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

Note that if $a=2b+1$ is an odd integer then $a^2=(2b+1)^2=8\frac {b(b+1)}2+1$
It won't do for $mn$ to be even, but if $mn$ is odd its square will be one more than a multiple of $8$.
