let n be Nat; ( n in 4k+3_Primes implies for x being even Nat
for i being Nat holds not n divides (x |^|^ (i + 2)) + 1 )
assume A1:
n in 4k+3_Primes
; for x being even Nat
for i being Nat holds not n divides (x |^|^ (i + 2)) + 1
then A2:
n >= 3
by Th4;
let x be even Nat; for i being Nat holds not n divides (x |^|^ (i + 2)) + 1
let i be Nat; not n divides (x |^|^ (i + 2)) + 1
reconsider i2 = i + 2 as Nat ;
consider m being Element of NAT such that
A3:
( i2 = 2 * m or i2 = (2 * m) + 1 )
by SCHEME1:1;
per cases
( x is zero or not x is zero )
;
suppose
not
x is
zero
;
not n divides (x |^|^ (i + 2)) + 1then consider m being
Nat such that A7:
x |^|^ (i + 1) = 2
* m
by ABIAN:def 2;
A8:
x |^|^ ((i + 1) + 1) =
x |^|^ (Segm ((i + 1) + 1))
.=
x |^|^ (succ (Segm (i + 1)))
by NAT_1:38
.=
exp (
x,
(x |^|^ (i + 1)))
by ORDINAL5:14
.=
x |^ (m * 2)
by A7
.=
(x |^ m) |^ 2
by NEWTON:9
;
assume
n divides (x |^|^ (i + 2)) + 1
;
contradictionthen
(((x |^ m) |^ 2) - (- 1)) mod n = 0
by A8, PEPIN:6, A2;
then A9:
(((x |^ m) ^2) - (- 1)) mod n = 0
by WSIERP_1:1;
consider k being
Nat such that A10:
(
n = (4 * k) + 3 &
n is
prime )
by A1, Def1;
A11:
n mod 4 =
3
mod 4
by A10, NAT_D:21
.=
3
by NAT_D:24
;
n > 2
by A2, XXREAL_0:2;
hence
contradiction
by A11, A9, INT_5:def 2, A10, INT_5:38;
verum end; end;