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
$$\frac{1}{8}(16k^2 l^2 +4(k+l)^2 +8kl(k+l)+4kl+2(k+l))$$
 A: 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.
A: While I think SBareS's answer should be clear enough, and the other answers are certainly valid, I'd like to present a clearer proof:
(nm)2 - 1 = (nm + 1)(nm - 1)
Note that, since both n and m are odd integers, (nm + 1) and (nm - 1) are both even integers, one of which is a multiple of 4. Their product is therefore a multiple of 8. QED
A: $$((2k+1)(2l+1))^2-1=16k^2l^2+4k^2+16kl^2+16kl+4k+4l^2+16lk^2+4l.$$
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 $8$.
A: 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 $8$.
A: 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 $4k(k+1)$.)
A: It is no big deal to perform exhaustive search i.e. compute $(nm)^2-1$ for all moduli
$$\begin{align}(1,1)&\to0\\
(3,1)&\to8\\
(5,1)&\to24\\
(7,1)&\to48\\
(3,3)&\to80\\
(5,3)&\to224\\
(7,3)&\to440\\
(5,5)&\to624\\
(7,5)&\to1224\\
(7,7)&\to2400.\end{align}$$
A: $$(mn)^2-1=[(2r+1)(2s+1)+1][(2r+1)(2s+1)-1)]=4(2rs+r+s+1)(2rs+r+s)$$ thus divisible by $8$ since it is the product of $4$ by two consecutive numbers.
A: 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.
A: 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.
