Finding an equation of solutions of an equation given constraints $$\left(\frac{ax}{p_1}\right)^2 - \left(\frac{bx}{p_2}\right)^2 = cx \quad\text{and}\quad a\neq p_1, \, \, b \neq p_2$$where $a,b,c$ are nonzero integers, and $p_1$ and $p_2$ are distinct prime factors of a nonzero natural number $x$. 
We assume that $x = p_1p_2p_3...$ where each prime factor is distinct.
The question is, how does one find such solution for any possible $x$? Does a solution exist?
 A: Using $p=p_1$ and $q=p_2$ for the two primes, multiplication by $p^2q^2$ brings it to
$$(a^2q^2-b^2p^2)x=cp^2q^2.$$
Since you want $x=pq...$ a product of distinct primes including $p,q$, write $x=pqy$ where $y$ is to be squarefree and not divisible by either of $p,q$. This brings it to
$$(a^2q^2-b^2p^2)pqy=cp^2q^2,$$
$$(a^2q^2-b^2p^2)y=cpq.$$
From this, since neither of $p,q$ divide $y$, we see that $p|a$ and $q|b$. Now put $a=pa'$ and $b=qb'$, and it becomes
$$(a'^2p^2 q^2-b'^2q^2p^2)pqy=cp^2q^2,$$
$$(a'^2-b'^2)pqy=c.$$
From this we see that $pq|c$ and may put $c=pqc'$ and arrive at
$$(a'^2-b'^2)y=c'.$$
Now for solutions, it all depends on what $a',b',c'$ happen to be at this point. Certainly $c'$ would have to be divisible by $y$,and so we could write $c'=yc''$ and then get to
$$(a'^2-b'^2)=c''.$$
At this point it seems the equation only boils down to the initial values of $a,b,c$, and may end up with no solutions, or with any number of them, by tracing backward through the substitutions above. 
