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$$\left(\frac{ax}{p_1}\right)^2 - \left(\frac{bx}{p_2}\right)^2 = cx \quad\text{and}\quad a\neq p_1, \, \, b \neq p_2$$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?

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Are you asking, given $a,b,c,p_1,p_2$, how do you find $x$? or are you asking, given $x$, how do you find $a,b,c,p_1,p_2$? or are you asking something else entirely? Please edit the question to clarify what's given and what's to be solved for. – Gerry Myerson Nov 25 '12 at 23:59
The answer below answers my question. Thanks anyway. – Numerio Nov 26 '12 at 2:35
up vote 1 down vote accepted

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.

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