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Consider the expression $$ a^2b^2-ab-ap\qquad (a,b\in \mathbb{Z}^+), $$ where $p\equiv1\pmod{4}$.

Question. For every prime $p\equiv1\pmod{4}$ do there exist $a,b\in\mathbb{Z}^+$ such that $b\equiv3\pmod{4}$ and the above expression is a perfect square?

Thanks!

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    $\begingroup$ There must be a story behind all this. What is it? $\endgroup$
    – Włodzimierz Holsztyński
    Commented Jan 18, 2015 at 19:29
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    $\begingroup$ After some thinking, I arrived at the following equivalent condition: there are positive integers $\ell$ and $m$ such that $m\mid\ell^2$ and $p+m\equiv 0\pmod{4\ell-1}$. I think it is possible that a counterexample exists, but justifying it might be very difficult. By the way, your title is misleading as being squareful is different from being a square. $\endgroup$
    – GH from MO
    Commented Jan 19, 2015 at 18:38
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    $\begingroup$ @GHfromMO Your criterion is very close to one of the known conditions (due to Nakayama) for $4/p$ to be expressible as the sum of three reciprocals (Erdos-Straus conjecture), namely that $p + 4a^2 d \equiv 0 (\operatorname{mod} 4aed-1)$. I suspect that the original poster is attempting to resolve the Erdos-Straus conjecture and is discovering one of its many equivalent formulations (see my paper with Elsholtz at arxiv.org/abs/1107.1010 for a summary). $\endgroup$
    – Terry Tao
    Commented Jan 20, 2015 at 3:18
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    $\begingroup$ @GH from MO A quick computer search shows that your condition holds for all primes $p\equiv 1\pmod{4}$, $p\leq 10^7$. $\endgroup$
    – Kasper Andersen
    Commented Jan 20, 2015 at 20:31
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    $\begingroup$ Is it verified that OP's formulation is (almost) equivalent to the conjecture of Erdős-Straus? It is true for all $p<10^9$, $p\equiv 1(\!\!\mod 4)$ that such $a,b$ exists. $\endgroup$
    – Lehs
    Commented Jan 24, 2015 at 12:15

1 Answer 1

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This is just an extended comment.

The problem is equivalent to finding $b\equiv 3\pmod{4}$ and $c\equiv 0\pmod{4}$ such that $$(\star)\qquad 4b^2c\mid (b+c+p)^2.$$

In fact, when $(\star)$ holds, setting $a$ equal to the quotient, i.e. $a:=\frac{(b+c+p)^2}{4b^2c}$, gives $$a^2b^2 - ab - ap = \left(\frac{(p+b)^2 - c^2}{4bc}\right)^2.$$

Now, $(\star)$ implies that $\gcd(b,c)|p$.

If $\gcd(b,c)=p$, then $b=pb'$ and $c=pc'$, implying that $$4pb'^2c'\mid (c'+b'+1)^2$$ and thus $b' < c'$ and $4pb'^2c' \leq (c'+b'+1)^2 \leq 4c'^2$, i.e. $pb'^2\leq c'$. On the other hand, since $c'|(b'+1)^2$, we have $pb'^2\leq c'\leq (b'+1)^2$, which is impossible since $p\geq 5$.

Hence, $\gcd(b,c)=1$ and $(\star)$ is equivalent to existence of integers $b,c$ such that $$\begin{cases}b\mid c+p,\\ 4c\mid (b+p)^2.\end{cases}$$

Let's consider a partial case when $c=4d$ and $d$ is squarefree. Then $(\star)$ implies that $4bd\mid b+4d+p$, and setting $q:=\frac{b+4d+p}{4bd}$ we get $$(qb-1)(4qd-1) = qp+1.$$ In particular, we have a positive answer to the original question as soon as there is an integer $q>0$ such that $qp+1$ has a divisor $m\equiv -1\pmod{4q}$ and $\frac{m+1}{4q}$ is squarefree.

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