
theorem
   for a,b be odd Integer, m be odd Nat holds
   Parity (a|^m + b|^m) = Parity (a + b)
   proof
     let a,b be odd Integer, m be odd Nat;
     per cases;
     suppose
       A1: a + b = 0; then
       a = -b; then
       a|^m = -(b|^m) by POWER:2;
       hence thesis by A1;
     end;
     suppose
       A1: a + b <> 0; then
       a <> -b; then
       a|^m <> (-b)|^m by NEWTON03:13; then
       a|^m <> -(b|^m) by POWER:2; then
       a|^m + b|^m <> 0; then
       Parity (a|^m + b|^m) = 2|^(2|-count (a|^m + b|^m)) &
         Parity (a+b) = 2|^(2|-count (a + b)) by A1,Def1;
       hence thesis by NEWTON03:81;
     end;
   end;
