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 $x,y,z$ ?
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2$\begingroup$ 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. $\endgroup$ – OR. Jul 19 '13 at 4:37
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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.
