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

I will use a method similar to yours. Letting $m,n$ respectively equal $2k+1$ and $2p+1$ and subsequently expanding the expression, we get $2k^2p^2+ 2k^2 p+ \frac{k^2}{2} + 2kp^2 +2kp + \frac{k}{2} +\frac{p^2}{2} + \frac{p}{2}$. Of course, $k$ and $p$ are both integers, so the only terms we have to worry about are the fractions. 
Note $\frac{p}{2} + \frac{p^2}{2} = \frac{p(p+1)}{2}$ which is clearly integer since one of $p$ and $p+1$ must be even. Hence $\frac{p}{2} + \frac{p^2}{2}$ is integer; with similar reasoning, $\frac{k}{2} + \frac{k^2}{2}$ is integer. 
This implies the expression is integer.
