theorem
  not 3 divides a & not 3 divides b implies
    3 divides a|^(2*n+1) + b|^(2*n+1) or 3 divides a|^(2*n+1) - b|^(2*n+1)
  proof
    assume not 3 divides a & not 3 divides b; then
    per cases by Th90;
    suppose
      3 divides a+1 & 3 divides b+1; then
      3 divides (a+1) - (b+1) & a - b divides a|^(2*n+1) - b|^(2*n+1)
        by INT_5:1, NEWTON01:33;
      hence thesis by INT_2:9;
    end;
    suppose
      3 divides a - 1 & 3 divides b - 1; then
      3 divides (a-1) - (b-1) & a - b divides a|^(2*n+1) - b|^(2*n+1)
        by INT_5:1, NEWTON01:33;
      hence thesis by INT_2:9;
    end;
    suppose
      3 divides a - 1 & 3 divides b + 1; then
      3 divides (a-1) + (b+1) & a + b divides a|^(2*n+1) + b|^(2*n+1)
        by WSIERP_1:4, NEWTON01:35;
      hence thesis by INT_2:9;
    end;
    suppose
      3 divides a + 1 & 3 divides b - 1; then
      3 divides (a+1) + (b-1) & a + b divides a|^(2*n+1) + b|^(2*n+1)
        by WSIERP_1:4, NEWTON01:35;
      hence thesis by INT_2:9;
    end;
  end;
