If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer I am trying to solve:

If $n$ and $m$ are odd integers, show that $ \frac{(nm)^2 -1}8$ is an integer.

If I write $n=2k+1$ and $m=2l+1$ I get stuck at
$$\frac{1}{8}(16k^2 l^2 +4(k+l)^2 +8kl(k+l)+4kl+2(k+l))$$

Expanding the whole thing, you get
$$
(2k+1)^2(2\ell+1)^2 -1 = 16 k^2 \ell^2+16 k^2 \ell+4 k^2+16 k \ell^2+16 k \ell+4 k+4 \ell^2+4 \ell
$$
Modulo 8, the terms multiplied by 16 disappear, and this becomes
$$
4 k^2+4 k+4 \ell^2+4 \ell = 4k(k+1) + 4\ell(\ell+1)
$$
and it only remains to see that both $\ell(\ell+1)$ and $k(k+1)$ are even to conclude. (since then $8$ divides both $4\ell(\ell+1)$ and $4k(k+1)$.)
