reserve X for set;
reserve a,b,c,k,m,n for Nat;
reserve i,j for Integer;
reserve r for Real;
reserve p,p1,p2 for Prime;

theorem Th78:
  k > 0 & n = 28*k+1 implies (2|^(2*n)+1)|^2 + 2|^2 is composite
  proof
    assume that
A1: k > 0 and
A2: n = 28*k+1;
    set x = (2|^(2*n)+1)|^2+2|^2;
A3: (2|^(2*n)+1)|^2 = (2|^(2*n)+1)^2 by WSIERP_1:1;
A4: 2|^(2*n) = 4|^n by Lm1,NEWTON:9;
    k >= 0+1 by A1,NAT_1:13;
    then 28*k >= 28*1 by XREAL_1:64;
    then 28*k+1 >= 28+1 by XREAL_1:6;
    then 28*k+1 > 1 by XXREAL_0:2;
    then 4|^n > 4|^1 by A2,PEPIN:66;
    then 2|^(2*n)+1 > 4+1 by A4,XREAL_1:8;
    then (2|^(2*n)+1)^2 > 5*5 by XREAL_1:96;
    then
A5: x > 25+4 by A3,Lm1,XREAL_1:8;
    assume x is non composite;
    then x is prime by A5,XXREAL_0:2,NUMBER02:def 1;
    hence thesis by A2,A5,Th77;
  end;
