What can we say if over the integers $aX^2 = bY^2$, that involves $\text{lcm}(a,b)$? Let $a,b$ be fixed integers and if $aX = bY$ for some integers $X,Y$, then we can say that $aX = bY = \text{lcm}(a,b)Z$ for some integer $Z$. So that if $U_c := \{cX: X \in \Bbb{Z}\}$, then $U_a \cap U_b \supset U_{\text{lcm}(a,b)} \ni x$ for any $x$ in the intersection.  That is sort of used used in the construction of the evenly-spaced integer topology used in Furstenberg's proof, but of course their basis sets have an offset.  In other words, $\{U_c : c \in \Bbb{Z}\}$ is a basis for a topology on $\Bbb{Z}$.
But what can we do if we define $U_c :=\{ cX^2 : X \in \Bbb{Z}\}$?  Can we say something like if $z = aX^2 = bY^2$, then $z = \text{lcm}(a,b)Z^2$ for some $Z$?
I'm lacking the tools to peer into such questions.  Any hints?
 A: we can say that $a,b$ are in the same squareclass; the easiest way to phrase that is to say that $ab$ is a square.
You wanted some with the lcm. Another way to write this is to write
$g =\gcd(a,b)$ and then say that both $|a/g|$ and $|b/g|$ are squares, along with $ab > 0.$ It is traditional to take the gcd positive.
You like lcm; take
$$ a = g \alpha,  $$
$$ b = g \beta,  $$
with $g >0$ the gcd and 
$$ \gcd(\alpha,\beta) = 1, $$
which you should confirm for yourself. 
Then either $\alpha = v^2, \beta = w^2,$ or $\alpha = -v^2, \beta = -w^2.$
Indeed
$$ \operatorname{lcm} (a,b) = g v^2 w^2  $$ because
$$ \operatorname{lcm} (a,b) = ab/g  $$ and
$$ ab = g^2 v^2 w^2  $$
A: Working by example first:
$z = (5\cdot 2^2 \cdot 3)(3\cdot7)^2 = (5 \cdot 7^2 \cdot 3)(2 \cdot 3)^2$.
In this case $z = \text{lcm}(a,b) \gcd(X,Y)^2$.  
I want to try and prove it in the general case by induction on the number of primes available to form numbers.
If either $a,b$ are $0$, then $aX^2 = bY^2 = 0 = \text{lcm}(a,b)\gcd(X,Y)^2 = 0$.  So Let $a,b$ both be nonzero.  If $a,b = \pm 1$.  The negative numbers bring trouble, so let's work with only $a,b \in \Bbb{Z}^+$.  Maybe multiplying by $\text{sgn}(a)$ would fix that.  Continuing, it's clearly true for $a=b = 1$.  So it's true when there are "zero primes involved in forming $a,b$".  
Now suppose that it's true when $a,b$ are composed of the first $n \geq 0$ primes only.  Now consider all $a', b'$ which are composed only of the first $n+1$ primes, say the new addition is called $p$.  Then if $a'X^2 = b'Y^2 = ap^{e_a} X^2 = bp^{e_b}Y^2$ where $e_b \geq e_a \geq 0$ WLOG, we have $aX^2 = bp^{k}Y^2$ where $k = e_b - e_a$. By our construction, $\gcd(a,p) = 1$ so that $p^{k/2} | X$ since $2$ must divde $k$ now.
Let $l = k/2$, we have then $z = aX^2 = b(p^l Y)^2$ and by the above inductive argument $z= \text{lcm}(a,b)\gcd(X,p^l Y)^2$. But since $p^l$ also divides $X$, we can factor it out: $z = \text{lcm}(a,b)p^{2l}\gcd(X,Y)^2$ and clearly $\text{lcm}(a,b)p^k = \text{lcm}(a', b')$.  But that's simply not true !  I fail...
