theorem Even:
  for a,b be odd Nat holds
    2|^m divides a|^n - b|^n implies 2|^(m+1) divides a|^(2*n) - b|^(2*n)
  proof
    let a,b be odd Nat;
A0: a|^(2*n) = (a|^n)|^2 & b|^(2*n) = (b|^n)|^2 by NEWTON:9;
    assume
A1: 2|^m divides a|^n - b|^n;
A2: 2 divides a|^n + b|^n by ABIAN:def 1;
    a|^(2*n) - b|^(2*n) = (a|^n - b|^n)*(a|^n + b|^n) by A0,NEWTON01:1; then
    2*2|^m divides a|^(2*n) - b|^(2*n) by A1,A2,NEWTON02:2;
    hence thesis by NEWTON:6;
  end;
