Prove that if a and b are both odd integer

[annoymage]

Homework Statement

prove that if a and b are both odd integer, then $$16|(a^2+b^2-2)$$

Homework Equations

n/a

The Attempt at a Solution

let $$a=2m+1 \ and \ b=2n+1$$, then $$a^2+b^2-2=4(m(m+1)+n(n+1))$$ so its divisible by 4, and also divisible 8 since $$m(m+1) \ and \ n(n+1)$$ are even.
so i only prove $$8|(a^2+b^2-2)$$, then how to continue? clue please T_T

 

[annoymage]

hey I've got counter example, a=1 and b=3, so the question is wrong?

 

[woodyallen1]

16\8=0 or am i wrong?

 

[woodyallen1]

A thought..Let a^2 +b^2-2<> 16k. a^2+b^2<>16k+2 => a^2+b^2<>2m => odd +odd<> even which is wrong. (<> different from)

 

[Dick]

hey I've got counter example, a=1 and b=3, so the question is wrong?

3^2+7^2-2=56. That isn't divisible by 16 either. Yes, there is something wrong with the question.