Q: Proof with Integers, Multiplicity and Perfect Squares 
Prove for every positive integer m that is a multiple of $8$, there exist two positive integers $a$ and $b$ that differ by $m$ such that $ab$ is a perfect square.

So far I'm having trouble setting up the proof as well as analyzing what it's actually asking me to do. How do I interpret the part that says "a and b differ by m"? 
Here's how I started:
Assume m is an arbitrary positive integer, and also assume that m is a multiple of 8.
WNTS: $\exists$a,b $\in$ $\mathbb{Z}$$^+$, (a-m) ^ (b-m), ab=x$^2$ for some arbitrary x $\in$ $\mathbb{Z}$.
m being a multiple of 8 $\implies$ 8|m $\implies$ m=8k$_1$ for some k$_1$ $\in$ $\mathbb{Z}$.
However, I don't think this is quite right, because I don't know what to do with the (a-m) ^ (b-m) portion.
If I use substitution I get a-m=a-8k$_1$ and b-m=b-8k$_1$. At this point I don't know what to do.
 A: Hint: Let $u$ be maximal with $u^2\mid a$, and $a=u^2w$. Then we must have $w\mid b$ and $\frac bw=v^2$ for some $v$ (and we end up with $ab=(uvw)^2$). Then $a-b=u^2w-v^2w=(u+v)(u-v)w$. Given $m$ with $8\mid m$, can you find suitable $u,v,w$?
A: 
How do I interpret the part that says "a and b differ by m"?

I'd interpret it as $a -b = m$ and we'll assume without loss of generality that $a > b$.
So we want $ab = x^2$
$a(a-m) = x^2$
$a(a - 8k_1) = x^2$
So we have to find some $a$ and $x$ that will allow this to work.
One way to do this is to assume $a$ and $a-8k_1$ are both perfect squares.
Now in general,  If $w = u+v$ then $w^2 = u^2 + 2uv + v^2$ so $w^2 - n^2 = 2uv+v^2$.
We  want to find some $w = \sqrt{a}; u= \sqrt{a-m}=\sqrt b; $ and $2uv + v^2 = m=8k_1$
Let $v$ be any even number small enough.  Say $v=2$. then we want $4u +4 = 8k_1$ so $u = 2k_1 -1$.  And $b = (2k_1 -1)^2$ and $a = (2k_1 - 1)^2 + 8k_1$
That'll do it if we put it together.
That is to say:  
Pf:
Let $m = 8k$
Let $b = (2k - 1)^2 $.  Let $a = (2k-1)^2 + 8k = 4k^2 - 4k + 1 + 8k = 4k^2 +4k + 1 = (2k + 1)^2$
Then $a - b = 8k = m$ and $ab = (2k-1)^2(2k+1)^2 = [(2k-1)(2k+1)]^2 = (4k^2 -1)^2$ which is a perfect square.
