coherence
proof
per cases;
suppose r > 0; then
r|^n > 0 by PREPOWER:6;
hence thesis;
end;
suppose r < 0; then
A1: |.r.| = -r & -r > 0 by ABSVALUE:def 1; then
r|^n = ((-1)*|.r.|)|^n
.= (-1)^n *|.r.||^n by NEWTON:7
.= 1*|.r.|^n;
hence thesis by A1,PREPOWER:6;
end;
end;
end;
:: some remarks on divisibility of the differences of like powers
theorem Th1:
t gcd z = -t gcd z
proof
t gcd z = |.t.| gcd |.z.| by INT_2:34
.= |.-t.| gcd |.z.| by COMPLEX1:52
.= -t gcd z by INT_2:34;
hence thesis;
end;
:: copied from GCD_1:3;
LmGCD: for a,b,c,d be Integer holds
b divides a & d divides c implies (b * d) divides (a * c)
proof
let a,b,c,d be Integer;
assume that
A1: b divides a and
A2: d divides c;
consider t such that
A3: b * t = a by A1;
consider z such that
A4: d * z = c by A2;
(b * d) * (t * z) = a * c by A3,A4;
hence thesis;
end;
::see also NAT_3:1
theorem
t divides z & u divides v implies t*u divides z*v by LmGCD;
theorem Th3:
t divides z iff t gcd z = |.t.|
proof
|.t.| in NAT & |.z.| in NAT by INT_1:3; then
consider k,l such that
A1: k = |.t.| & l = |.z.|;
k divides l iff k gcd l = k by NEWTON:49,INT_2:def 2;
hence thesis by A1,INT_2:34,INT_2:16;
end;
theorem
t*u divides z*u iff |.u.|*(t gcd z) = |.u.|*|.t.|
proof
A1: t*u divides z*u iff t*u gcd z*u = |.t*u.| by Th3;
t*u gcd z*u = |.u.|*(t gcd z) by INT_6:16;
hence thesis by A1,COMPLEX1:65;
end;
Lm0c: (a+u*b) gcd b = a gcd b
proof
b divides u*b; then
A0: b gcd u*b = |.b.| by Th3 .= b;
per cases;
suppose a <> 0;
hence thesis by EULER_1:16;
end;
suppose
a = 0;
hence thesis by A0;
end;
end;
Lm0d: (t+u*z) gcd z = t gcd z
proof
A0: z <= 0 implies (t + u*z) gcd z = t gcd z
proof
assume
A1: z <= 0; then
A1a: -z in NAT by INT_1:3;
per cases;
suppose t>=0; then
A2: t in NAT & -z in NAT by A1,INT_1:3;
(t+u*z) gcd z = t + (-u)*(-z) gcd -z by Th1
.= t gcd -z by A2,Lm0c;
hence thesis by Th1;
end;
suppose t<0; then
A3: -t in NAT by INT_1:3;
(t + u*z) gcd z = -(-t-u*z) gcd -z by Th1
.= (-t + u*(-z)) gcd -z by Th1
.= -t gcd -z by A1a,A3,Lm0c
.= -t gcd z by Th1
.= t gcd z by Th1;
hence thesis;
end;
end;
z > 0 implies (t + u*z) gcd z = t gcd z
proof
assume
A1: z > 0; then
A1a: z in NAT by INT_1:3;
per cases;
suppose t>=0; then
t in NAT & z in NAT by A1,INT_1:3;
hence thesis by Lm0c;
end;
suppose t<0; then
A3: -t in NAT by INT_1:3;
(t + u*z) gcd z = -(-t-u*z) gcd z
.= (-t + (-u)*z) gcd z by Th1
.= -t gcd z by A1a,A3,Lm0c
.= t gcd z by Th1;
hence thesis;
end;
end;
hence thesis by A0;
end;
theorem Th5:
(t+u*z) gcd z = t gcd z & (t-u*z) gcd z = t gcd z
proof
(t-u*z) gcd z = t gcd z
proof
(t-u*z) gcd z = ((t-u*z) + u*z) gcd z by Lm0d
.= t gcd z;
hence thesis;
end;
hence thesis by Lm0d;
end;
Lm1: b>0 & k>0 & a+b > k & a+b divides k*a|^n implies not a,b are_coprime
proof
assume
A1: b>0 & k>0 & a+b > k & a+b divides k*a|^n;
defpred P[Nat] means a+b divides k*a|^$1 implies not a,b are_coprime;
L1: P[0]
proof
k*a|^0 = k*1 by NEWTON:4;
hence thesis by A1,INT_2:27;
end;
L2: P[x] implies P[x+1]
proof
assume
B1: P[x];
B2: a|^(x+1) = a|^x*a by NEWTON:6;
a,b are_coprime & not (a+b) divides k*a|^x implies
not (a+b) divides k*a|^(x+1)
proof
assume
C1: a,b are_coprime & not (a+b) divides k*a|^x; then
1 = b+1*a gcd a by A1,EULER_1:16; then
a+b,a are_coprime; then
not a+b divides k*a|^x*a by C1,INT_2:25;
hence thesis by B2;
end;
hence thesis by B1;
end;
for c be Nat holds P[c] from NAT_1:sch 2(L1,L2);
hence thesis by A1;
end;
Lm2: t divides ((t+z)|^x - z|^x) implies t divides (t+z)|^x*z - z|^(x+1)
proof
assume t divides ((t+z)|^x - z|^x); then
t divides ((t+z)|^x - z|^x)*z by INT_2:2; then
t divides (t+z)|^x*z - z|^x*z;
hence thesis by NEWTON:6;
end;
theorem Th6:
n>0 implies t divides t|^n
proof
t in INT by INT_1:def 2; then
consider k such that
A1: t = k or t = -k by INT_1:def 1;
t|^n = k|^n or t|^n = -k|^n by A1,POWER:1,2; then
A2: |.k.| = |.t.| & |.k|^n.| = |.t|^n.| by A1,COMPLEX1:52;
assume n>0; then
k|^1 divides k|^n by NEWTON:89,NAT_1:14;
hence thesis by A2,INT_2:16;
end;
Lm5a: a is odd & a gcd b = 1 implies a gcd 2*b = 1
proof
assume
A1: a is odd & a gcd b = 1; then
A2: a,b are_coprime;
a,2|^1 are_coprime by A1,NAT_5:3; then
a,2*b are_coprime by A2,INT_2:26;
hence thesis;
end;
Lm5b: a is even & a gcd b = 1 implies a gcd 2*b = 2
proof
assume
A1: a is even & a gcd b = 1;
a,b are_coprime by A1; then
a gcd (b*2) = a gcd 2 by INT_6:19;
hence thesis by A1,NEWTON:49;
end;
Lm5: b,c are_coprime implies 2*b + c gcd c <= 2
proof
assume
A1: b,c are_coprime;
A3: (2*b + c*1) gcd c = 2*b gcd c by EULER_1:8;
per cases;
suppose
c is odd;
then 2*b gcd c = 1 by A1,Lm5a;
hence thesis by A3;
end;
suppose
c is even;
hence thesis by A1,A3,Lm5b;
end;
end;
Lm6: a>0 & a = (a gcd b)*k & b = (a gcd b)*l implies k gcd l = 1
proof
assume
A1: a>0 & a = (a gcd b)*k & b = (a gcd b)*l; then
A2: a gcd b > 0;
k > 0 by A1; then
(a gcd b)*1 = (a gcd b)*(k gcd l) by A1,EULER_1:15;
hence thesis by XCMPLX_1:5,A2;
end;
theorem Th7:
a|^n gcd b|^n = (a gcd b)|^n
proof
a gcd b = k implies a|^n gcd b|^n = k|^n
proof
assume
A0: a gcd b = k; then
consider l be Nat such that
A2: a = k * l by INT_2:21,NAT_D:def 3;
consider m be Nat such that
A3: b = k * m by A0,INT_2:21,NAT_D:def 3;
per cases;
suppose
A4: a > 0; then
l gcd m = 1 by A0,A2,A3,Lm6; then
A6: l|^n gcd m|^n = 1 by WSIERP_1:12;
a10: l > 0 by A2,A4;
k|^n = (k|^n)*(l|^n gcd m|^n) by A6
.= (k|^n)*(l|^n) gcd (k|^n)*(m|^n) by a10,EULER_1:15
.= k|^n*l|^n gcd (k*m)|^n by NEWTON:7
.= a|^n gcd b|^n by A2,A3,NEWTON:7;
hence thesis;
end;
suppose
A14: a = 0; then
A16: a|^n = 0 or n < 1 by NEWTON:11;
n = 0 implies a|^n = 1 & b|^n=1 & k|^n = 1 by NEWTON:4;
hence thesis by A0,A14,A16,NAT_1:14,NAT_D:32;
end;
end;
hence thesis;
end;
theorem Th8:
a > b & a,b are_coprime implies (a+b) gcd (a-b) <= 2
proof
assume
A1: a > b & a,b are_coprime; then
consider c such that
A2: a = b + c by NAT_1:10;
c+1*b,b are_coprime by A1,A2; then
c,b are_coprime by Th5; then
2*b + c gcd c <= 2 by Lm5;
hence thesis by A2;
end;
theorem Th9:
t gcd z is even iff t is even & z is even
proof
t gcd z is even implies t is even & z is even
proof
assume
A1: t gcd z is even;
t gcd z divides t & t gcd z divides z by INT_2:21;
hence thesis by A1;
end;
hence thesis by INT_2:def 2;
end;
theorem Th10:
t divides ((t+z)|^n - z|^n) & z divides ((t+z)|^n - t|^n)
proof
defpred P[Nat] means t divides ((t+z)|^$1 - z|^$1)
& z divides ((t+z)|^$1 - t|^$1);
A1: P[0]
proof
(t+z)|^0 = 1 & z|^0 = 1 & t|^0 = 1 by NEWTON:4;
hence thesis by INT_1:def 4,INT_1:11;
end;
A2: P[x] implies P[x+1]
proof
assume
A2a: P[x]; then
A4: t divides (t+z)|^x*z -z|^(x+1) by Lm2;
A5: z divides (t+z)|^x*t -t|^(x+1) by Lm2,A2a;
A6: t divides (t+z)|^x*t & z divides (t+z)|^x*z; then
A7: t divides (t+z)|^x*t+((t+z)|^x*z - z|^(x+1)) by WSIERP_1:4,A4;
A3: (t+z)|^(x+1) = (t+z)|^x*(t+z)|^1 by NEWTON:8
.= (t+z)|^x*t+(t+z)|^x*z;
z divides (t+z)|^x*z+((t+z)|^x*t - t|^(x+1)) by WSIERP_1:4,A5,A6;
hence thesis by A3,A7;
end;
for m holds P[m] from NAT_1:sch 2(A1,A2);
hence thesis;
end;
theorem Th11:
u divides (u+z)|^n iff u divides z|^n
proof
A0: u divides (u+z)|^n - z|^n by Th10;
consider t such that
A1: t = - z|^n;
A2: u divides (u+z)|^n + t by A0,A1; then
A3: u divides t implies u divides (u+z)|^n by INT_2:1;
u divides (u+z)|^n implies u divides t by INT_2:1,A2;
hence thesis by A1,A3,INT_2:10;
end;
theorem
t divides (t+z)|^n implies t divides (t+z)|^n + z|^n
proof
assume
A0: t divides (t+z)|^n;
then t divides z|^n by Th11;
hence thesis by WSIERP_1:4,A0;
end;
theorem
(t+u) divides ((t+2*u)|^n - u|^n)
proof
(t+u) divides ((t+u+u)|^n -u|^n) by Th10;
hence thesis;
end;
theorem Th14:
l>0 & t divides z implies t divides z|^l
proof
assume
A0: l>0 & t divides z;
then consider n be Nat such that
A1: l = 1+n by NAT_1:10,14;
z|^(1+n) = z|^1*z|^n by NEWTON:8
.= z*z|^n;
hence thesis by A0,A1,INT_2:2;
end;
theorem Th15:
t divides z implies t|^n divides z|^n
proof
assume
A1: t divides z;
|.t.| in NAT & |.z.| in NAT by INT_1:3; then
consider a,b such that
A2: a = |.t.| & b = |.z.|;
A3: |.t.||^n = |.t|^n.| & |.z.||^n = |.z|^n.| by TAYLOR_2:1;
(a gcd b)|^n = |.t.||^n gcd |.z.||^n by A2,Th7
.= t|^n gcd z|^n by A3,INT_2:34; then
t|^n gcd z|^n = |.a.||^n by A1,A2,INT_2:16,Th3
.= |.t|^n.| by A2,TAYLOR_2:1;
hence thesis by Th3;
end;
theorem
n>0 & not t divides (t+z)|^n implies not t divides z
proof
assume n>0 & not t divides (t+z)|^n;
then not t divides t+z by Th14;
hence thesis by WSIERP_1:4;
end;
theorem Th17:
m>0 implies t*z divides (t+z)|^m - (t|^m + z|^m)
proof
assume m>0; then
consider n such that
A0: m = 1+n by NAT_1:14,10;
t*z divides (t+z)|^(n+1) - (t|^(n+1) + z|^(n+1))
proof
defpred P[Nat] means
t*z divides ((t+z)|^($1+1) - (t|^($1+1) + z|^($1+1)));
A1: P[0] by INT_2:12;
A2: P[x] implies P[x+1]
proof
assume
P[x]; then
B2: t*z divides ((t+z)|^(x+1) - (t|^(x+1) + z|^(x+1)))*(t+z) by INT_2:2;
B3: ((t+z)|^(x+1) - (t|^(x+1) + z|^(x+1)))*(t+z) =
(t+z)|^(x+1)*(t+z) - t|^(x+1)*t - z*t|^(x+1) - t*z|^(x+1) - z|^(x+1)*z
.= (t+z)|^(x+1+1) - t|^(x+1)*t- z*t|^(x+1) - t*z|^(x+1) - z|^(x+1)*z
by NEWTON:6
.= (t+z)|^(x+1+1) - t|^(x+1+1)- z*t|^(x+1) - t*z|^(x+1) - z|^(x+1)*z
by NEWTON:6
.= (t+z)|^(x+1+1) - t|^(x+1+1)- z*(t|^x*t) - t*z|^(x+1) - z|^(x+1)*z
by NEWTON:6
.= (t+z)|^(x+1+1) - t|^(x+1+1)- z*t|^x*t - t*(z|^x*z) - z|^(x+1)*z
by NEWTON:6
.= (t+z)|^(x+1+1) - t|^(x+1+1) - z*(t|^x*t) - t*(z|^x*z) - z|^(x+1+1)
by NEWTON:6
.= (t+z)|^(x+1+1) -t|^(x+1+1)- z|^(x+1+1) + t*z*(-t|^x - z|^x);
t*z divides t*z*(-t|^x - z|^x);
hence thesis by B2,B3,INT_2:1;
end;
for m holds P[m] from NAT_1:sch 2(A1,A2);
hence thesis;
end;
hence thesis by A0;
end;
theorem Th18:
t-z divides t|^m - z|^m
proof
(t-z) divides ((t-z)+z)|^n iff (t-z) divides z|^n by Th11;
hence thesis by Th10;
end;
theorem Th19:
n>0 implies t*z divides (t-z)|^n - (t|^n + (-z)|^n)
proof
n>0 implies t*(-z) divides (t+(-z))|^n - (t|^n + (-z)|^n) by Th17; then
n>0 implies -(t*z) divides (t+(-z))|^n - (t|^n + (-z)|^n);
hence thesis by INT_2:10;
end;
theorem Th20:
t*z divides (t+z)|^n - (t-z)|^n + ((-z)|^n - z|^n)
proof
A1:(t+z)|^n - (t|^n + z|^n) =
(t+(-z))|^n - (t|^n + (-z)|^n) + ((t+z)|^n -(t-z)|^n + ((-z)|^n - z|^n));
per cases;
suppose n = 0; then
1*(t+z)|^n - 1*(t-z)|^n + (1*(-z)|^n - 1*z|^n) = 0;
hence thesis by INT_2:12;
end;
suppose n>0; then
t*z divides (t-z)|^n - (t|^n + (-z)|^n) & t*z divides
(t+z)|^n - (t|^n + z|^n) by Th17,Th19;
hence thesis by A1,INT_2:1;
end;
end;
theorem
n>0 implies t divides (t+z)|^n + (t|^n - z|^n)
proof
assume n>0; then
A1: t divides -t|^n by INT_2:10,Th6;
t divides (t+z)|^n - z|^n + t|^n +(-t|^n) by Th10;
hence thesis by A1,INT_2:1;
end;
theorem Th22:
u divides t+z & u divides t-z implies u divides 2*t & u divides 2*z
proof
A0: t + z = (t + t) + (z-t) & t + z = (t-z) + (z + z);
assume
A1: u divides t+z & u divides t-z; then
u divides -(t-z) by INT_2:10;
hence thesis by A0,A1,INT_2:1;
end;
theorem
t*z divides (t+z)|^(2*n) - (t-z)|^(2*n)
proof
(-z)|^(2*n) = z|^(2*n) by POWER:1; then
(-z)|^(2*n) - z|^(2*n) = 0; then
t*z divides (t+z)|^(2*n) - (t-z)|^(2*n) + 0 by Th20;
hence thesis;
end;
theorem
n>0 implies t*z divides (t-z)|^(2*n) - (t|^(2*n) + z|^(2*n))
proof
(-z)|^(2*n) = z|^(2*n) by POWER:1;
hence thesis by Th19;
end;
theorem
t*z divides (t-z)|^(2*n+1) - (t|^(2*n+1) - z|^(2*n+1))
proof
A1: (-z)|^(2*n+1) = -z|^(2*n+1) by POWER:2;
t*z divides (t-z)|^(2*n+1) - (t|^(2*n+1) + (-z)|^(2*n+1)) by Th19;
hence thesis by A1;
end;
theorem
k>0 & x divides a+k & x divides a-k implies x <= 2*k
proof
assume
A1: k>0;
x divides a+k & x divides a-k implies x divides 2*a & x divides 2*k
by Th22;
hence thesis by A1,NAT_D:7;
end;
:: gcd
theorem
k>0 implies a gcd b <= a gcd b*k
proof
assume
k>0; then
b*k = 0 iff b = 0; then
A1: a gcd b*k = 0 implies a gcd b = 0 by INT_2:5;
A2: a gcd b divides b & a gcd b divides a by INT_2:def 2; then
a gcd b divides b*k by INT_2:2;
hence thesis by A1,A2,INT_2:22,27;
end;
theorem
n > 0 implies (a gcd b) gcd b|^n = a gcd b
proof
assume n > 0; then
A2: n >= 1 by NAT_1:14;
(a gcd b) gcd b|^n = a gcd (b gcd b|^n) by NEWTON:48
.= a gcd b|^1 by A2,NEWTON:49,89
.= a gcd b;
hence thesis;
end;
Lm6: t+z,z are_coprime implies t+z,t are_coprime
proof
assume t+z, z are_coprime; then
1 = t+1*z gcd z .= t gcd z by Th5
.= z+1*t gcd t by Th5;
hence thesis;
end;
theorem
t+z,t are_coprime iff t+z,z are_coprime by Lm6;
theorem
a,b are_coprime & a*b = c|^n implies ex k st k|^n = a
proof
assume
A1: a,b are_coprime & a*b = c|^n;
consider k such that
A3: k = a gcd c;
per cases;
suppose
B1: n = 0; then
a = 1|^0 by A1,NAT_1:15,NEWTON:4;
hence thesis by B1;
end;
suppose
n> 0 & a = 0; then
a = a|^n by NEWTON:11,NAT_1:14;
hence thesis;
end;
suppose b = 0; then
a = 1|^n by A1;
hence thesis;
end;
suppose
B1: n > 0 & a > 0 & b > 0; then
consider m such that
B2: n = 1 + m by NAT_1:10,14;
B3: a|^m,b are_coprime by A1,WSIERP_1:10;
k|^n = a|^n gcd c|^n by A3,Th7
.= a|^m*a gcd a*b by A1,B2,NEWTON:6
.= a*1 by B3,B1,EULER_1:15;
hence thesis;
end;
end;
theorem Th31:
for a,b st a,b are_coprime & a + b > 2 holds
a+b divides a|^n+b|^n iff not a+b divides a|^n-b|^n
proof
let a,b such that
A1: a,b are_coprime & a + b > 2;
A2: b > 0
proof
assume not thesis; then
b = 0;
hence contradiction by A1;
end;
A3: not (a+b) divides 2*a|^n by A1,A2,Lm1;
(a+b) divides a|^n-b|^n implies not (a+b divides a|^n+b|^n)
proof
assume not thesis; then
a+b divides (a|^n+b|^n +(a|^n -b|^n)) by WSIERP_1:4;
hence contradiction by A3;
end;
hence thesis by NEWTON01:38;
end;
theorem
a,b are_coprime & a + b > 2 & n is odd implies not a+b divides a|^n-b|^n
proof
assume
A1: a,b are_coprime & a + b > 2 & n is odd; then
consider k such that
A2: n = 2*k+1 by ABIAN:9;
a+b divides a|^n+b|^n by A2,NEWTON01:35;
hence thesis by A1,Th31;
end;
theorem
a,b are_coprime & a + b > 2 & n is even implies not a+b divides a|^n+b|^n
proof
assume
A1: a,b are_coprime & a + b > 2 & n is even; then
a+b divides a|^n-b|^n by NEWTON01:36;
hence thesis by A1,Th31;
end;
theorem Th27:
a,b are_coprime implies a*b,(a|^(n+1) + b|^(n+1)) are_coprime
proof
assume a,b are_coprime; then
a|^(n+1), b|^(n+1) are_coprime by WSIERP_1:11; then
(a|^(n+1)+1*b|^(n+1)) gcd b|^(n+1) = 1 &
(b|^(n+1)+1*a|^(n+1)) gcd a|^(n+1) = 1 by Th5; then
(a|^(n+1)+b|^(n+1))*1, b*b|^n are_coprime &
(a|^(n+1)+b|^(n+1))*1, a*a|^n are_coprime by NEWTON:6; then
(a|^(n+1)+b|^(n+1)), b are_coprime & (a|^(n+1)+b|^(n+1)), a are_coprime
by NEWTON01:41;
hence thesis by INT_2:26;
end;
theorem Th28:
a,b are_coprime implies a*b,(a|^(n+1) - b|^(n+1)) are_coprime
proof
A1: b*b|^n = |.b|^(n+1).| by NEWTON:6;
A2: a*a|^n = |.a|^(n+1).| by NEWTON:6;
A3: |.(a|^(n+1)-b|^(n+1)).| = |.-(a|^(n+1)-b|^(n+1)).| by COMPLEX1:52;
assume a,b are_coprime; then
a|^(n+1), b|^(n+1) are_coprime by WSIERP_1:11; then
(a|^(n+1)-1*b|^(n+1)) gcd b|^(n+1) = 1 & (b|^(n+1)-1*a|^(n+1)) gcd
a|^(n+1) = 1 by Th5; then
|.a|^(n+1)-b|^(n+1).|*1, b*b|^n are_coprime &
|.a|^(n+1)-b|^(n+1).|*1, a*a|^n are_coprime by A1,A2,A3,INT_2:34; then
|.a|^(n+1)-b|^(n+1).|, b are_coprime & |.a|^(n+1)-b|^(n+1).|, a
are_coprime by NEWTON01:41; then
|.a|^(n+1)-b|^(n+1).|,a*b are_coprime by INT_2:26;
hence thesis by INT_6:14;
end;
theorem Th29:
q>0 & n>0 implies ex r st q = r|^n
proof
assume q>0 & n>0; then
q>0 & n>=1 by NAT_1:14; then
A2: q = (n -Root q)|^n by PREPOWER:19;
n -Root q in REAL by XREAL_0:def 1;
hence thesis by A2;
end;
theorem
k>0 & a+b > k & a+b divides k*a implies not a,b are_coprime
proof
assume
A1: k>0 & a+b > k & a+b divides k*a;
per cases;
suppose
B1: b > 0;
B2: not a+b, a are_coprime by A1,INT_2:25,27;
assume not thesis; then
1 = b+1*a gcd a by B1,EULER_1:16;
hence contradiction by B2;
end;
suppose
b = 0;
hence thesis by A1,NAT_1:14;
end;
end;
theorem
k>1 implies not k divides (k+1)|^n
proof
assume k>1; then
not k divides 1|^n by NAT_D:7;
hence thesis by Th11;
end;
theorem
a>1 & b>0 & a gcd b = 1 implies not a divides (a+b)|^n
proof
assume
A0: a>1 & b>0 & a gcd b =1;
then a gcd b|^n = 1 by WSIERP_1:12;
then not a divides a gcd b|^n by WSIERP_1:15,A0;
then not a divides b|^n by INT_2:def 2;
hence thesis by Th11;
end;
:: inequalities (see also SERIES_3)
theorem Th40:
for c st c > 0
for r,s be non negative Real holds r < s iff r|^c < s|^c
proof
let c such that
A0: c > 0;
let r,s be non negative Real;
(r < s implies r|^c < s|^c) & (r|^c < s|^c implies r < s)
proof
A0a: r < s implies r|^c < s|^c
proof
assume
A1: r < s; then
s|^c > 0 by PREPOWER:6; then
A2: r = 0 implies s|^c > r|^c by A0,NAT_1:14,NEWTON:11;
r>0 implies r to_power c < s to_power c by A0,A1,POWER:37;
hence thesis by A2;
end;
s = 0 implies s|^c = 0 by A0,NAT_1:14,NEWTON:11; then
r > 0 & s = 0 implies r|^c > s|^c by PREPOWER:6;
hence thesis by A0a,PREPOWER:9;
end;
hence thesis;
end;
theorem
for r,s be non negative Real holds
r >= s implies r|^n >= s|^n
proof
let r,s be non negative Real;
n = 0 implies r|^n = 1 & s|^n = 1 by NEWTON:4;
hence thesis by Th40;
end;
theorem
a>0 & n>0 implies ex r st a|^n + b|^n = r|^n by Th29;
theorem
for a holds ex b st b|^(n+1) <= a & a < (b+1)|^(n+1)
proof
defpred P[Nat] means ex b st b|^(n+1) <= $1 & $1 < (b+1)|^(n+1);
A1: P[0]
proof
0|^(n+1) <= 0 & 0 < (0+1)|^(n+1);
hence thesis;
end;
A2: P[k] implies P[k+1]
proof
assume P[k]; then consider b such that
A3: b|^(n+1) <= k & k < (b+1)|^(n+1);
A4: b + 1 > b + 0 & (b+1)+1 > (b+1) + 0 by XREAL_1:6;
k+1 <= (b+1)|^(n+1) by A3,NAT_1:13; then
per cases by XXREAL_0:1;
suppose
A5: k+1 < (b+1)|^(n+1);
k+1 > b|^(n+1) by A3,NAT_1:13;
hence thesis by A5;
end;
suppose
A5: k+1 = (b+1)|^(n+1); then
k+1 < (b+1+1)|^(n+1) by A4,Th40;
hence thesis by A5;
end;
end;
for x holds P[x] from NAT_1:sch 2(A1,A2);
hence thesis;
end;
theorem
n>0 & a>b & a,b are_coprime implies (a|^n+b|^n) gcd (a|^n -b|^n) <= 2
proof
assume n>0 & a>b & a,b are_coprime; then
a|^n > b|^n & a|^n, b|^n are_coprime by Th40,WSIERP_1:11;
hence thesis by Th8;
end;
theorem
a+b divides c & a,c are_coprime implies a,b are_coprime
proof
assume
A1: a+b divides c & a,c are_coprime; then
consider k such that
A2: c = (a+b)*k by NAT_D:def 3;
1 = a gcd ((b*k) + k*a) by A1,A2
.= a gcd b*k by Th5; then
a*1, k*b are_coprime;
hence thesis by NEWTON01:41;
end;
Lm20: t is even & t,u are_coprime implies u is odd
proof
A1: 2*0+1 is odd;
assume not thesis;
hence contradiction by A1,Th9;
end;
theorem
t,z are_coprime & t,u are_coprime & t is even implies
u + z is even & u - z is even & u*z is odd
proof
assume t,z are_coprime & t,u are_coprime & t is even; then
z is odd & u is odd by Lm20;
hence thesis;
end;
theorem
a,b are_coprime & c is even & a|^n + b|^n = c|^n implies
a+b is even & a-b is even
proof
assume
A1: a,b are_coprime & c is even & a|^n + b|^n = c|^n;
n = 0 implies 1*a|^n +1*b|^n > 1*c|^n; then
(a is even & b is even) or (a is odd & b is odd) by A1;
hence thesis;
end;
theorem
a is even & a,b are_coprime implies (a-b),(a+b) are_coprime
proof
assume
A1: a is even & a,b are_coprime; then
not 2 divides (a gcd b) by NAT_D:7; then
A2: b is odd by A1,INT_2:def 2; then
A3: ((a+b) gcd (a-b)) is odd by A1,Th9;
A4: (a+b) gcd (a-b) <> 0 by A2,INT_2:5;
per cases by A1,A2,XXREAL_0:1;
suppose
a > b; then
(a+b) gcd (a-b) <= 2 by A1,Th8; then
(a+b) gcd (a-b) < 2 by A3,XXREAL_0:1;
hence thesis by A4,NAT_1:23;
end;
suppose
a < b; then
(a+b) gcd (b-a) <= 2 by A1,Th8; then
(a+b) gcd -(b-a) <= 2 by Th1; then
(a+b) gcd (a-b) < 2 by A3,XXREAL_0:1;
hence thesis by A4,NAT_1:23;
end;
end;
theorem
a,b are_coprime implies (a+b),(a*b) are_coprime
proof
assume a,b are_coprime; then
a + 1*b gcd b = 1 & b+1*a gcd a = 1 by Th5;
hence thesis by WSIERP_1:7;
end;
theorem Th50:
not 3 divides a*b implies 3 divides (a+b)*(a-b)
proof
assume not 3 divides a*b; then
not 3 gcd a = |.3.| & not 3 gcd b = |.3.| by INT_2:2,Th3; then
A3: 3,a|^2 are_coprime & 3,b|^2 are_coprime by PEPIN:2,41,WSIERP_1:10; then
3 gcd a|^2 <> |.3.|; then
not a|^2-0,0 are_congruent_mod 3 by Th3; then
A4: a|^2, 1 are_congruent_mod 3 by NAT_6:15;
3 gcd b|^2 <> |.3.| by A3; then
not b|^2-0,0 are_congruent_mod 3 by Th3; then
b|^2, 1 are_congruent_mod 3 by NAT_6:15; then
a|^2 - b|^2, 1-1 are_congruent_mod 3 by A4,INT_1:17;
hence thesis by NEWTON01:1;
end;
Lm33: 3 divides (a+b)*(a-b)+a*b implies 3 divides a
proof
assume
A1: 3 divides (a+b)*(a-b) + a*b;
L1: 3 divides a*b
proof
assume
A2: not 3 divides a*b; then
3 divides (a+b)*(a-b) by Th50;
hence contradiction by A1,A2,INT_2:1;
end;
L2: 3 divides (a+b)*(a-b)
proof
assume
A2: not 3 divides (a+b)*(a-b); then
3 divides a*b by Th50;
hence contradiction by A1,A2,INT_2:1;
end;
assume
C1: not 3 divides a; then
C2: 3 divides b by L1,NAT_6:7,PEPIN:41; then
C3: 3 divides -b by INT_2:10;
per cases by L2,INT_5:7,PEPIN:41;
suppose 3 divides (a+b);
hence contradiction by C1,C2,INT_2:1;
end;
suppose 3 divides (a+(-b));
hence contradiction by C1,C3,INT_2:1;
end;
end;
Lm34: 3 divides (a+b)*(a-b)+a*b implies 3 divides b
proof
assume
A1: 3 divides (a+b)*(a-b) + a*b;
L1: 3 divides a*b
proof
assume
A2: not 3 divides a*b; then
3 divides (a+b)*(a-b) by Th50;
hence contradiction by A1,A2,INT_2:1;
end;
3 divides (a+b)*(a-b)
proof
assume
A2: not 3 divides (a+b)*(a-b); then
3 divides a*b by Th50;
hence contradiction by A1,A2,INT_2:1;
end; then
3 divides a+b or 3 divides a-b by INT_5:7,PEPIN:41; then
L3: 3 divides (a+b) or 3 divides -(a-b) by INT_2:10;
assume
C1: not 3 divides b; then
C2: 3 divides a by L1,NAT_6:7,PEPIN:41; then
C3: 3 divides -a by INT_2:10;
per cases by L3;
suppose 3 divides (a+b);
hence contradiction by C1,C2,INT_2:1;
end;
suppose 3 divides (b+(-a));
hence contradiction by C1,C3,INT_2:1;
end;
end;
theorem Th51:
3 divides (a+b)*(a-b)+a*b iff 3 divides a & 3 divides b
proof
3 divides a & 3 divides b implies 3 divides (a+b) by WSIERP_1:4;
hence thesis by Lm33,Lm34,WSIERP_1:5;
end;
theorem
b|^2 = a*(a-b) implies 3 divides a & 3 divides b
proof
assume b|^2 = a*(a-b); then
0 = b|^2 - a*a + a*b
.= b|^2 - a|^2 + a*b by NEWTON:81
.= (b-a)*(b+a) + a*b by NEWTON01:1; then
3 divides (b-a)*(b+a) + a*b by INT_2:12;
hence thesis by Th51;
end;
theorem
a,b are_coprime implies not 3 divides (a+b)*(a-b)+ a*b
proof
assume a,b are_coprime; then
not 3 divides a or not 3 divides b by PYTHTRIP:def 1;
hence thesis by Th51;
end;
theorem
a > b & a + b >= 2|^(n+1) implies a > 2|^n
proof
assume
A1: a > b & a + b >= 2|^(n+1); then
a + a > a + b by XREAL_1:6; then
2*a > 2|^(n+1) by A1,XXREAL_0:2; then
2*a > 2*2|^n by NEWTON:6;
hence thesis by XREAL_1:66;
end;
theorem Th55: :: see SERIES_3:6
a<>b implies 2*a*b < a|^2 + b|^2
proof
A0: a*a = a|^2 & b*b = b|^2 by NEWTON:81;
assume a <> b; then
(a - b) is non zero real & 2 is even; then
A1: (a - b)|^2 > 0;
assume not thesis; then
2*a*b -2*a*b >= a|^2 + b|^2 - 2*a*b by XREAL_1:9; then
0 >= (a-b)*(a-b) by A0;
hence contradiction by A1,NEWTON:81;
end;
theorem
n>0 & a<>b implies 2*(a|^n)*(b|^n) < a|^(2*n) + b|^(2*n)
proof
A0: (a|^n)|^2 = a|^(2*n) & (b|^n)|^2 = b|^(2*n) by NEWTON:9;
assume n>0 & a<>b; then
n>=1 & (a>b or b>a) by NAT_1:14,XXREAL_0:1; then
a|^n <> b|^n by PREPOWER:10;
hence thesis by A0,Th55;
end;
theorem
b>0 implies ex n st b >= 2|^n & b < 2|^(n+1)
proof
assume b > 0; then
consider a such that
A0: b = 1+a by NAT_1:10,14;
ex n st (a+1) >= 2|^n & (a+1) < 2|^(n+1)
proof
defpred P[Nat] means ex n st ($1+1) >= 2|^n & ($1+1) <2|^(n+1);
(0+1) >= 1*2|^0 & (0+1) < 2|^(0+1); then
A1: P[0];
A2: P[k] implies P[k+1]
proof
assume P[k]; then consider n such that
B1: (k+1) >= 2|^n & k+1 < 2|^(n+1);
per cases;
suppose
C1: k+1+1 < 2|^(n+1);
k+1+1 > k+1+0 by XREAL_1:6; then
k+1+1 >= 2|^n by B1,XXREAL_0:2;
hence thesis by C1;
end;
suppose
C1: k+1+1 >= 2|^(n+1);
2*(k+1) < 2*2|^(n+1) by B1,XREAL_1:68; then
C2: 2*k + 2 < 2|^(n+1+1) by NEWTON:6;
(k+2)+0 <= k+2 + k by XREAL_1:6; then
k+1+1 < 2|^(n+1+1) by C2,XXREAL_0:2;
hence thesis by C1;
end;
end;
for c be Nat holds P[c] from NAT_1:sch 2(A1,A2);
hence thesis;
end;
hence thesis by A0;
end;
:: division by 4
theorem Th58:
for a,b be odd Nat holds (4 divides a + b iff not 4 divides a - b)
proof
let a,b be odd Nat;
a+b is even & a-b is even; then
2 divides a+b & 2 divides a - b; then
consider t,z such that
A3: a + b = 2*t & a - b = 2*z;
A5: t is odd iff z is even
proof
a = t + z & b = t-z by A3;
hence thesis;
end;
A6: 2*(2 gcd z) = |.2.|*(2 gcd z)
.= 2*2 gcd 2*z by INT_6:16;
A7: 2*2 divides (a+b) implies not 2*2 divides (a-b)
proof
assume (2*2) divides a+b; then
|.(2*2).| = 2*2 gcd (2*t) by A3,Th3
.=|.2.|*(t gcd 2) by INT_6:16; then
2*2 gcd 2*z <> |.4.| by A5,A6,Th3;
hence thesis by A3,Th3;
end;
(not 2*2 divides (a+b)) implies 2*2 divides (a-b)
proof
assume not (2*2) divides a+b; then
not |.(2*2).| = 2*2 gcd (2*t) by A3,Th3; then
not |.(2*2).| =|.2.|*(t gcd 2) by INT_6:16; then
2*2 gcd 2*z = |.4.| by A5,A6,Th3;
hence thesis by A3,Th3;
end;
hence thesis by A7;
end;
theorem
b+c gcd b = 1 & c is odd implies 2*b + c gcd c = 1
proof
assume
A1: b+c gcd b =1 & c is odd; then
c + b*1 gcd b = 1; then
A2: c gcd b = 1 by EULER_1:8;
(2*b + c*1) gcd c = 2*b gcd c by EULER_1:8;
hence thesis by A1,A2,Lm5a;
end;
theorem
a + b = k*a + k*b & a*b > 0 implies k = 1
proof
assume
A0: a + b = k*a + k*b & a*b > 0; then
(k-1)*(a + b) = 0; then
a + b = 0 or k -1 = 0; then
a = 0 or k = 1;
hence thesis by A0;
end;
theorem
t*z = t+z implies t=z
proof
assume t*z = t+z; then
0 = (t-1)*(z-1) - 1; then
(t-1 = 1 & z-1 = 1) or (t-1 = -1 & z-1 = -1) by INT_1:9;
hence thesis;
end;
theorem Th62:
(2*n+1)|^2 = 4*n*(n+1)+1
proof
thus (2*n+1)|^2 = (2*n+1)*(2*n+1) by NEWTON:81
.= 4*n*(n+1)+1;
end;
theorem
a is odd & b is odd implies 8 divides a|^2 - b|^2
proof
assume
A0: a is odd & b is odd; then consider k such that
A1: a = 2*k + 1 by ABIAN:9;
consider n such that
A1a: b = 2*n + 1 by A0,ABIAN:9;
(k is even or (k+1) is even) & (n is even or (n+1) is even); then
k*(k+1) - n*(n+1) is even; then
A2: 2*4 divides 4*(k*(k+1) - n*(n+1)) by INT_6:8;
a|^2 - b|^2 = (2*k+1)|^2 -(4*n*(n+1)+1) by A1,A1a,Th62
.= (4*k*(k+1)+1) - (4*n*(n+1)+1) by Th62
.= 4*(k*(k+1) - n*(n+1));
hence thesis by A2;
end;
Lm60:
for a,b,m be odd Nat holds 4 divides a + b implies 4 divides a|^m + b|^m
proof
let a,b,m be odd Nat;
consider n such that
A0: m = 2*n+1 by ABIAN:9;
assume
A1: 4 divides a + b;
a+b divides a|^(2*n+1)+b|^(2*n+1) by NEWTON01:35;
hence thesis by A0,A1,INT_2:9;
end;
theorem Th64:
for a,b be odd Nat holds 4 divides a - b implies 4 divides a|^n-b|^n
proof
let a,b be odd Nat;
assume
A1: 4 divides a - b;
a-b divides a|^n - b|^n by NEWTON01:33;
hence thesis by A1,INT_2:9;
end;
theorem Th65:
for a,b be odd Nat, m be even Nat holds 4 divides a|^(m) - b|^(m)
proof
let a,b be odd Nat, m be even Nat;
consider n such that
L0: m = 2*n by ABIAN:def 2;
L1: 4 divides a + b implies 4 divides a|^(m)-b|^(m)
proof
assume
A1: 4 divides a + b;
a+b divides a|^(m)-b|^(m) by NEWTON01:36,L0;
hence thesis by A1,INT_2:9;
end;
4 divides a - b implies 4 divides a|^(m)-b|^(m) by Th64;
hence thesis by L1,Th58;
end;
theorem
t is even & not 4 divides t implies ex u st u = t/2 & u is odd
proof
assume
A1: t is even & not 4 divides t; then
consider u such that
A2: t = 2*u by ABIAN:11;
not 2*2 divides 2*u by A1,A2; then
u = t/2 & u is odd by A2,LmGCD;
hence thesis;
end;
Lm63: a is even & not 4 divides a implies ex b st b = a/2 & b is odd
proof
assume
A1: a is even & not 4 divides a; then
consider b such that
A2: a = 2*b;
not 2*2 divides 2*b by A1,A2; then
b = a/2 & b is odd by A2,LmGCD;
hence thesis;
end;
theorem Th67:
a is odd & 2|^n divides a*b implies 2|^n divides b
proof
assume
A1: a is odd & 2|^n divides a*b; then
2|^n,a are_coprime by NAT_5:3;
hence thesis by A1,WSIERP_1:29;
end;
theorem Th68:
for a,b,m be odd Nat holds 4 divides a|^m + b|^m iff 4 divides a + b
proof
let a,b,m be odd Nat;
consider n such that
L0: m = 2*n+1 by ABIAN:9;
4 divides a|^(2*n+1)+b|^(2*n+1) implies 4 divides a + b
proof
B2: 2|^2 = 2*2 by NEWTON:81;
A1: 4 divides a|^(2*n) - b|^(2*n) by Th65;
A2: 4 divides (a|^(2*n) - b|^(2*n))*((a|^1 - b|^1)/2) by Th65,INT_2:2;
consider l such that
A3: l = (a|^(2*n)+b|^(2*n))/2 & l is odd by A1,Th58,Lm63;
A4: a|^(2*n+1)+b|^(2*n+1)=((a|^(2*n)+b|^(2*n))*(a|^1+b|^1)+
(a|^(2*n)-b|^(2*n))*(a|^1-b|^1))/2 by NEWTON01:8
.= (a|^(2*n)+b|^(2*n))*(a|^1+b|^1)/2 +
(a|^(2*n)-b|^(2*n))*(a|^1-b|^1)/2;
assume 4 divides a|^(2*n+1)+b|^(2*n+1); then
2|^2 divides ((a|^(2*n) + b|^(2*n))/2)*(a|^1+b|^1) by B2,A2,A4,INT_2:1;
hence thesis by B2,A3,Th67;
end;
hence thesis by L0,Lm60;
end;
theorem
for a,b,m be odd Nat holds 4 divides a-b iff not 4 divides a|^m + b|^m
proof
let a,b,m be odd Nat;
thus 4 divides a-b implies not 4 divides a|^m + b|^m
proof
assume 4 divides a-b; then
not 4 divides a + b by Th58;
hence thesis by Th68;
end;
assume not 4 divides a|^m + b|^m; then
not 4 divides a + b by Th68;
hence thesis by Th58;
end;
theorem
a|^2 + b|^2 = c|^2 implies ex t st b|^2 = (2*a+t)*t
proof
assume a|^2+b|^2 = c|^2; then
b|^2 = c|^2 - a|^2
.= (c-a)*(c+a) by NEWTON01:1
.= (c-a)*((c-a)+2*a);
hence thesis;
end;
theorem
b|^2 = (2*a+t)*t implies ex c st a|^2 + b|^2 = c|^2
proof
assume b|^2 = (2*a+t)*t; then
A1: b|^2 = ((a+t)+ a)*((a+t)-a)
.= (a+t)|^2 - a|^2 by NEWTON01:1;
|.a+t.| in NAT by INT_1:3; then
consider c such that
A2: c =|.a+t.|;
c|^2 = (a+t)|^2 or c|^2 = (-(a+t))|^2 by A2,ABSVALUE:1; then
c|^2 = (a+t)|^2 by WSIERP_1:1;
hence thesis by A1;
end;
Lm40: a is odd & b is odd implies a|^2+b|^2 <> c|^2
proof
a is odd & b is odd & c is even implies a|^2+b|^2 <> c|^2
proof
assume
A1: a is odd & b is odd & c is even; then
4 divides a|^1 - b|^1 or 4 divides (a|^1+b|^1) by Th58; then
4 divides (a|^1-b|^1)*(a|^1+b|^1) by INT_2:2; then
4 divides a|^2 - b|^2 by NEWTON01:1; then
A2: not 4 divides a|^2 + b|^2 by A1,Th58;
2*2 divides c*c by A1,LmGCD;
hence thesis by A2,NEWTON:81;
end;
hence thesis;
end;
theorem
a is odd & b is odd & m is even implies a|^m + b|^m <> c|^m
proof
a is odd & b is odd implies a|^(2*n) + b|^(2*n) <> c|^(2*n)
proof
A1: a|^(2*n) = (a|^n)|^2 & b|^(2*n) = (b|^n)|^2 & c|^(2*n) = (c|^n)|^2
by NEWTON:9;
assume a is odd & b is odd;
hence thesis by A1,Lm40;
end;
hence thesis;
end;
theorem Th73:
t,z|^n are_coprime & n > 0 implies t,z are_coprime
proof
assume
A1: t,z|^n are_coprime & n > 0; then
A2: t divides t|^n & z divides z|^n by Th6;
t gcd z divides t & t gcd z divides z by INT_2:def 2; then
A3: t gcd z divides t|^n & t gcd z divides z|^n by A2,INT_2:2;
t|^n, z|^n are_coprime by A1,WSIERP_1:10;
hence thesis by A3,INT_2:13,22;
end;
theorem Th74:
a,b are_coprime implies
(a+b)|^2 gcd (a|^2 + b|^2 - (n-2)*a*b) = a|^2 + b|^2 - (n-2)*a*b gcd n
proof
A0: a|^2 = a*a & b|^2 = b*b by NEWTON:81;
assume
A1: a,b are_coprime;
A2: (a+b)|^2 = (a+b)*(a+b) by NEWTON:81
.= (a|^2+b|^2 - (n-2)*a*b) + n*a*b by A0;
a|^(0+1)+b|^(0+1),a*b are_coprime by A1,Th27; then
(a+b)|^2,a*b are_coprime by WSIERP_1:10; then
1 = (a+b)|^2 - n*(a*b) gcd a*b by Th5
.=(a|^2+b|^2 -(n-2)*a*b) + n*a*b - n*a*b gcd a*b by A2; then
A4: (a|^2+b|^2 -(n-2)*a*b),(a*b) are_coprime;
(a+b)|^2 gcd (a|^2+b|^2 -(n-2)*a*b) =
n*a*b+1*(a|^2+b|^2 -(n-2)*a*b) gcd (a|^2+b|^2 -(n-2)*a*b) by A2
.= n*(a*b) gcd (a|^2+b|^2 -(n-2)*a*b) by Th5
.= n gcd (a|^2+b|^2 - (n-2)*a*b) by A4,INT_6:19;
hence thesis;
end;
theorem
a,b are_coprime implies (a+b),(a|^2 + b|^2 + a*b) are_coprime
proof
assume
A1: a,b are_coprime;
A2: |. a|^2 + b|^2 - (1-2)*a*b .| in NAT by INT_1:3;
(a+b)|^2 gcd (a|^2 + b|^2 + a*b) =
|. a|^2 + b|^2 - (1-2)*a*b .| gcd |.1.| by A1,Th74
.= 1 by NEWTON:51,A2; then
(a+b)|^2, (a|^2 + b|^2 + a*b) are_coprime;
hence thesis by Th73;
end;
theorem
a,b are_coprime implies
(a-b)|^2 gcd (a|^2 + b|^2 + (n-2)*a*b) = a|^2 + b|^2 + (n-2)*a*b gcd n
proof
A0: a|^2 = a*a & b|^2 = b*b by NEWTON:81;
assume
A1: a,b are_coprime;
A2: (a-b)|^2 = (a-b)*(a-b) by NEWTON:81
.= (a|^2 + b|^2 + (n-2)*a*b) - n*a*b by A0;
a|^(0+1)-b|^(0+1),a*b are_coprime by A1,Th28; then
(a-b)|^2,a*b are_coprime by WSIERP_1:10; then
1 = (a-b)|^2 + n*(a*b) gcd a*b by Th5
.=(a|^2+b|^2 +(n-2)*a*b) - n*a*b + n*a*b gcd a*b by A2; then
A4: (a|^2+b|^2 +(n-2)*a*b),(a*b) are_coprime;
(a-b)|^2 gcd (a|^2+b|^2 +(n-2)*a*b) =
-n*a*b+1*(a|^2+b|^2 +(n-2)*a*b) gcd (a|^2+b|^2 +(n-2)*a*b) by A2
.= (-n)*(a*b) gcd (a|^2+b|^2 +(n-2)*a*b) by Th5
.= (-n) gcd (a|^2+b|^2 + (n-2)*a*b) by A4,INT_6:19;
hence thesis by Th1;
end;
theorem Th77:
a divides k*(a*n+1) iff a divides k
proof
a divides k*(a*n+1) implies a divides k
proof
assume a divides k*(a*n+1); then
A1: a divides k*a*n + k*1;
a divides a*(k*n);
hence thesis by A1,INT_2:1;
end;
hence thesis by INT_2:2;
end;
theorem
for n be positive Nat holds a divides k*(a|^n+1) iff a divides k
proof
let n be positive Nat;
consider k such that
A1: n = 1 + k by NAT_1:10,14;
a|^(1+k) = a*a|^k by NEWTON:6;
hence thesis by A1,Th77;
end;
theorem
for a,b be positive Nat holds a mod b = b mod a implies a = b
proof
let a,b be positive Nat;
per cases by XXREAL_0:1;
suppose
a < b; then
a mod b = a by NAT_D:24;
hence thesis by NAT_D:1;
end;
suppose
b < a; then
b mod a = b by NAT_D:24;
hence thesis by NAT_D:1;
end;
suppose b = a;
hence thesis;
end;
end;
theorem Th80:
k*(a*n+1) mod a = k mod a
proof
thus k*(a*n+1) mod a = (a*(k*n) + k) mod a
.= k mod a by NAT_D:21;
end;
theorem Th81:
for n be positive Nat holds k*(a|^n+1) mod a = k mod a
proof
let n be positive Nat;
consider k such that
A1: n = 1 + k by NAT_1:10,14;
a|^(1+k) = a*a|^k by NEWTON:6;
hence thesis by A1,Th80;
end;
theorem Th82:
for n be positive Nat holds k*(a|^n+1)|^m mod a = k mod a
proof
let n be positive Nat;
per cases by NAT_1:25;
suppose
C1: a > 1;
(a|^n+1)|^m mod a = ((a|^n+1) mod a)|^m mod a by GR_CY_3:30; then
k*(a|^n+1)|^m mod a = k*((1*(a|^n+1) mod a)|^m mod a) mod a by RADIX_2:3
.= k*((1 mod a)|^m mod a) mod a by Th81
.= k*(1|^m mod a) mod a by C1, NAT_D:14
.= k*1 mod a by RADIX_2:3;
hence thesis;
end;
suppose
a = 1; then
k*(a|^n+1)|^m mod a = 0 & k mod a = 0 by RADIX_2:1;
hence thesis;
end;
suppose
a = 0;
hence thesis;
end;
end;
theorem Th83:
for n be positive Nat holds
b*(a|^n+1)|^m + c*(a|^n+1)|^l mod a = b+c mod a
proof
let n be positive Nat;
A1: b*(a|^n+1)|^m mod a = b mod a & c*(a|^n+1)|^l mod a = c mod a by Th82;
b*(a|^n+1)|^m + c*(a|^n+1)|^l mod a
= ((b mod a) + (c mod a))mod a by A1,NAT_D:66
.= b+c mod a by NAT_D:66;
hence thesis;
end;
theorem
for a,n be positive Nat holds
a divides b*(a|^n+1)|^m + c*(a|^n+1)|^l iff a divides b+c
proof
let a,n be positive Nat;
b*(a|^n+1)|^m + c*(a|^n+1)|^l mod a = 0 iff b+c mod a = 0 by Th83;
hence thesis by PEPIN:6;
end;
theorem
|.t.| < a implies t mod a = |.t.| or t mod a = a -|.t.|
proof
assume |.t.| < a; then
A2: t < a & t > -a by SEQ_2:1;
per cases;
suppose
B1: t >= 0; then
t mod a = t by A2,NAT_D:63;
hence thesis by B1,ABSVALUE:def 1;
end;
suppose
B1: t < 0; then
t mod a = a + t by A2,NAT_D:63
.= a - (-t)
.= a -|.t.| by B1,ABSVALUE:def 1;
hence thesis;
end;
end;
theorem
-t mod a = (u*a - (t mod a)) mod a
proof
thus (u*a - (t mod a)) mod a =
((((0+u*a)mod a) -((t mod a)mod a))mod a) by INT_6:7
.= ((0 mod a)-(t mod a))mod a
.= (0-t) mod a by INT_6:7
.= -t mod a;
end;
:: division by 3
Lm3: t mod 3 = 0 or t mod 3 = 1 or t mod 3 = 2
proof
A1: (t mod 3) in NAT by INT_1:3,57;
(t mod 3) mod 3 = t mod 3; then
t mod 3 < 2+1 by A1,NAT_D:1;
hence thesis by A1,NAT_1:22,23;
end;
LmMod: for n be odd Nat holds 2|^n mod 3 = 2
proof
let n be odd Nat;
consider m such that
L0: n = 2*m +1 by ABIAN:9;
defpred P[Nat] means 2|^(2*$1+1) mod 3 = 2;
L1: P[0] by NAT_D:63;
L2: P[k] implies P[k+1]
proof
assume
A1: P[k];
A2: 1 = 1 mod 3 by NAT_D:63
.= 1*3 + 1 mod 3 by NAT_D:21
.= 2*2 mod 3
.= 2|^2 mod 3 by NEWTON:81;
2 = (2*1 mod 3) by NAT_D:63
.= 2|^(2*k+1)*2|^2 mod 3 by A1,A2,NAT_D:67
.= 2|^(2*k+1+2) mod 3 by NEWTON:8
.= 2|^(2*(k+1)+1) mod 3;
hence thesis;
end;
for c be Nat holds P[c] from NAT_1:sch 2(L1,L2);
hence thesis by L0;
end;
theorem Th87:
for n be odd Nat holds t|^n mod 3 = t mod 3
proof
let n be odd Nat;
per cases by Lm3;
suppose
A1: t mod 3 = 0; then
0 = (t mod 3)|^n mod 3;
hence thesis by A1,GR_CY_3:30;
end;
suppose
A1: t mod 3 = 1; then
1 = (t mod 3)|^n mod 3;
hence thesis by A1,GR_CY_3:30;
end;
suppose
A1: t mod 3 = 2; then
2 = (t mod 3)|^n mod 3 by LmMod;
hence thesis by A1,GR_CY_3:30;
end;
end;
theorem Th88:
(t + (u mod z))mod z = (t+u) mod z
proof
thus (t+(u mod z)) mod z = ((t mod z + ((u mod z) mod z)) mod z) mod z
by NAT_D:66
.= (t+u) mod z by NAT_D:66;
end;
LmSum: a+b-c mod 3 = ((a mod 3) + (b mod 3) + 2*(c mod 3)) mod 3
proof
thus (a+b-c) mod 3 = ((a+b-c)+c*3) mod 3 by EULER_1:12
.= (a+(b+2*c)) mod 3
.= ((a mod 3)+((b+2*c) mod 3)) mod 3 by NAT_D:66
.= (((a mod 3) mod 3) +(((b mod 3) + (c+c mod 3)) mod 3)) mod 3 by NAT_D:66
.= (((a mod 3) mod 3) +(((b mod 3) + (((c mod 3)+(c mod 3))
mod 3)) mod 3)) mod 3 by NAT_D:66
.= (((a mod 3) mod 3) +(((b mod 3) + (((c mod 3)+(c mod 3)))) mod 3))
mod 3 by Th88
.= (((a mod 3)) +(((b mod 3) + (((c mod 3)+(c mod 3)))))) mod 3 by Th88
.= ((a mod 3) + (b mod 3) + 2*(c mod 3)) mod 3;
end;
theorem Th89:
for n be odd Nat holds (a+b-c) mod 3 = (a|^n+b|^n-c|^n) mod 3
proof
let n be odd Nat;
A1: (a|^n mod 3) = (a mod 3) & (b|^n mod 3) = (b mod 3) & (c|^n mod 3) =
(c mod 3) by Th87;
(a+b-c) mod 3 = ((a mod 3) + (b mod 3) + 2*(c mod 3)) mod 3 by LmSum
.= (a|^n+b|^n-c|^n) mod 3 by A1,LmSum;
hence thesis;
end;
theorem Th90:
for k be positive Nat holds t mod k = k - 1 iff (t+1) mod k = 0
proof
let k be positive Nat;
thus t mod k = k - 1 implies (t+1) mod k = 0
proof
assume
A1: t mod k = k - 1;
0 = (0 + 1*k) mod k
.= (1 + (t mod k)) mod k by A1
.= (1 + t) mod k by Th88;
hence thesis;
end;
assume
A1: (t+1) mod k = 0;
(k-1)+1 > (k-1)+0 by XREAL_1:6; then
k - 1 = ((k -1) + ((t+1) mod k)) mod k by A1,NAT_D:24
.= ((k -1) + (t +1)) mod k by Th88
.= (t + 1*k) mod k
.= t mod k by EULER_1:12;
hence thesis;
end;
LmABC: 3 divides u + t + z implies 3 divides u*t*z or
(u mod 3 = t mod 3 & t mod 3 = z mod 3)
proof
assume
A1: 3 divides u + t + z; then
A2: (u+t+z) mod 3 = 0 by INT_1:62;
A3: (u+t) mod 3 = ((u mod 3) + (t mod 3)) mod 3 by NAT_D:66;
per cases by Lm3;
suppose u mod 3 = 0; then
3 divides u by INT_1:62; then
3 divides u*(t*z) by INT_2:2;
hence thesis;
end;
suppose
B1: u mod 3 = 1;
per cases by Lm3;
suppose t mod 3 = 0; then
3 divides t by INT_1:62; then
3 divides t*(u*z) by INT_2:2;
hence thesis;
end;
suppose
C1: t mod 3 = 1; then
(u+t) mod 3 = 3-1 by B1,A3,NAT_D:24; then
((u+t)+1) mod 3 = ((u+t)+z) mod 3 by A2,Th90; then
(u+t+z),(u+t+1) are_congruent_mod 3 by NAT_D:64; then
z,1 are_congruent_mod 3; then
z mod 3 = 1 mod 3 by NAT_D:64
.= 1 by NAT_D:14;
hence thesis by C1,B1;
end;
suppose
t mod 3 = 2; then
u+t mod 3 = 0 by B1,A3,NAT_D:25; then
3 divides (u+t) by INT_1:62; then
3 divides z by A1,INT_2:1;
hence thesis by INT_2:2;
end;
end;
suppose
B1: u mod 3 = 2;
per cases by Lm3;
suppose t mod 3 = 0; then
3 divides t by INT_1:62; then
3 divides t*(u*z) by INT_2:2;
hence thesis;
end;
suppose
t mod 3 = 1; then
u+t mod 3 = 0 by B1,A3,NAT_D:25; then
3 divides (u+t) by INT_1:62; then
3 divides z by A1,INT_2:1;
hence thesis by INT_2:2;
end;
suppose
C1: t mod 3 = 2; then
C2: (u+t) mod 3 = (1+3) mod 3 by B1,A3
.= ((1 mod 3) + (3 mod 3)) mod 3 by NAT_D:66
.= 1+(3 mod 3) mod 3 by RADIX_2:2
.= 1+ 0 mod 3 by NAT_D:25
.= 1 by NAT_D:14;
((u+t)+2) mod 3 = ((2 mod 3)+((u+t)mod 3))mod 3 by NAT_D:66
.= 2+1 mod 3 by NAT_D:24,C2
.= ((u+t)+z) mod 3 by A2,NAT_D:25; then
(u+t+z),(u+t+2) are_congruent_mod 3 by NAT_D:64; then
z,2 are_congruent_mod 3; then
z mod 3 = 2 mod 3 by NAT_D:64
.= 2 by NAT_D:24;
hence thesis by B1,C1;
end;
end;
end;
LmAB3: u + t = z implies 3 divides u*t*z or 3 divides u - t
proof
assume u + t = z; then
3 divides (u+t-z)+3*z; then
3 divides u + t + 2*z; then
3 divides u*t*(2*z) or (u mod 3) = (t mod 3) by LmABC; then
3 divides 2*(u*t*z) or u,t are_congruent_mod 3 by NAT_D:64;
hence thesis by PEPIN:41,INT_2:25,28,30;
end;
theorem
a|^2 + b|^2 = c|^2 implies 3 divides a*b*c
proof
A0: a|^2*b|^2*c|^2 = (a*b)|^2*c|^2 by NEWTON:7
.= (a*b*c)|^2 by NEWTON:7;
assume
A1: a|^2 + b|^2 = c|^2; then
per cases by LmAB3;
suppose
3 divides a|^2*b|^2*c|^2;
hence thesis by A0,NAT_3:5,PEPIN:41;
end;
suppose
3 divides a|^2 - b|^2 & not 3 divides a|^2*b|^2*c|^2; then
B2: not 3 divides (a|^2)*(b|^2*c|^2) & not 3 divides (b|^2)*(a|^2*c|^2)
& not 3 divides (c|^2)*(b|^2*a|^2); then
not 3 divides a|^2 & not 3 divides b|^2 & not 3 divides c|^2
by INT_2:2; then
not 3 divides a & not 3 divides b & not 3 divides c by Th14; then
not 3 divides (c*a) by INT_5:7,PEPIN:41; then
3 divides (c+a)*(c-a) by Th50; then
3 divides c|^2 - a|^2 by NEWTON01:1;
hence thesis by A1,B2,INT_2:2;
end;
end;
theorem
for a,n be non zero Nat holds t mod a = z mod a implies
t|^n mod a = z|^n mod a
proof
let a,n be non zero Nat;
assume t mod a = z mod a; then
t|^n,z|^n are_congruent_mod a by NAT_D:64,GR_CY_3:34;
hence thesis by NAT_D:64;
end;
theorem
3 divides t-z implies 3 divides t|^n - z|^n
proof
3 divides (t-z) & (t-z) divides (t|^n - z|^n) implies
3 divides (t|^n - z|^n) by INT_2:9;
hence thesis by Th18;
end;
theorem
for n be odd Nat holds 3 divides (a+b-c) iff 3 divides (a|^n+b|^n-c|^n)
proof
let n be odd Nat;
thus 3 divides (a+b-c) implies 3 divides (a|^n+b|^n-c|^n)
proof
assume 3 divides a+b-c; then
a+b-c mod 3 = 0 by INT_1:62; then
(a|^n+b|^n-c|^n) mod 3 = 0 by Th89;
hence thesis by INT_1:62;
end;
assume 3 divides a|^n+b|^n-c|^n; then
a|^n+b|^n-c|^n mod 3 = 0 by INT_1:62; then
(a+b-c) mod 3 = 0 by Th89;
hence thesis by INT_1:62;
end;
theorem
(t+u-z)|^2,(t|^2+u|^2+z|^2) are_congruent_mod 2
proof
(t-z+u)|^2=t|^2+z|^2+u|^2-2*t*z+2*t*u-2*z*u by SERIES_4:3; then
(t-z+u)|^2 - (t|^2+z|^2+u|^2) = 2*(t*u-t*z-z*u);
hence thesis;
end;
LmTUZ: (t+u+z)|^3,(t|^3+u|^3+z|^3) are_congruent_mod 3
proof
(t+u+z)|^3 = (t|^3+u|^3+z|^3) + (3*t|^2*u+3*t|^2*z) +
(3*u|^2*t+3*u|^2*z)+(3*
z|^2*t+3*z|^2*u)+6*t*u*z by SERIES_4:8; then
(t+u+z)|^3 - (t|^3+u|^3+z|^3) =
3*((t|^2*u+t|^2*z)+ (u|^2*t+u|^2*z) + (z|^2*t+z|^2*u) + 2*t*u*z);
hence thesis;
end;
theorem
(t+u-z)|^3,(t|^3+u|^3-z|^3) are_congruent_mod 3
proof
2*1+1 is odd; then
A1: (-z)|^3 = -z|^3 by POWER:2;
(t+u+(-z))|^3,(t|^3+u|^3+(-z)|^3) are_congruent_mod 3 by LmTUZ;
hence thesis by A1;
end;
theorem
6 divides a|^3 - a
proof
3 divides a*1 or 3 divides (a+1)*(a-1) by Th50; then
A1: 3 divides (a-1)*(a+1)*(a*1) by INT_2:2;
a is even or (a+1) is even; then
A2: a*(a+1)*(a-1) is even;
2*1+1 is odd; then
A3: 3,2|^1 are_coprime by NAT_5:3;
a|^(2+1) - a = a|^2*a - a by NEWTON:6
.= (a|^2 -1|^2)*a*1
.= (a-1)*(a+1)*a*1 by NEWTON01:1; then
|.3.| divides |.a|^3 - a.| & |.2.| divides |.a|^3 - a.|
by A1,A2,INT_2:16; then
|.3*2.| divides |.a|^3 - a.| by PEPIN:4,A3;
hence thesis by INT_2:16;
end;
theorem
for a,b,c be odd Nat holds
3 divides t|^a + t|^b + t|^c
proof
let a,b,c be odd Nat;
A1: t|^a mod 3 = t mod 3 & t|^b mod 3 = t mod 3 & t|^c mod 3 = t mod 3
by Th87;
(t|^a + t|^b + t|^c) mod 3 = (t|^a + t|^b + (t|^c mod 3)) mod 3 by Th88
.= ((t|^c mod 3) + t|^a + t|^b) mod 3
.= ((t|^c mod 3) + t|^a + (t|^b mod 3))mod 3 by Th88
.= ((t|^c mod 3) + (t|^b mod 3) + t|^a)mod 3
.= ((t|^c mod 3) + (t|^b mod 3) + (t|^a mod 3))mod 3 by Th88
.= (0+3*(t mod 3)) mod 3 by A1
.= 0;
hence thesis by INT_1:62;
end;
theorem
2|^m - 1 divides 2|^(2*m+1) - 2 & 2|^m + 1 divides 2|^(2*m+1) - 2
proof
2|^(2*m+1) - 2 = 2*2|^(2*m) - 2 by NEWTON:6
.= (2|^(2*m) - 1|^m)*2
.= ((2|^m)|^2 - (1|^m)|^2)*2 by NEWTON:9
.= ((2|^m) - (1|^m))*((2|^m) + (1|^m))*2 by NEWTON01:1
.= (2|^m - 1)*((2|^m + 1)*2);
hence thesis by INT_1:def 3,INT_2:2;
end;
theorem Th100:
(u + t + z) is even implies u*t*z is even
proof
assume
A1: u + t + z is even;
per cases;
suppose u is even;
hence thesis;
end;
suppose
B1: u is odd;
per cases;
suppose t is even;
hence thesis;
end;
suppose
t is odd; then
z is even by A1,B1;
hence thesis;
end;
end;
end;
theorem
t|^n + u|^n = z|^n implies 2|^n divides (t*u*z)|^n
proof
assume t|^n + u|^n = z|^n; then
t|^n + u|^n +(-z|^n) = 0; then
t|^n + u|^n +(-z|^n) is even; then
t|^n*u|^n*(-z|^n) is even by Th100; then
-(t|^n*u|^n*z|^n) is even; then
t|^n*u|^n*z|^n is even by INT_2:10; then
t is even or u is even or z is even; then
t*u*z is even;
hence thesis by Th15;
end;
theorem
t|^n,t|^m are_congruent_mod t-1
proof
per cases;
suppose n >= m;
then consider k such that
B2: n = m + k by NAT_1:10;
t|^(m+k) = t|^m * t|^k by NEWTON:8; then
t|^(m+k) - t|^m = t|^m*(t|^k-1|^k);
hence thesis by B2,Th18,INT_2:2;
end;
suppose m >= n;
then consider k such that
B2: m = n + k by NAT_1:10;
t|^(n+k) = t|^n * t|^k by NEWTON:8; then
t|^(n+k) - t|^n = t|^n*(t|^k-1|^k); then
(t-1) divides t|^m - t|^n by B2,Th18,INT_2:2; then
(t-1) divides -(-t|^n + t|^m) by INT_2:10;
hence thesis;
end;
end;
begin :: Fermat's Little Theorem Revisited
reserve a,b,c,d,m,x,n,k,l for Nat,
t,z for Integer,
f,F,G for FinSequence of REAL;
reserve q,r,s for real number;
reserve D for set;
theorem Th1:
for f be FinSequence holds
f is D-valued iff f is FinSequence of D
proof
let f be FinSequence;
f is D-valued implies f is FinSequence of D
proof
assume f is D-valued; then
rng f c= D by RELAT_1:def 19;
hence thesis by FINSEQ_1:def 4;
end;
hence thesis;
end;
:: Seg vs Segm
theorem Th2:
k+1 in Seg n iff k < n
proof
thus k+1 in Seg n implies k < n
proof
assume k+1 in Seg n; then
k + 1 <= n by FINSEQ_1:1;
hence k < n by NAT_1:13;
end;
thus thesis by FINSEQ_3:11;
end;
theorem Th3: :: AFINSEQ_1:Lm1
n+1 <= len f iff n+1 in dom f
proof
thus n+1 <= len f implies n+1 in dom f
proof assume n+1 <= len f; then
n < len f by NAT_1:13;
then n+1 in Seg len f by Th2;
hence n+1 in dom f by FINSEQ_1:def 3;
end;
assume n+1 in dom f;
then n+1 in Seg len f by FINSEQ_1:def 3; then
n < len f by Th2;
hence n+1 <= len f by NAT_1:13;
end;
theorem
k in Segm n iff k+1 in Seg n by Th2,NAT_1:44;
theorem Th7:
n in dom f & 1 <= m & m <= n implies f.m = (f|n).m
proof
assume
A1: n in dom f & 1 <= m & m <= n; then
m in Seg n;
hence thesis by A1, RFINSEQ:6;
end;
theorem
f is FinSequence of D implies
f|n is FinSequence of D & f/^n is FinSequence of D
proof
assume
A1: f is FinSequence of D;
(f|n)^(f/^n) = f by RFINSEQ:8;
hence thesis by A1,FINSEQ_1:36;
end;
theorem Th9:
n in dom f implies (f|n).1 = f.1
proof
assume
A1: n in dom f; then
n >= 1 by FINSEQ_3:24,NAT_1:14;
hence thesis by A1,Th7;
end;
theorem Th10:
for f be FinSequence of REAL st n in dom f holds len(f|n) = n
proof
let f be FinSequence of REAL;
assume n in dom f; then
n <= len f by FINSEQ_3:25;
hence thesis by FINSEQ_1:59;
end;
registration
let s;
cluster <*s*> -> REAL-valued;
coherence by TOPREALC:19;
end;
registration
let D;
let f be D-valued FinSequence;
let n;
cluster (f|n) -> D-valued;
coherence;
end;
registration
let D;
let f be FinSequence of D;
let n;
cluster (f/^n) -> D-valued;
coherence;
end;
theorem for f be FinSequence of COMPLEX holds
k in dom (f|n) implies k in dom f
proof
let f be FinSequence of COMPLEX;
assume k in dom (f|n); then
k in dom ((f|n)^(f/^n)) by FINSEQ_2:15;
hence thesis by RFINSEQ:8;
end;
registration
let n;
cluster {}/^n -> empty;
coherence
proof
reconsider f = <*>NAT as FinSequence of NAT;
A1: f/^(len f) = {} by RFINSEQ:27;
per cases;
suppose
n = 0;
hence thesis by A1;
end;
suppose
n > 0; then
n > len {};
hence thesis by RFINSEQ:def 1;
end;
end;
end;
registration
let f,n;
cluster (f|n)/^n -> empty;
coherence
proof
per cases;
suppose
A1: n > 0;
len (f|n) = n or not n in dom f by Th10; then
len (f|n) = n or n < 1 or n > len f by FINSEQ_3:25; then
len (f|n) <= n by FINSEQ_1:58,A1,NAT_1:14;
hence thesis by FINSEQ_5:32;
end;
suppose
n = 0; then
(f|n) = {};
hence thesis;
end;
end;
end;
registration
let D;
let f be D-valued FinSequence;
let n;
cluster (f/^n) -> D-valued;
coherence
proof
rng f c= D by RELAT_1:def 19; then
f is FinSequence of D by FINSEQ_1:def 4;
hence thesis;
end;
end;
registration
let f be FinSequence of NAT,n;
cluster f.n -> natural;
coherence;
end;
registration
let f be FinSequence of NAT,n,k;
cluster (f|n).k -> natural;
coherence;
cluster ((f|n)/^1).k -> natural;
coherence;
end;
theorem Th14:
Sum(f^F) = Sum(f) + Sum(F)
proof
f is FinSequence of COMPLEX & F is FinSequence of COMPLEX by RVSUM_1:146;
hence thesis by MATRIXC1:46;
end;
theorem Th15:
for f be FinSequence of REAL holds
k in dom (f/^n) & n in dom f implies n+k in dom f
proof
let f be FinSequence of REAL;
assume
A1: k in dom (f/^n) & n in dom f; then
len (f|n) + k in dom ((f|n)^(f/^n)) by FINSEQ_1:28; then
n+k in dom ((f|n)^(f/^n)) by A1,Th10;
hence thesis by RFINSEQ:8;
end;
theorem Th16:
for k be positive Nat holds n+k in dom f implies k in dom (f/^n)
proof
let k be positive Nat;
A0: k >= 1 by NAT_1:14;
assume n+k in dom f; then
A2: n+k <= len f by FINSEQ_3:25; then
A3: n+k-n <= len f - n by XREAL_1:9;
n + k > n +0 by XREAL_1:6; then
len f > n by A2,XXREAL_0:2; then
k <= len (f/^n) by A3,RFINSEQ:def 1;
hence thesis by A0,FINSEQ_3:25;
end;
theorem Th17:
for n be positive Nat st n+1 = len f holds
Sum f = Sum((f|n)/^1) + f.1 + f.(n+1)
proof
let n be positive Nat such that
A0: n+1 = len f;
A2: f =(f|n)^<*f.(n+1)*> by A0,RFINSEQ:7;
n >= 1 & len f >= n by A0,NAT_1:13,14; then
A2a: n in dom f by FINSEQ_3:25;
n+1 > n+0 by XREAL_1:6; then
A3: len (f|n) > 0 by A0,FINSEQ_1:59;
<*f.(n+1)*> is FinSequence of REAL by RVSUM_1:145; then
Sum f = Sum (f|n) + Sum <*f.(n+1)*> by Th14,A2
.= Sum (f|n) + f.(n+1) by RVSUM_1:73
.= ((f|n).1 + Sum((f|n)/^1)) + f.(n+1) by A3,IRRAT_1:17;
hence thesis by A2a,Th9;
end;
theorem
n+1 = len f implies (f/^n) = <*f.(n+1)*>
proof
assume n+1 = len f; then
(f|n)^<*f.(n+1)*> = f by RFINSEQ:7
.= (f|n)^(f/^n) by RFINSEQ:8;
hence thesis by FINSEQ_1:33;
end;
theorem
((f|n)/^1) is not empty implies len ((f|n)/^1) <= len f - 1
proof
assume (f|n)/^1 is not empty; then
(f|n) is not empty; then
A2: 1 in dom (f|n) by FINSEQ_5:6;
(f|n) = ((f|n)|1)^((f|n)/^1) by RFINSEQ:8; then
len (f|n) = len ((f|n)|1) + len((f|n)/^1) by FINSEQ_1:22; then
A3: len ((f|n)/^1) + 1 = len (f|n) by A2,Th10;
len (f|n)-1 <= len f -1 by XREAL_1:9,FINSEQ_5:16;
hence thesis by A3;
end;
theorem Th20:
((f|n)/^1) is not empty implies len ((f|n)/^1) < n
proof
assume (f|n)/^1 is not empty; then
(f|n) is not empty; then
A2: 1 in dom (f|n) by FINSEQ_5:6;
(f|n) = ((f|n)|1)^((f|n)/^1) by RFINSEQ:8; then
len (f|n) = len ((f|n)|1) + len((f|n)/^1) by FINSEQ_1:22; then
A3: len ((f|n)/^1) + 1 = len (f|n) by A2,Th10;
len (f|n) <= n by FINSEQ_5:17;
hence thesis by A3,NAT_1:13;
end;
:: n choose k
theorem Th21:
n is prime & k <> 0 & k <> n implies n divides (n choose k)
proof
assume
A1: n is prime & k <> 0 & k <> n; then
per cases by XXREAL_0:1;
suppose
k > n; then
n choose k = 0 by NEWTON:def 3;
hence thesis by NAT_D:6;
end;
suppose
B1: k < n; then
consider m such that
B1a: n = 1 + m by NAT_1:14,10;
B1b: k!,n are_coprime by A1,B1,NAT_4:19,GROUP_17:5;
consider l such that
B2: n = k + l by B1,NAT_1:10;
k+l > 0+l by A1,XREAL_1:6; then
B3a: l!,n are_coprime by A1,B2,NAT_4:19,GROUP_17:5;
l = n - k by B2; then
n choose k = (n!)/((k!) * (l!)) by B1,NEWTON:def 3; then
(n choose k)*((k!)*(l!)) = n! by XCMPLX_1:87; then
(n choose k)*((k!)*(l!)) = m!*n by B1a,NEWTON:15; then
n divides ((n choose k)*(k!))*(l!); then
n divides (n choose k)*(k!) by B3a,INT_2:25;
hence thesis by B1b,INT_2:25;
end;
end;
:: factorial
theorem Th22:
b >= 2 implies (b+1)! > 2|^b
proof
defpred P[Nat] means ($1+1)! > 2|^$1;
A1: P[2]
proof
B1: (2+1)! = 2*3 by NEWTON:14,15;
2|^(2) = 2*2 by NEWTON:81;
hence thesis by B1;
end;
A2: for k be Nat st k >= 2 & P[k] holds P[k+1]
proof
let k be Nat;
assume
B1: k>=2 & (k+1)! > 2|^k; then
k+2 > 2+0 by XREAL_1:6; then
(k+1)!*(k+1+1) > 2|^k*2 by B1,XREAL_1:96; then
(k+1)!*(k+1+1) > 2|^(k+1) by NEWTON:6;
hence thesis by NEWTON:15;
end;
for x be Nat st x >= 2 holds P[x] from NAT_1:sch 8(A1,A2);
hence thesis;
end;
Lm3: for b be positive Nat holds b divides (b+c)!
proof
let b be positive Nat;
b+c >= b+0 by XREAL_1:6;
hence thesis by NEWTON:41;
end;
Lm4: for a,b be positive Nat holds a,b! are_coprime implies a,b are_coprime
proof
let a,b be positive Nat;
assume
A1: a,b! are_coprime;
b divides (b+0)! by Lm3;
hence thesis by A1,WSIERP_1:15,NEWTON:57;
end;
theorem Th23:
b > 1 iff b! > 1
proof
L0: b is Element of NAT by ORDINAL1:def 12;
thus b > 1 implies b! > 1
proof
assume b > 1; then
b >= 1+1 by NAT_1:13;
hence thesis by L0, ASYMPT_1:55;
end;
thus thesis by L0,ASYMPT_1:56,NEWTON:13;
end;
theorem Th24:
b >= 2 implies b! < b|^b
proof
defpred P[Nat] means $1! < $1|^$1;
A1: P[2]
proof
2|^(2) = 2*2 by NEWTON:81;
hence thesis by NEWTON:14;
end;
A2: for k be Nat st k >= 2 & P[k] holds P[k+1]
proof
let k be Nat such that
B1: k >= 2 & k! < k|^k;
B2: k >= 1 by B1,XXREAL_0:2;
k+1 > k +0 by XREAL_1:6; then
(k+1)|^k > k|^k by B2,PREPOWER:10; then
B3: (k+1)|^k*(k+1) > k|^k*(k+1) by XREAL_1:68;
k!*(k+1) < k|^k*(k+1) by B1,XREAL_1:68; then
k!*(k+1) < (k+1)|^k*(k+1) by B3,XXREAL_0:2;then
(k+1)! < (k+1)|^k*(k+1) by NEWTON:15;
hence thesis by NEWTON:6;
end;
for x be Nat st x >= 2 holds P[x] from NAT_1:sch 8(A1,A2);
hence thesis;
end;
theorem
(b+1)! >= 2|^b
proof
per cases by NAT_1:23;
suppose b >=2;
hence thesis by Th22;
end;
suppose
b = 0;
hence thesis by NEWTON:4,13;
end;
suppose
b = 1;
hence thesis by NEWTON:14;
end;
end;
theorem
b! <= b|^b
proof
per cases by NAT_1:23;
suppose b >=2;
hence thesis by Th24;
end;
suppose
b = 0;
hence thesis by NEWTON:4,12;
end;
suppose
b = 1;
hence thesis by NEWTON:13;
end;
end;
theorem
b>0 & a,b! are_coprime implies a,b are_coprime
proof
assume
A1: b > 0 & a,b! are_coprime;
per cases;
suppose
a = 0; then
b <= 1 by A1,Th23; then
b < 1+1 by NAT_1:13; then
b = 1 or b = 0 by NAT_1:23;
hence thesis by A1,NEWTON:51;
end;
suppose a>0;
hence thesis by A1,Lm4;
end;
end;
theorem
a,(a+b)! are_coprime implies a = 1 or (a = 0 & (b = 0 or b = 1))
proof
assume
A1: a,(a+b)! are_coprime;
per cases;
suppose
B1: a = 0; then
b <= 1 by A1,Th23; then
b < 1+1 by NAT_1:13;
hence thesis by B1,NAT_1:23;
end;
suppose a>0;
hence thesis by A1,Lm3,NEWTON:49;
end;
end;
theorem Th29:
for n st n in dom f &
m in dom ((f|n)/^1) holds ((f|n)/^1).m = f.(m+1)
proof
let n such that
A0: n in dom f;
set F = (f|n);
assume
A1: m in dom ((f|n)/^1);
A1a: m+1 >= 0+1 by XREAL_1:6;
(f|n) <> {} by A1; then
A2: 1 in dom (f|n) by FINSEQ_5:6; then
m+1 <= len (f|n) by A1,Th15,Th3; then
A4: m+1 <= n by A0,Th10;
(((f|n))/^1).m = (((f|n))/^1)/.m by A1, PARTFUN1:def 6
.= (f|n)/.(1+m) by A1,FINSEQ_5:27
.= (f|n).(1+m) by A1,A2,Th15,PARTFUN1:def 6
.= f.(1+m) by A0,Th7,A1a,A4;
hence thesis;
end;
:: Newton_Coeff
registration
let n;
cluster Newton_Coeff n -> non empty;
coherence
proof
len (Newton_Coeff n) = n + 1 by NEWTON:def 5;
hence thesis;
end;
end;
registration
let n,m;
cluster (Newton_Coeff n).m -> natural;
coherence
proof
per cases;
suppose
A0: m in dom (Newton_Coeff n); then
consider k such that
A1: m = 1+k by NAT_1:10, FINSEQ_3:25;
k = m - 1 by A1; then
(Newton_Coeff n).m = n choose k by A0,NEWTON:def 5;
hence thesis;
end;
suppose
not m in dom (Newton_Coeff n);
hence thesis by FUNCT_1:def 2;
end;
end;
end;
registration
let n;
cluster Newton_Coeff n -> NAT-valued;
coherence
proof
for k st k in dom(Newton_Coeff n) holds (Newton_Coeff n).k in NAT
by ORDINAL1:def 12;
hence thesis by FINSEQ_2:12;
end;
end;
registration
let h be FinSequence of NAT;
cluster Sum h -> natural;
coherence
proof
A1: for n st n in dom h holds h.n >= 0;
Sum h is integer; then
Sum h is Element of NAT by A1,RVSUM_1:84,INT_1:3;
hence thesis;
end;
end;
theorem Th30:
n > 0 implies n in dom (Newton_Coeff n)
proof
assume n>0; then
A1: n in Seg n by FINSEQ_1:3;
dom (Newton_Coeff n) = Seg (len (Newton_Coeff n)) by FINSEQ_1:def 3
.= Seg (n+1) by NEWTON:def 5;
hence thesis by A1,FINSEQ_2:8;
end;
theorem
(Newton_Coeff n) is FinSequence of NAT
proof
for k st k in dom(Newton_Coeff n) holds (Newton_Coeff n).k in NAT;
hence thesis by FINSEQ_2:12;
end;
theorem Th32:
(Newton_Coeff n).(n+1) = 1
proof
A1: len (Newton_Coeff n) = n+1 by NEWTON:def 5; then
A2: n = len(Newton_Coeff n) - 1;
n+1 in dom (Newton_Coeff n) by A1,Th3; then
(Newton_Coeff n).(n+1) = n choose n by A2,NEWTON:def 5;
hence thesis by NEWTON:21;
end;
theorem Th33:
(Newton_Coeff k).1 = 1
proof
((1,1) In_Power k).1 = 1
proof
thus ((1,1) In_Power k).1 = 1|^k by NEWTON:28
.= 1;
end;
hence thesis by NEWTON:31;
end;
theorem Th34:
for n be positive Nat holds
Sum(Newton_Coeff n) =
Sum(((Newton_Coeff n)|n)/^1) + (Newton_Coeff n).1 + (Newton_Coeff n).(n+1)
proof
let n be positive Nat;
A1: n+1 > 0+1 & len(Newton_Coeff n) = n+1 by XREAL_1:6,NEWTON:def 5;
A2: (Newton_Coeff n) =((Newton_Coeff n)|n)^<*(Newton_Coeff n).(n+1)*>
by NEWTON:def 5,RFINSEQ:7;
A2a: n in dom (Newton_Coeff n) by Th30;
n+1 > n+0 by XREAL_1:6; then
A3: len ((Newton_Coeff n)|n) > 0 by A1,FINSEQ_1:59;
<*(Newton_Coeff n).(n+1)*> is FinSequence of REAL by RVSUM_1:145; then
Sum(Newton_Coeff n) = Sum ((Newton_Coeff n)|n) +
Sum <*(Newton_Coeff n).(n+1)*> by Th14,A2
.= Sum ((Newton_Coeff n)|n) + (Newton_Coeff n).(n+1) by RVSUM_1:73
.= (((Newton_Coeff n)|n).1 + Sum(((Newton_Coeff n)|n)/^1)) +
(Newton_Coeff n).(n+1) by A3,IRRAT_1:17;
hence thesis by A2a,Th9;
end;
theorem Th35:
for n be positive Nat holds
Sum(Newton_Coeff n) = Sum(((Newton_Coeff n)|n)/^1) + 2
proof
let n be positive Nat;
Sum(Newton_Coeff n) = Sum(((Newton_Coeff n)|n)/^1) +
(Newton_Coeff n).1 + (Newton_Coeff n).(n+1) by Th34
.= Sum(((Newton_Coeff n)|n)/^1) + (Newton_Coeff n).1 + 1 by Th32
.= Sum(((Newton_Coeff n)|n)/^1) + 1 + 1 by Th33;
hence thesis;
end;
theorem
Sum(Newton_Coeff n) = Sum ((Newton_Coeff n)|n) + 1
proof
A0: Newton_Coeff n is FinSequence of NAT by Th1;
A1: (Newton_Coeff n).(n+1) = ((1,1)In_Power n).(n+1) by NEWTON:31
.= 1^n by NEWTON:29;
A2: n+1 = len (Newton_Coeff n) by NEWTON:def 5;
n+1 >= 0+1 by XREAL_1:6; then
A3: n+1 in dom (Newton_Coeff n) by FINSEQ_3:25,A2;
(Newton_Coeff n)
= ((Newton_Coeff n)|n)^<*(Newton_Coeff n)/.(n+1)*> by A0,A2,FINSEQ_5:21
.= ((Newton_Coeff n)|n)^<*(Newton_Coeff n).(n+1)*> by A3,PARTFUN1:def 6;
hence thesis by A1,RVSUM_1:74;
end;
theorem
len((Newton_Coeff n)|n) = n
proof
len (Newton_Coeff n) = n+1 by NEWTON:def 5; then
n < len (Newton_Coeff n) by NAT_1:13;
hence thesis by FINSEQ_1:59;
end;
theorem Th38:
m in dom (((Newton_Coeff n)|n)/^1) implies
(((Newton_Coeff n)|n)/^1).m = (Newton_Coeff n).(m+1)
proof
A1: n = 0 implies ((Newton_Coeff n)|n) = {};
n>0 implies n in dom (Newton_Coeff n) by Th30;
hence thesis by A1,Th29;
end;
theorem Th39:
n is prime implies n divides (((Newton_Coeff n)|n)/^1).k
proof
L1: n is prime & k >= n implies n divides (((Newton_Coeff n)|n)/^1).k
proof
assume
A1: n is prime & k >= n; then
n in dom (Newton_Coeff n) by Th30; then
k >= len ((Newton_Coeff n)|n) by A1,Th10; then
k+1 > len ((Newton_Coeff n)|n) by NAT_1:13; then
not k+1 in dom ((Newton_Coeff n)|n) by FINSEQ_3:25; then
not k in dom (((Newton_Coeff n)|n)/^1) by FINSEQ_5:26; then
(((Newton_Coeff n)|n)/^1).k = {} by FUNCT_1:def 2;
hence thesis by NAT_D:6;
end;
n is prime & k < n implies n divides (((Newton_Coeff n)|n)/^1).k
proof
A0: k = (k+1) - 1;
assume
A1: n is prime & k < n;
per cases;
suppose
A1aa: k > 0; then
A1a: k+1 > 0+1 & k+1 < n+1 & len(Newton_Coeff n) = n+1
by A1,XREAL_1:8,NEWTON:def 5; then
A2: k+1 in dom (Newton_Coeff n) by FINSEQ_3:25;
n in dom (Newton_Coeff n) by A1,Th30; then
A3: len((Newton_Coeff n)|n) = n by Th10;
k+1 <= n by A1,NAT_1:13; then
(k)+1 in dom ((Newton_Coeff n)|n) by A1a,A3,FINSEQ_3:25; then
k in dom (((Newton_Coeff n)|n)/^1) by A1aa,Th16; then
(((Newton_Coeff n)|n)/^1).(k) = (Newton_Coeff n).(k+1) by Th38
.= (n) choose (k) by A0,A2,NEWTON:def 5;
hence thesis by A1,A1aa,Th21;
end;
suppose k = 0; then
not k in dom (((Newton_Coeff n)|n)/^1) by FINSEQ_3:25; then
(((Newton_Coeff n)|n)/^1).k = {} by FUNCT_1:def 2;
hence thesis by NAT_D:6;
end;
end;
hence thesis by L1;
end;
theorem Th40:
for n be prime Nat holds n divides 2|^n - 2
proof
let n be prime Nat;
reconsider h = (Newton_Coeff n) as FinSequence of NAT by Th1;
reconsider f = (((Newton_Coeff n)|n)/^1) as FinSequence of NAT by Th1;
A2: for k st k in dom f holds n divides f.k by Th39;
Sum h = Sum f + 2 by Th35; then
2|^n = Sum f +2 by NEWTON:32;
hence thesis by A2,INT_4:36;
end;
registration
let k be positive Nat;
let n;
cluster n|^k - n -> natural;
coherence
proof
k >= 1 by NAT_1:14; then
n|^k >= n or n < 1 by PREPOWER:12; then
n|^k - n >= n - n or n = 0 by XREAL_1:9,NAT_1:14; then
n|^k - n is Element of NAT by INT_1:3;
hence thesis;
end;
end;
theorem
for k,n be prime Nat holds n*k divides (2|^n - 2)*(2|^k -2)
proof
let k,n be prime Nat;
n divides 2|^n - 2 & k divides 2|^k -2 by Th40;
hence thesis by NAT_3:1;
end;
theorem
for n be odd Prime st n = 2*k + 1
holds n divides (2|^k - 1) iff not n divides (2|^k + 1)
proof
let n be odd Prime such that
A1: n = 2*k + 1;
not 2 divides n; then
A2: not n divides 2 by INT_2:28,30,PYTHTRIP:def 2;
A3: n divides (2|^k - 1) or n divides (2|^k + 1)
proof
n divides (2|^(2*k+1) - 2) by A1,Th40; then
n divides (2|^(2*k)*2 - 2) by NEWTON:6; then
n divides 2*(2|^(2*k) - 1|^2); then
n divides (2|^(2*k) - 1) by A2,INT_5:7; then
n divides (2|^k)|^2 - 1|^2 by NEWTON:9; then
n divides (2|^k-1)*(2|^k+1) by NEWTON01:1;
hence thesis by INT_5:7;
end;
(2|^k + 1) - (2|^k - 1) = 2;
hence thesis by A2,INT_5:1,A3;
end;
definition
let n be natural number;
func n\ -> FinSequence of REAL equals
(1,1) In_Power n;
coherence;
end;
registration
let n;
identify n\ with Newton_Coeff n;
correctness by NEWTON:31;
identify Newton_Coeff n with n\;
correctness;
end;
:: The use of this registration would make most of the above code unnecessary. I leave it as an illustration.
:: In_Power
theorem Th43:
n > 0 implies n in dom ((a,b) In_Power n)
proof
assume n>0; then
A1: n in Seg n by FINSEQ_1:3;
dom ((a,b) In_Power n) = Seg (len ((a,b) In_Power n)) by FINSEQ_1:def 3
.= Seg (n+1) by NEWTON:def 4;
hence thesis by A1,FINSEQ_2:8;
end;
registration
let a,b,n,m;
cluster ((a,b) In_Power n).m -> natural;
coherence
proof
per cases;
suppose
A0: m in dom ((a,b) In_Power n); then
consider k such that
A1: m = 1+k by NAT_1:10,FINSEQ_3:25;
len ((a,b) In_Power n) = n+1 by NEWTON:def 4; then
n >= k by A0,FINSEQ_3:25,A1,XREAL_1:6; then
consider l such that
A2: n = k+l by NAT_1:10;
k = m - 1 & l = n - k by A1,A2; then
((a,b) In_Power n).m = (n choose k)*a|^l*b|^k by A0,NEWTON:def 4;
hence thesis;
end;
suppose
not m in dom ((a,b) In_Power n);
hence thesis by FUNCT_1:def 2;
end;
end;
end;
registration
let a,b,n;
cluster (a,b) In_Power n -> NAT-valued;
coherence
proof
for k st k in dom ((a,b) In_Power n) holds
((a,b) In_Power n).k in NAT by ORDINAL1:def 12;
hence thesis by FINSEQ_2:12;
end;
end;
Lm1: ((a,b) In_Power (k+l)).(k+1) = ((k+l) choose k)*(a|^l)*(b|^k)
proof
k+l >= k+0 by XREAL_1:6; then
((a,b) In_Power (k+l)).(k+1) = ((k+l) choose k) * (a^((k+l)-k)) * (b^k)
by IRRAT_1:19
.= ((k+l) choose k) * (a|^l) * (b|^k) by POWER:41;
hence thesis;
end;
theorem
(k+l) is prime & k > 0 & l > 0 implies
(k+l) divides ((a,b) In_Power (k+l)).(k+1)
proof
assume
A1: (k+l) is prime & k > 0 & l > 0; then
k+l > k+0 by XREAL_1:6; then
A2: (k+l) divides ((k+l) choose k) by A1,Th21;
((k+l)choose k) divides ((k+l) choose k)*((a|^l)*(b|^k)); then
consider x such that
A3: ((k+l) choose k)*((a|^l)*(b|^k)) = (k+l)*x by A2,INT_2:9,NAT_D:def 3;
((a,b) In_Power (k+l)).(k+1) = ((k+l) choose k)*(a|^l)*(b|^k) by Lm1
.=(k+l)*x by A3;
hence thesis;
end;
Lm5:
m in dom ((((a,b) In_Power n)|n)/^1) implies
((((a,b) In_Power n)|n)/^1).m = ((a,b) In_Power n).(m+1)
proof
A1: n = 0 implies (((a,b) In_Power n)|n) = {};
n>0 implies n in dom ((a,b) In_Power n) by Th43;
hence thesis by A1,Th29;
end;
theorem Th45:
a<>0 implies ((a,b) In_Power m).1 <> 0
proof
assume
A1: a<>0;
((a,b) In_Power m).1 = a |^m by NEWTON:28;
hence thesis by A1;
end;
theorem
for m be non zero Nat holds a = 0 iff ((a,b) In_Power m).1 = 0
proof
for m be non zero Nat holds a = 0 implies ((a,b) In_Power m).1 = 0
proof
let m be non zero Nat;
assume
A1: a = 0;
((a,b) In_Power m).1 = a|^m by NEWTON:28;
hence thesis by A1;
end;
hence thesis by Th45;
end;
theorem
((a,b) In_Power m).1=0 implies m<>0
proof
assume
A1: ((a,b) In_Power m).1=0;
a|^0 = 1 by NEWTON:4;
hence thesis by A1, NEWTON:28;
end;
theorem Th48:
for m be positive Nat holds
Sum (a,b) In_Power m = a|^m + b|^m + Sum ((((a,b) In_Power m)|m)/^1)
proof
let m be positive Nat;
len ((a,b) In_Power m) = m + 1 by NEWTON:def 4; then
Sum ((a,b) In_Power m) = Sum(((a,b) In_Power m)|m/^1) +
((a,b) In_Power m).1 + ((a,b) In_Power m).(m+1) by Th17
.= Sum(((a,b) In_Power m)|m/^1) + a|^m + ((a,b) In_Power m).(m+1)
by NEWTON:28
.= Sum(((a,b) In_Power m)|m/^1) + a|^m + b|^m by NEWTON:29;
hence thesis;
end;
theorem
Sum((a,b) In_Power (m+n)) = Sum((a,b) In_Power m)*Sum((a,b) In_Power n)
proof
Sum((a,b) In_Power (m+n)) = (a+b)|^(m+n) by NEWTON:30
.=(a+b)|^m*(a+b)|^n by NEWTON:8
.= Sum((a,b) In_Power m)*(a+b)|^n by NEWTON:30
.= Sum((a,b) In_Power m)*Sum((a,b) In_Power n) by NEWTON:30;
hence thesis;
end;
theorem Th50:
l>0 implies ex x st ((a,b) In_Power (k+l)).(k+1) = a*x
proof
assume l>0;
then consider n such that
A3: l = 1+n by NAT_1:10,14;
consider x such that
A4: x = ((k+l) choose k)*(a|^n)*(b|^k);
((a,b) In_Power (k+l)).(k+1)=((k+l) choose k)*(a|^l)*(b|^k) by Lm1
.= ((k+l) choose k)*(a|^1*a|^n)*(b|^k) by A3,NEWTON:8
.= a*x by A4;
hence thesis;
end;
theorem
m>0 implies ex k st ((a,b) In_Power m).1 = a*k
proof
m>0 implies ex k st ((a,b) In_Power (m+0)).(0+1) = a*k by Th50;
hence thesis;
end;
theorem
l>0 implies ex x st ((a,b) In_Power (k+l)).l = a*x
proof
assume l>0;
then consider n be Nat such that
A3: l = 1+n by NAT_1:10,14;
set m=k+1;
consider x such that
A4: x = ((m+n) choose n)*(a|^k)*(b|^n);
((a,b) In_Power (k+l)).l = ((m+n) choose n)*(a|^m)*(b|^n) by Lm1,A3
.= ((m+n) choose n)*(a|^1*a|^k)*(b|^n) by NEWTON:8
.= a*x by A4;
hence thesis;
end;
theorem
for n st n = ((a,b) In_Power (k+l)).(k+1) & l>0 holds a divides n
proof
let n;
assume n = ((a,b) In_Power (k+l)).(k+1) & l>0; then
consider x such that
A2: n = a*x by Th50;
thus thesis by A2;
end;
theorem Th54:
for n be prime Nat, a,b be positive Nat holds
n*a*b divides (((a,b) In_Power n)|n/^1).k
proof
let n be prime Nat, a,b be positive Nat;
L1: not k in dom (((a,b) In_Power n)|n/^1) implies
n*a*b divides ((((a,b)In_Power n)|n)/^1).k
proof
assume not k in dom (((a,b) In_Power n)|n/^1); then
((((a,b) In_Power n)|n)/^1).k = {} by FUNCT_1:def 2;
hence thesis by NAT_D:6;
end;
n is prime & k in dom (((a,b) In_Power n)|n/^1) implies
n*a*b divides ((((a,b) In_Power n)|n)/^1).k
proof
A0: k = (k+1) - 1;
assume
A1: n is prime & k in dom (((a,b) In_Power n)|n/^1); then
A2:(((a,b) In_Power n)|n/^1) is not empty;
A3: k >= 1 & k <= len (((a,b) In_Power n)|n/^1) by FINSEQ_3:25,A1; then
A3a: k < n by A2,Th20,XXREAL_0:2; then
A4: k+1 < n+1 & k+1 > 1 by A1,FINSEQ_3:25,NAT_1:13,XREAL_1:6;
A4a: n-k > k-k by A3a,XREAL_1:9;
consider l such that
A4b: n = k + l by A3a,NAT_1:10;
A4d: l = n-k by A4b;
len ((a,b) In_Power n) = n+1 by NEWTON:def 4; then
A6: k+1 in dom ((a,b) In_Power n) by A4, FINSEQ_3:25;
A7: ((((a,b) In_Power n)|n)/^1).(k) = ((a,b) In_Power n).(k+1)
by A1,Lm5
.= (n choose k) * a|^l * b|^k by A0,A4d,A6,NEWTON:def 4;
A8: n divides (n choose k) by A3,A3a,Th21;
a divides a|^l & b divides b|^k by A4a,A4b,A3,NAT_3:3; then
a*b divides a|^l*b|^k by NAT_3:1; then
n*(a*b) divides (n choose k) *(a|^l * b|^k) by A8,NAT_3:1;
hence thesis by A7;
end;
hence thesis by L1;
end;
theorem Th55:
for n be prime Nat, a,b be positive Nat holds
n*a*b divides (a+b)|^n - (a|^n + b|^n)
proof
let n be prime Nat, a,b be positive Nat;
reconsider g = ((a,b) In_Power n) as FinSequence of NAT by Th1;
reconsider h = (((a,b) In_Power n)|n)/^1 as FinSequence of NAT by Th1;
A1: for k st k in dom h holds n*a*b divides h.k by Th54;
Sum g = a|^n + b|^n + Sum h by Th48; then
(a+b)|^n = Sum h + a|^n + b|^n by NEWTON:30;
hence thesis by A1,INT_4:36;
end;
theorem Th56:
for n be prime Nat holds n*a divides (a+1)|^n - (a|^n + 1)
proof
let n be prime Nat;
L1: a > 0 implies n*a*1 divides (a+1)|^n - (a|^n + 1|^n) by Th55;
a = 0 implies n*a divides (a+1)|^n - (a|^n + 1);
hence thesis by L1;
end;
theorem
for a,b be positive Nat holds
2*a*b divides (a+b)|^2 - (a|^2 + b|^2) by Th55,INT_2:28;
theorem Th58:
for n be prime Nat holds n divides a|^n - a
proof
let n be prime Nat;
defpred P[Nat] means n divides $1|^n - $1;
L1: P[0] by INT_2:12;
L2: P[k] implies P[k+1]
proof
assume
A0: P[k];
n*k divides (k+1)|^n - (k|^n + 1) by Th56; then
n divides (k+1)|^n - (k|^n + 1) by INT_2:2,INT_1:def 3; then
n divides k|^n - k + ((k+1)|^n - (k|^n + 1)) by A0,WSIERP_1:4;
hence thesis;
end;
for x be Nat holds P[x] from NAT_1:sch 2(L1,L2);
hence thesis;
end;
theorem Th59:
for k be Nat st k+1 is prime & not k+1 divides a
holds k+1 divides a|^k - 1
proof
let k be Nat such that
A1: k+1 is prime and
A2: not k+1 divides a;
k+1 divides a|^(k+1) - a by A1,Th58; then
k+1 divides a|^k*a - a by NEWTON:6; then
k+1 divides a*(a|^k - 1);
hence thesis by A1,A2,INT_5:7;
end;
theorem Th60:
for n be prime Nat holds n divides a+b iff n divides a|^n + b|^n
proof
let n be prime Nat;
n divides a|^n - a & n divides b|^n - b by Th58; then
A1: n divides (a|^n - a) + (b|^n - b) by WSIERP_1:4;
A2: a|^n + b|^n = (a+b) + ((a|^n + b|^n) - (a + b));
hence n divides a+b implies n divides a|^n + b|^n by A1,WSIERP_1:4;
assume n divides a|^n + b|^n;
hence thesis by A1,A2,INT_2:1;
end;
theorem
163 divides (a+b) iff 163 divides a|^163 + b|^163 by NAT_4:35,Th60;
theorem Th62:
for n be prime Nat holds n divides a iff n divides a|^n
proof
let n be prime Nat;
a divides a|^n by NAT_3:3;
hence thesis by NAT_D:4, NAT_3:5;
end;
theorem Th63:
for n be prime Nat holds n divides a|^n + 1 iff n divides a+1
proof
let n be prime Nat;
n divides a|^n + 1|^n iff n divides a+1 by Th60;
hence thesis;
end;
theorem
for n be prime Nat holds
n divides a|^n + b|^n iff n divides (a+b)|^n by Th60,Th62;
theorem
7 divides a|^7 + 1 iff 7 divides a+1 by Th63,NAT_4:26;
theorem
not 7 divides a implies 7 divides a|^6 - 1
proof
assume not 7 divides a; then
A1: not a gcd 7 = 7 by INT_2:def 2;
7 divides a|^(6+1) - a by NAT_4:26,Th58; then
7 divides a|^6*a - a by NEWTON:6; then
7 divides a*(a|^6 - 1);
hence thesis by A1,NAT_4:26,PEPIN:2,INT_2:25;
end;
theorem
for n be prime Nat, a,b be positive Nat holds
n*a*b divides (a+b)|^(k*n) - (a|^n + b|^n)|^k
proof
let n be prime Nat, a,b be positive Nat;
(a+b)|^n - (a|^n + b|^n) divides ((a+b)|^n)|^k - (a|^n + b|^n)|^k
by NEWTON01:33; then
A1: (a+b)|^n - (a|^n + b|^n) divides (a+b)|^(k*n) - (a|^n + b|^n)|^k
by NEWTON:9;
n*a*b divides (a+b)|^n - (a|^n + b|^n) by Th55;
hence thesis by A1,INT_2:9;
end;
theorem
for n be prime Nat, a,b be positive Nat holds
n*a*b divides (a+t)|^n - (a|^n + b|^n) implies
n*a*b divides (a+b)|^n - (a+t)|^n
proof
let n be prime Nat, a,b be positive Nat;
assume
A1: n*a*b divides (a+t)|^n - (a|^n + b|^n);
n*a*b divides -((a+b)|^n - (a|^n + b|^n)) by INT_2:10,Th55; then
n*a*b divides ((a|^n + b|^n) - (a+b)|^n) + ((a+t)|^n - (a|^n + b|^n))
by A1,WSIERP_1:4; then
n*a*b divides -((a+b)|^n - (a+t)|^n);
hence thesis by INT_2:10;
end;
theorem
for n be prime Nat, a,b,c be positive Nat holds
n*a*b divides c-b implies n*a*b divides (a|^n + b|^n) - (a+c)|^n
proof
let n be prime Nat, a,b,c be positive Nat;
assume n*a*b divides c-b; then
n*a*b divides -(b-c); then
A1: n*a*b divides b-c by INT_2:10;
(a+b) - (a+c) divides (a+b)|^n - (a+c)|^n by NEWTON01:33; then
A3: n*a*b divides (a+b)|^n - (a+c)|^n by A1,INT_2:9;
n*a*b divides -((a+b)|^n - (a|^n + b|^n)) by INT_2:10,Th55; then
n*a*b divides (a+b)|^n - (a+c)|^n + -((a+b)|^n - (a|^n + b|^n))
by A3,WSIERP_1:4;
hence thesis;
end;
theorem Th70:
for p be prime Nat st p = 2*n+1 holds
p divides a or p divides a|^n-1 or p divides a|^n+1
proof
let p be prime Nat such that
A1: p = 2*n+1;
a|^(2*n+1) - a = a|^(2*n)*a - a by NEWTON:6
.= a*(a|^(2*n) - 1)
.= a*((a|^n)|^2 - 1|^2) by NEWTON:9
.= a*((a|^n - 1)*(a|^n + 1)) by NEWTON01:1; then
p divides a or p divides ((a|^n - 1)*(a|^n + 1)) by A1,Th58,INT_5:7;
hence thesis by INT_5:7;
end;
theorem
for p be prime Nat st not p divides a holds
ex n st p divides a|^n - 1 & 0 < n & n < p
proof
let p be prime Nat such that
A0: not p divides a;
p > 1 by INT_2:def 4; then
p >= 1+1 by NAT_1:13; then
consider k such that
A1: p = 2 + k by NAT_1:10;
p = (k+1) + 1 by A1; then
p divides (a|^(k+1) - 1) by A0,Th59;
hence thesis by A1,XREAL_1:6;
end;
theorem
5 divides a|^3 - 1 iff 5 divides a - 1
proof
L1: 5 divides a|^3 - 1 implies 5 divides a - 1
proof
A1: 5 divides (a|^5 - a) by PEPIN:59,Th58;
assume
A2: 5 divides a|^3 - 1; then
a2b: a|^3 <> 0 by INT_2:13;
A3: a divides a|^3 by NAT_3:3;
5 divides (-a|^2)*(a|^3 - 1) by A2,INT_2:2; then
5 divides -a|^2*a|^3 + a|^2; then
5 divides -a|^(2+3) + a|^2 by NEWTON:8;then
5 divides (-a|^5 + a|^2) + (a|^5 - a) by WSIERP_1:4,A1; then
5 divides a|^2 - a; then
5 divides a*a - a by NEWTON:81; then
A4: 5 divides a*(a-1);
not 5 divides (a|^3-1) + 1 by A2,a2b,NEWTON:39; then
not 5 divides a by A3,INT_2:9;
hence thesis by A4,PEPIN:59,INT_5:7;
end;
a - 1 divides a|^3 - 1|^3 by NEWTON01:33;
hence thesis by L1,INT_2:9;
end;
theorem Th73:
for k st k+1 is prime holds k+1 divides a|^(n*k+1) - a
proof
let k such that
A1: k+1 is prime;
per cases;
suppose k+1 divides a; then
k+1 divides a*(a|^(n*k) - 1) by INT_2:2; then
k+1 divides a*a|^(n*k) - a;
hence thesis by NEWTON:6;
end;
suppose not k+1 divides a; then
B1: k+1 divides a|^k - 1 by A1,Th59;
a|^k - 1|^k divides a|^k|^n - 1|^k|^n by NEWTON01:33; then
k+1 divides a|^k|^n - 1|^k|^n by B1,INT_2:9; then
k+1 divides (a|^(n*k) - 1) by NEWTON:9; then
k+1 divides a*(a|^(n*k) - 1) by INT_2:2; then
k+1 divides a*a|^(n*k) - a;
hence thesis by NEWTON:6;
end;
end;
theorem Th74:
2 divides a|^(n+1) - a
proof
1+1 divides a|^(n*1+1) - a by INT_2:28,Th73;
hence thesis;
end;
theorem Th75:
3 divides a|^(2*n+1) - a
proof
2+1 divides a|^(2*n+1) - a by PEPIN:41,Th73;
hence thesis;
end;
theorem Th76:
5 divides a|^(4*n+1) - a
proof
4+1 divides a|^(4*n+1) - a by PEPIN:59,Th73;
hence thesis;
end;
theorem Th77:
7 divides a|^(6*n+1) - a
proof
6+1 divides a|^(6*n+1) - a by NAT_4:26,Th73;
hence thesis;
end;
theorem Th78:
for k,l st k <> l & k+1 is odd prime & l+1 is odd prime
holds 2*(k+1)*(l+1) divides a|^(k*l+1) - a
proof
let k,l such that
A0: k <> l and
A1: k+1 is odd prime & l+1 is odd prime;
A2: k+1 <> l+1 by A0;
k+1 divides a|^(k*l+1) - a & l+1 divides a|^(k*l+1) - a by A1,Th73; then
A5: (k+1)*(l+1) divides a|^(k*l+1) - a by A1,A2,INT_2:30,PEPIN:4;
A6: (k+1)*(l+1),2|^1 are_coprime by A1,NAT_5:3;
2 divides a|^(1*(k*l)+1) - a by Th74; then
2*((k+1)*(l+1)) divides a|^(k*l+1) - a by A5,A6,PEPIN:4;
hence thesis;
end;
theorem
154 divides a|^61 - a
proof
6+1 is odd prime & 10+1 is odd prime by NAT_4:26,27; then
2*(6+1)*(10+1) divides a|^(6*10+1) - a by Th78;
hence thesis;
end;
theorem
6 divides a|^(2*n+1) - a
proof
A1: 2 divides a|^(2*n+1) - a by Th74;
3 divides a|^(2*n+1) - a by Th75; then
2*3 divides a|^(2*n+1) - a by INT_2:28,30,PEPIN:41,A1,PEPIN:4;
hence thesis;
end;
theorem
30 divides a|^(4*n+1) - a
proof
A0: 2,5 are_coprime & 3,5 are_coprime by INT_2:28,30,PEPIN:41,59;
A1: 2 divides a|^(4*n+1) - a by Th74;
3 divides a|^(2*(2*n)+1) - a by Th75; then
A2: 2*3 divides a|^(4*n+1) - a by INT_2:28,30,PEPIN:41,A1,PEPIN:4;
5 divides a|^(4*n+1) - a by Th76; then
5*(2*3) divides a|^(4*n+1) - a by A0,EULER_1:14,A2,PEPIN:4;
hence thesis;
end;
theorem
42 divides a|^(6*n+1) - a
proof
A0: 2,7 are_coprime & 3,7 are_coprime by INT_2:28,30,PEPIN:41,NAT_4:26;
A1: 2 divides a|^(6*n+1) - a by Th74;
3 divides a|^(2*(3*n)+1) - a by Th75; then
A2: 2*3 divides a|^(6*n+1) - a by INT_2:28,30,PEPIN:41,A1,PEPIN:4;
7 divides a|^(6*n+1) - a by Th77; then
7*(2*3) divides a|^(6*n+1) - a by A0,EULER_1:14,A2,PEPIN:4;
hence thesis;
end;
theorem
for n be prime Nat holds n divides a|^(n+k) - a|^(k+1)
proof
let n be prime Nat;
a|^(n+k) - a|^(k+1) = a|^n *a|^k - a|^(k+1) by NEWTON:8
.= a|^n * a|^k - a|^k *a by NEWTON:6
.= a|^k*(a|^n - a);
hence thesis by Th58,INT_2:2;
end;
theorem Th84:
for n st 2*n+1 is prime for k st 2*n > k & k > 1
holds not 2*n+1 divides a|^n - k & not 2*n+1 divides a|^n + k
proof
let n such that
A1: 2*n + 1 is prime;
set p = 2*n+1;
2*n <> 0 by A1; then
A1a: n > 0;
let k such that
A1b: 2*n > k & k > 1;
A1c: k-1 > 1-1 by A1b,XREAL_1:9;
A1d: (k-1)+2 > (k-1)+1 > (k-1)+0 by XREAL_1:6;
2*n+1 > k+1 by A1b, XREAL_1:6; then
2*n+1 > k+1 & 2*n+1 > k by A1d,XXREAL_0:2; then
A2: 2*n+1 > k+1 & 2*n+1 > k & 2*n+1 > k-1 by A1d,XXREAL_0:2;
A3: p divides a or p divides a|^n - 1 or p divides a|^n + 1 by A1,Th70;
a divides a|^n by A1a,NAT_3:3; then
p divides a implies p divides a|^n by INT_2:9; then
A4: p divides (-1)*(a|^n) or p divides (-1)*(a|^n -1) or
p divides (-1)*(a|^n + 1) by A3,INT_2:2;
assume not thesis; then
p divides (-a|^n) + (a|^n - k) or
p divides (-a|^n) + (a|^n + k) or
p divides (-a|^n -1) + (a|^n - k) or
p divides (-a|^n -1) + (a|^n + k) or
p divides (-a|^n +1) + (a|^n - k) or
p divides (-a|^n +1) + (a|^n + k) by A4,WSIERP_1:4; then
p divides -k or p divides k or p divides -(1+k) or
p divides -1+k or p divides -(k-1) or p divides 1+k;
hence contradiction by A1b,A1c,A2,INT_2:27,INT_2:10;
end;
theorem Th85:
not 5 divides a|^2 - 2 & not 5 divides a|^2 + 2 &
not 5 divides a|^2 - 3 & not 5 divides a|^2 + 3
proof
5 = 2*2 + 1;
hence thesis by Th84,PEPIN:59;
end;
:: Pythagorean triples
theorem
a|^2 + b|^2 = c|^2 implies 5 divides a or 5 divides b or 5 divides c
proof
assume
A1: a|^2 + b|^2 = c|^2;
A1a: a|^2 + (b|^2 - 1) = (c|^2 -1) &
(a|^2 - 2) + (b|^2 + 1) = (c|^2 -1) by A1;
A1b: a|^2 + (b|^2 + 1) = (c|^2 +1) &
(a|^2 + 2) + (b|^2 - 1) = (c|^2 +1) by A1;
A2: 2*2+1 = 5; then
per cases by Th70,PEPIN:59;
suppose
5 divides c;
hence thesis;
end;
suppose
B1: 5 divides c|^2 - 1;
per cases by A2,Th70,PEPIN:59;
suppose 5 divides b;
hence thesis;
end;
suppose 5 divides b|^2 - 1; then
5 divides a|^2 by A1a,B1,INT_2:1;
hence thesis by NAT_3:5,PEPIN:59;
end;
suppose 5 divides b|^2 + 1;
hence thesis by A1a,B1,INT_2:1,Th85;
end;
end;
suppose
B2: 5 divides c|^2 + 1;
per cases by A2,Th70,PEPIN:59;
suppose 5 divides b;
hence thesis;
end;
suppose 5 divides b|^2 + 1; then
5 divides a|^2 by A1b,B2,INT_2:1;
hence thesis by NAT_3:5,PEPIN:59;
end;
suppose 5 divides b|^2 - 1;
hence thesis by A1b,B2,INT_2:1,Th85;
end;
end;
end;
theorem Th87:
not 7 divides a|^3 - 2 & not 7 divides a|^3 + 2 &
not 7 divides a|^3 - 3 & not 7 divides a|^3 + 3
& not 7 divides a|^3 - 4 & not 7 divides a|^3 + 4 &
not 7 divides a|^3 - 5 & not 7 divides a|^3 + 5
proof
7 = 2*3 + 1;
hence thesis by Th84,NAT_4:26;
end;
theorem Th88:
2 divides 2|^n - 1 iff n = 0
proof
L1: 2 divides 2|^n - 1 implies n = 0
proof
assume
A1: not thesis; then
2 divides 2|^n & not 2 divides -1 by NAT_3:3,INT_2:13;
hence contradiction by A1;
end;
n = 0 implies 2|^n - 1 = 1 - 1 by NEWTON:4;
hence thesis by INT_2:12,L1;
end;
theorem
2|^(k+l) divides 2|^(n+k) - 2|^k implies l = 0 or n = 0
proof
A0: 2|^(k+l) = 2|^k*2|^l & 2|^(n+k) = 2|^n*2|^k by NEWTON:8;
assume 2|^(k+l) divides 2|^(n+k) - 2|^k; then
A2: (2|^k)*(2|^l) divides (2|^k)*(2|^n - 1) by A0;
reconsider a = 2|^n - 1 as Element of NAT by INT_1:3;
reconsider b = 2|^l as Element of NAT by ORDINAL1:def 12;
reconsider c = 2|^k as Element of NAT by ORDINAL1:def 12;
A3: b divides a or c = 0 by A2,PYTHTRIP:7;
2 divides 2|^l or l = 0 by NAT_3:3;
hence thesis by Th88,INT_2:9,A3;
end;
theorem Th90:
3 divides b or 3 divides b-1 or 3 divides b+1
proof
2*1+1 = 3 & b|^1 = b;
hence thesis by Th70,PEPIN:41;
end;
theorem Th91:
not 3 divides b implies not 3 divides b|^2 + c|^2
proof
L1: not 3 divides b & not 3 divides c implies not 3 divides b|^2 + c|^2
proof
assume
A0: not 3 divides b & not 3 divides c;
2+1 = 3; then
3 divides b|^2 - 1 & 3 divides c|^2 - 1 by A0,Th59,PEPIN:41; then
A1: 3 divides (b|^2 -1) + (c|^2 - 1) by WSIERP_1:4;
b|^2 + c|^2 = (b|^2 + c|^2 - 2) + 2;
hence thesis by A1,INT_2:1,27;
end;
not 3 divides b & 3 divides c implies not 3 divides b|^2 + c|^2
proof
assume
A1: not 3 divides b & 3 divides c;
c divides c|^2 by NAT_3:3; then
not 3 divides b|^2 & 3 divides c|^2 by INT_2:9,A1,NAT_3:5,PEPIN:41;
hence thesis by INT_2:1;
end;
hence thesis by L1;
end;
theorem Th92:
not 3 divides b|^2 + 1 & not 3 divides b|^2 - 2
proof
A1: (b|^2 + 1) = 3 + (b|^2 - 2);
not 3 divides 1|^2 + b|^2 by INT_2:27,Th91;
hence thesis by A1,WSIERP_1:4;
end;
theorem
not 3 divides b|^3 + b|^2 - b + 1
proof
A0: (b|^2 + 1) + (b|^3 - b) = b|^3 + b|^2 - b + 1;
not 3 divides b|^2 + 1 by Th92;
hence thesis by A0,PEPIN:41,Th58,INT_2:1;
end;
theorem Th94: for a be positive Nat holds
b,c are_coprime & a+1 divides b implies not a+1 divides c
proof
let a be positive Nat;
assume
A1: b,c are_coprime & a+1 divides b;
A2: a+1 > 0+1 by XREAL_1:6;
assume not thesis;
hence contradiction by A1,A2,PYTHTRIP:def 1;
end;
theorem Th95: b,c are_coprime implies not 3 divides b|^2 + c|^2
proof
assume
A1: b,c are_coprime;
2+1 divides b implies not 2+1 divides c by A1,Th94;
hence thesis by Th91;
end;
theorem Th96: for p be prime Nat holds p divides a implies p divides a|^(n+1)
proof
let p be prime Nat;
assume p divides a; then
p divides a*a|^n by INT_2:2;
hence thesis by NEWTON:6;
end;
theorem
b,c are_coprime & b|^2 + c|^2 = a|^2 implies not 3 divides a
proof
b,c are_coprime & b|^2 + c|^2 = a|^2 implies
not 3 divides a|^(1+1) by Th95;
hence thesis by Th96,PEPIN:41;
end;
theorem Th98:
for p be prime Nat holds p divides a + b implies
p divides a|^(2*n+1) + b|^(2*n+1)
proof
let p be prime Nat;
A1: a+b divides a|^(2*n+1) + b|^(2*n+1) by NEWTON01:35;
assume p divides a + b;
hence thesis by A1,INT_2:9;
end;
theorem
for p be prime Nat holds
not p divides a|^(2*n+1) + b|^(2*n+1) & p divides a|^2 - b|^2
implies p divides a - b
proof
let p be prime Nat;
assume not p divides a|^(2*n+1) + b|^(2*n+1) & p divides a|^2 - b|^2; then
not p divides a + b & p divides (a+b)*(a-b) by NEWTON01:1,Th98;
hence thesis by INT_5:7;
end;
theorem 3 divides a*b or 3 divides a+b or 3 divides a-b
proof
per cases;
suppose 3 divides a or 3 divides b;
hence thesis by INT_2:2;
end;
suppose not 3 divides a & not 3 divides b; then
per cases by Th90;
suppose
3 divides a+1 & 3 divides b+1; then
3 divides (a+1)-(b+1) by INT_5:1;
hence thesis;
end;
suppose
3 divides a - 1 & 3 divides b - 1; then
3 divides (a-1) - (b-1) by INT_5:1;
hence thesis;
end;
suppose
3 divides a - 1 & 3 divides b + 1;then
3 divides (a-1)+(b+1) by WSIERP_1:4;
hence thesis;
end;
suppose
3 divides a + 1 & 3 divides b - 1;then
3 divides (a+1)+ (b-1) by WSIERP_1:4;
hence thesis;
end;
end;
end;
theorem
not 3 divides a & not 3 divides b implies
3 divides a|^(2*n+1) + b|^(2*n+1) or 3 divides a|^(2*n+1) - b|^(2*n+1)
proof
assume not 3 divides a & not 3 divides b; then
per cases by Th90;
suppose
3 divides a+1 & 3 divides b+1; then
3 divides (a+1) - (b+1) & a - b divides a|^(2*n+1) - b|^(2*n+1)
by INT_5:1, NEWTON01:33;
hence thesis by INT_2:9;
end;
suppose
3 divides a - 1 & 3 divides b - 1; then
3 divides (a-1) - (b-1) & a - b divides a|^(2*n+1) - b|^(2*n+1)
by INT_5:1, NEWTON01:33;
hence thesis by INT_2:9;
end;
suppose
3 divides a - 1 & 3 divides b + 1; then
3 divides (a-1) + (b+1) & a + b divides a|^(2*n+1) + b|^(2*n+1)
by WSIERP_1:4, NEWTON01:35;
hence thesis by INT_2:9;
end;
suppose
3 divides a + 1 & 3 divides b - 1; then
3 divides (a+1) + (b-1) & a + b divides a|^(2*n+1) + b|^(2*n+1)
by WSIERP_1:4, NEWTON01:35;
hence thesis by INT_2:9;
end;
end;
theorem
a|^3 + b|^3 = c|^3 implies 7 divides a or 7 divides b or 7 divides c
proof
assume
A1: a|^3 + b|^3 = c|^3;
A1a: a|^3 + (b|^3 - 1) = (c|^3 -1) & (a|^3 - 2) + (b|^3 + 1) =
(c|^3 -1) by A1;
A1b: a|^3 + (b|^3 + 1) = (c|^3 +1) & (a|^3 + 2) + (b|^3 - 1) =
(c|^3 +1) by A1;
A2: 2*3+1 = 7; then
per cases by Th70,NAT_4:26;
suppose
7 divides c;
hence thesis;
end;
suppose
B1: 7 divides c|^3 - 1;
per cases by A2,Th70,NAT_4:26;
suppose 7 divides b;
hence thesis;
end;
suppose
7 divides b|^3 - 1; then
7 divides a|^3 by A1a,B1,INT_2:1;
hence thesis by NAT_3:5,NAT_4:26;
end;
suppose 7 divides b|^3 + 1;
hence thesis by A1a,B1,INT_2:1,Th87;
end;
end;
suppose
B2: 7 divides c|^3 + 1;
per cases by A2,Th70,NAT_4:26;
suppose 7 divides b;
hence thesis;
end;
suppose 7 divides b|^3 + 1; then
7 divides a|^3 by A1b,B2,INT_2:1;
hence thesis by NAT_3:5,NAT_4:26;
end;
suppose 7 divides b|^3 - 1;
hence thesis by A1b,B2,INT_2:1,Th87;
end;
end;
end;