proofs involving congruence of integers Let n, m be integers. Prove that if n ≡ 1( mod 2) and m ≡ 3( mod 4), then n^2 +m ≡ 0(mod4)
My thought process is  that n=3 and m=7 so n^2 would be 9+7 but then i do not believe that would equal 0(mod4) which i think would equal 4. 
Any direction would help on where to start or how to attack this problem.

$n \equiv 1( \mod 2)$ means $\exists a \in \mathbb{Z}$ such that $n = 2\cdot a + 1$, so 
$$ n^2 = (2\cdot a + 1)^2 = 4a^2 + 4a +1.$$
So you can calculate $n^2 +m \mod 4$ now.
