
theorem
  for a be Integer, b be non zero Integer holds a|^2 mod b = (b-a)|^2 mod b
  proof
    let a be Integer, b be non zero Integer;
    reconsider x = -a as Integer;
    (b - a)|^2 mod b = ((x + 1*b) mod b)|^2 mod b by GRCY330
    .= (x mod b)|^2 mod b by NAT_D:61
    .= ((-a)|^(2*1)) mod b by GRCY330
    .= a|^2 mod b by POWER:1;
    hence thesis;
  end;
