Show that for any two integers a, b , (a+b)^2 ≡ a^2 + b^2 (mod 2) I have my solution below i wanted someone to help chekc if i have done anything wrong. Thank You for your help. The thing that is going on here is that 2x = 0 (mod 2) for any x. If x = ab, then 2ab = 0 (mod 2). We see that (a+b)^2 = a^2 + 2ab + b^2. Then this equals a^2 + 0 + b^2 (mod 2). Therfore, the Proof: By definition, showing (a+b)^2 = a^2 + b^2 (mod 2) is equivalent to showing that 2 divides [(a+b)^2 - (a^2+b^2)] 2 divides [(a^2 + 2ab + b^2 - a^2 - b^2] 2 divides 2ab To show 2 divides 2ab we need to find an integer k such that 2k = 2ab. Take k = ab. Thus it is proved.