let n be Nat; :: thesis: ( 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 ; :: thesis: 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; :: thesis: for i being Nat holds not n divides (x |^|^ (i + 2)) + 1
let i be Nat; :: thesis: 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 A4: x is zero ; :: thesis: not n divides (x |^|^ (i + 2)) + 1
per cases ( i + 2 = 2 * m or i + 2 = (2 * m) + 1 ) by A3;
suppose i + 2 = 2 * m ; :: thesis: not n divides (x |^|^ (i + 2)) + 1
end;
suppose i + 2 = (2 * m) + 1 ; :: thesis: not n divides (x |^|^ (i + 2)) + 1
end;
end;
end;
suppose not x is zero ; :: thesis: not n divides (x |^|^ (i + 2)) + 1
then 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 ; :: thesis: contradiction
then (((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; :: thesis: verum
end;
end;