Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

share|improve this question
    
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

1 Answer 1

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.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.