
theorem
  for k,l be odd square Integer holds (k+l) mod 8 = 2
  proof
    let k,l be odd square Integer;
    1|^2 is square & 2*0+1 is odd & 1 is integer; then
    A1: (k - 1) mod 8 = 0 & (l - 1) mod 8 = 0 by N0319;
    (k+l) mod 8 = ((k-1)+((l-1)+2)) mod 8
    .= (((k-1) mod 8) + (((l-1)+2) mod 8)) mod 8 by NAT_D:66
    .= (((l-1) mod 8) + (2 mod 8)) mod 8 by NAT_D:66,A1
    .= 2 mod (2+6) by A1
    .= 2;
    hence thesis;
  end;
