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
$$\frac{1}{8}(16k^2 l^2 +4(k+l)^2 +8kl(k+l)+4kl+2(k+l))$$

$$((2k+1)(2l+1))^2-1=16k^2l^2+4k^2+16kl^2+16kl+4k+4l^2+16lk^2+4l.$$
Dropping all the terms with coefficient $16$,
$$4(k^2+k+l^2+l)=4(k(k+1)+l(l+1))$$ must be a multiple of $8$.

With a slightly simpler evaluation:
$$((2k+1)(2l+1))^2-1=(4kl+2k+2l)(4kl+2k+2l+2)=4(2kl+k+l)(2kl+k+l+1).$$
This is a multiple of $8$.
