theorem
  for a,b be odd Integer, m be odd Nat holds
    2 |-count (a|^m + b|^m) = 2 |-count (a+b)
  proof
    let a,b be odd Integer, m be odd Nat;
    reconsider c = -b as odd Integer;
    reconsider n = (m-1)/2 as Nat;
    2 |-count (a|^m + (-c)|^m) =
      2 |-count (a|^(2*n+1) +(-c|^(2*n+1))) by POWER:2
    .= 2 |-count (a|^(2*n+1) - c|^(2*n+1))
    .= 2 |-count (a -(-b)) by Odd;
    hence thesis;
  end;
