theorem SC1:
  for a,b be odd Nat, m be even Nat holds
    2 |-count (a|^m + b|^m) = 1
proof
  let a,b be odd Nat, m be even Nat;
  A1: 4 = 2*2 .= 2|^2 by NEWTON:81;
  4 divides a|^m - b|^m by NEWTON02:65; then
  A2: not 2|^(1+1) divides a|^m + b|^m by A1,NEWTON02:58;
  a|^m + b|^m is even; then
  2|^1 divides (a|^m + b|^m);
  hence thesis by A2,NAT_3:def 7;
end;
