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In the following paper (http://www.ams.org/journals/jams/1991-04-04/S0894-0347-1991-1119199-X/), C.L. Stewart showed that there exist infinitely many integers $h$ such that the equation

(1) \begin{equation} xy(x+y) = h \end{equation}

has at least 18 solutions in co-prime integers $x$ and $y$. In the same paper, he also conjectured (his conjecture covers far more) that there exists positive numbers $r,c$ such that for all integers $h$ with $|h| \geq r$, the equation (1) has at most $c$ solutions in co-prime integers $x$ and $y$.

My question is, is it reasonable to conjecture that one may take $c = 18$? That is, does there exist infinitely many $h$ for which (1) has at least 19 solutions in co-prime integers $x$ and $y$?

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    $\begingroup$ You probably already know this, but letting $E_h$ be the elliptic curve (1), there is a bound $$\#\text{(coprime integer solutions)}\le C^{1+\text{rank }E_h(\mathbb Q)}$$ for an absolute constant $C$. So if the current conjecture that elliptic curve ranks are bounded is true, then Stewart's conjecture would follow. $\endgroup$ Commented Jan 10, 2018 at 1:12
  • $\begingroup$ Also that if you drop the coprimality condition then the number of solutions is unbounded. $\endgroup$ Commented Jan 11, 2018 at 2:30
  • $\begingroup$ @NoamD.Elkies As Noam says, without the gcd condition, the number is unbounded, but I believe that it is not known if for all $\epsilon>0$, the number of solutions can be greater than $O_\epsilon(|h|^{1-\epsilon})$ for infinitely many $h$. This seems to be a very interesting, and also very difficult, problem. $\endgroup$ Commented Jan 11, 2018 at 4:31
  • $\begingroup$ Do you mean $O_\epsilon(\log|h|)^{1-\epsilon}$? $\endgroup$ Commented Jan 11, 2018 at 4:36
  • $\begingroup$ (That's what one would get from elliptic curves of unbounded rank in this family.) $\endgroup$ Commented Jan 11, 2018 at 4:52

2 Answers 2

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There seem to be lots of $h$ which have $20$ solutions. Among squarefree $47$-smooth numbers I found $$101623830, 363993630, 455885430, 1418488890, 15427730010, 31983962010, 322640788470, 1087394017710, 11406069164490, 304250263527210$$ with $20$ solutions and $903210$ and $23730036330$ with $24$.

EDIT: Oops: my program was counting only solutions with $x > 0$ (but $y$ of any sign). Those numbers $101623830, \ldots, 304250263527210$ actually have $30$ solutions each, and $903210$ and $23730036330$ have $36$.

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  • $\begingroup$ Yes, the number of solutions should be a multiple of 6 since whenever $(x,y)$ is a solution then so are $(y,x), (y, - x - y), (-x-y, x), (-x-y, y), (x,-x-y)$, and these should typically be co-prime whenever $(x,y)$ are co-prime with only finitely many exceptions. $\endgroup$ Commented Jan 12, 2018 at 10:35
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There are infinitely many such numbers with at least $12$ solutions. Here is the new proof:

If $(x,y)$ is a solution to

$$xy(x+y)=h,$$

then

$$(y,x), (x,-x-y),(-x-y,x),(y,-x-y),(-x-y,y)$$

are also solutions to

$$xy(x+y)=h.$$

In fact, one can take simultaneously $x=1$ and $x=2$ to get two positive integer solutions $(1,a)$ and $(2,b)$ provided

$$h=a(a+1)=2b(b+2).$$

If $2a+1=k$ and $b+1=l$, one has:

$$k^2-8l^2=-7,$$

a generalized Pell equation with infinitely many positive integer solutions $(k,l)$, and obviously $k$ must be odd. In fact, $h=\frac{k^2-1}{4}=2l^2-2$.

This brings up at least $2\cdot6=12$ solutions for infinitely many $h$.

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