1
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ 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
1
$\begingroup$

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 $d=qs$. $\quad\blacksquare$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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