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.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Do there exist $3$ positive integers $a,b,c~(a<b<c),$ such that the equation $$ax^2+a=by^2+b=cz^2+c$$ has infinitely many integer solutions $x,y,z$ ?

share|cite|improve this question
Something tells me that if there is going to be it should be something like $a=r$, $b=rs$, $c=rs^4$, for $s$ not a square. This is such that the general solutions of the Pell equations related to these are computed by powers $M+\sqrt{s}N$ for the two equations. Otherwise different radicals appear and it seems unlikely one gets infinitely many solutions. But I thought about it without being careful. – Mlazhinka Shung Gronzalez LeWy Jul 19 '13 at 4:37
up vote 1 down vote accepted

This would require $$ ax^2-cz^2=c-a\\ (ax)^2-(ac)z^2=a(c-a) $$ and likewise $$ (by)^2-(bc)z^2=b(c-b) $$

According to Bennett "On the number of solutions of simultaneous Pell equations" this can only have finitely many solutions unless $$ acb(c-b) = bca(c-a) $$ that is, unless $b=a$, so the answer to your question is negative.

share|cite|improve this answer

Your Answer


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.