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

theorem Th15:
  for n being even Nat st n divides 2|^n+2 & n-1 divides 2|^n+1 holds
  for n1 being Nat st n1 = 2|^n+2 holds
   n1-1 divides 2|^n1+1 & n1 divides 2|^n1+2
  proof
    let n be even Nat such that
A1: n divides 2|^n+2 and
A2: n-1 divides 2|^n+1;
    let n1 be Nat such that
A3: n1 = 2|^n+2;
    reconsider n as non zero Nat by A1;
    consider k being non zero odd Nat such that
A4: 2|^n+2 = n*k by A1,Th14;
    2|^n+1 divides 2|^(n*k)+1 by Th13;
    hence n1-1 divides 2|^n1+1 by A3,A4;
    consider m being Nat such that
A5: 2|^n+1 = (n-1)*m by A2,NAT_D:def 3;
A6: m is odd by A5;
A7: 2*(2|^(n-1)+1) = 2*(2|^(n-1))+2*1
    .= (2|^(n-1+1))+2 by NEWTON:6
    .= n1 by A3;
A8: 2|^((n-1)*m+1) = 2|^((2|^n+1)+1) by A5
    .= 2|^n1 by A3;
    2*(2|^((n-1)*m)+1) = 2*(2|^((n-1)*m))+2
    .= 2|^((n-1)*m+1)+2 by NEWTON:6;
    hence n1 divides 2|^n1+2 by A6,A7,A8,Th13,INT_6:8;
  end;
