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
$$\frac{1}{8}(16k^2 l^2 +4(k+l)^2 +8kl(k+l)+4kl+2(k+l))$$
All odd numbers $a$ satisfy:

$$a^2 \equiv 1 \pmod8$$

So $8 | (nm)^2 - 1$, since $nm$ is an odd number. To see that the above holds, it suffices to check that $1^2 \equiv 3^2\equiv 1\pmod8$, since all odd numbers are either congruent to $1$, $3$, $-3$ or $-1$ modulo 8.
Okay, I'm not the best at writing proofs because I'm only a senior in high school and haven't ever written one except about calculus theorems that my teacher didn't explain, so excuse any terrible structure.

Axiom: Two integers multiplied by each other produces a third integer.
An integer that has a factor of 8 is divisible by 8.

Let n and m be odd integers defined by the following:
n = 2k + 1 and
m = 2l + 1,
where k and l are integers. Therefore, the equation follows:
((nm)^2 - 1) = (((2k + 1)(2l + 1))^2 - 1) = (2k + 1)^2 * (2l + 1)^2 - 1 =
(4k^2 + 4k + 1) * (4l^2 + 4l + 1) - 1.
Note that thus far, I have excluded the 1/8. And this is becoming a pain to type like that and it will become even more confusing, but the proof will continue as follows: Multiply the factors of the two trinomial expressions. Simplify. Each term will have a factor of 8 except + 1 and - 1, which cancel each other. When multiplied by 1/8, this leave factors of 2, k, k^2, l, and l^2. Each of these is an integer (and k and l need not be odd), therefore, the result must also be an integer.