
theorem
  for a,b be odd Nat holds ((a - b)/2)|^2 mod (a*b) = ((a + b)/2)|^2 mod (a*b)
  proof
    let a,b be odd Nat;
    A1: ((a - b)/2)|^2 = ((a - b)/2)*((a -b)/2) &
    ((a + b)/2)|^2 = ((a + b)/2)*((a + b)/2) by NEWTON:81;
    (((a - b)/2)|^2 + a*b) mod (a*b) =
    ((((a - b)/2)|^2 mod (a*b)) + ((a*b) mod (a*b))) mod (a*b) by NAT_D:66;
    hence thesis by A1;
  end;
