I have the Diophantine equation $$3a^2(4a^2+1)=b(b+1). \tag{$\star$}$$

Each side can evidently be “separated” into two [integer] factors as $$3a^2 \cdot (4a^2+1) = b \cdot (b+1).$$ Now I believe I can claim that, for some integers $p,q,r,s$, the system of equations \begin{align} 3a^2 &= pq, &&& b &= pr, \\ 4a^2+1 &= rs, &&& b+1 &= qs \end{align} must cover all possibilities. Is this correct, or am I missing some case(s)?

EDIT: If this does cover all possibilities, then this proof should be valid to show that ($\star$) has no non-trivial integer solutions.

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    $\begingroup$ Looks good to me. $\endgroup$ – vadim123 Nov 6 '14 at 21:49
  • $\begingroup$ Is there any solution of this equation? Give at least one example. In addition to the trivial. $\endgroup$ – individ Nov 7 '14 at 5:07
  • $\begingroup$ @individ: See edited question, and in particular the link. $\endgroup$ – Kieren MacMillan Nov 7 '14 at 14:46
  • $\begingroup$ That's why I wondered where the solutions can be. This equation has not tried to solve it? mathoverflow.net/questions/133348/… $\endgroup$ – individ Nov 7 '14 at 15:17

The answer is Yes: it’s an easy consequence of the Fundamental Theorem of Arithmetic (FTA). Here’s the proof of the general case.

Proposition: If $a$, $b$, $c$, and $d$ are positive integers with $ab=cd$, then there exist positive integers $p$, $q$, $r$, and $s$ such that $a=pq$, $b=rs$, $c=pr$, and $d=qs$.

Proof. We are given $ab=cd$. Set $p=\gcd(a,c)\ge1$, so that $a=pq$ and $c=pr$ for positive integers $q$ and $r$ with $\gcd(q,r)=1$. Hence $pqb = prd$, and dividing both sides by $p \ne 0$ leaves $qb = rd$. As $\gcd(q,r)=1$, the FTA implies $r \mid b$, say $b=rs$ for an integer $s \ge 1$. ‎Then $qrs=rd$, so $d=qs$. $\quad\blacksquare$

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