Finding integer points on elliptic curves via divisibility conditions like $(a+b)^2 \mid (2b^3+6ab^2-1)$ Is the following conjecture correct? 
Conjecture. The divisibility condition $(\alpha+\beta)^2 \mid (2\beta^3+6\alpha\beta^2-1)$ has no solutions in positive integers $1 \le \beta < \alpha < 2\beta$.
This question is related to finding integer points on a Mordell curve. A computer search outside the range indicated turned up the expected single solution $(a,b)=(4,1)$ [corresponding to the single integer point on the original Mordell curve], as well as an unexpected solution $(\alpha,\beta)=(11364,46061)$, which I can't explain. Any insights would be appreciated. 
I believe a method of solution — particularly by descent — to this special case would be immediately applicable to a large class of elliptic curves.
FWIW, I've developed a partial proof which I include below. The Vieta jump implies that, for any solution $(a,b)$, there is a rational solution $(b,\tfrac{6b^2}{k}-2b-a)$ with the same $k$; in the case of the one known solution $(a,b)=(4,1)$, we do have $k=1$ in the original equation
$$
  (5b-a+1)(a+b)^2 = 2(2b^3+6ab^2-1),
$$
which makes the second Vieta root degenerate at $(1,0)$. 

Proof (incomplete). The divisibility hypothesis implies
\begin{equation}
 k(\alpha+\beta)^2 = 2\beta^3+6\alpha\beta^2-1,  \tag{1}
\end{equation}
for an integer $k \ge 1$. Rearranging (1) and replacing $\alpha$ with the variable $\xi$ yields
\begin{equation*}
 k\xi^2 + 2\beta(k-3\beta)\xi + (k\beta^2-2\beta^3+1) = 0.%  \label{EQ: solve this}
\end{equation*}
One root of this equation is $\xi_1 = \alpha$. By Vieta's formulas, the other root may be written as
\begin{align}
 \xi_2 &= \frac{2\beta(3\beta-k)}{k} - \alpha = \frac{\beta^2(k-2\beta)+1}{k \alpha}.  \tag{2}
\end{align}
First, assume $\xi_2$ is an integer. Since $\alpha$ is an integer, the first relation in (2) implies that the fraction must also be an integer. Hence $k \mid 6\beta^2$. But (1) implies both that $k$ is odd, and that $\gcd(k,\beta)=1$. Hence $k \mid 3$, so $k = 1$ or $3$. If $k = 3$, then the second relation in (2) implies that $\xi_2$ is positive when $\beta = 1$, and negative when $\beta > 1$. On the other hand, the first relation with $k=3$ gives $\xi_2 = 2\beta(\beta-1) - \alpha$. When $\beta = 1$, this implies $0 < \xi_2 = 2\beta(\beta-1) - \alpha = -\alpha$, contradicting $\alpha > 0$. When $\beta > 1$, we have $0 > \xi_2 = 2\beta(\beta-1) - \alpha$. Hence $\alpha > 2\beta(\beta-1)$, contradicting $\alpha < 2\beta$.
END OF PARTIAL PROOF
Note that the unexpected solution $(\alpha,\beta)=(11364,46061)$ implies
\begin{equation*}
 k = \frac{2\beta^3+6\alpha\beta^2-1}{(\alpha+\beta)^2} = \frac{340107729770625}{3297630625} = 3 \cdot 31 \cdot 1109,
\end{equation*}
and then we have
\begin{equation*}
 \frac{2\beta(3\beta-k)}{k} - \alpha = \frac{\beta^2(k-2\beta)+1}{k \alpha} = \frac{685486248}{34379}
\end{equation*}
not an integer.
 A: I think that you can, except for your one case of $(b,j)=(1,2)$, change the bounds to $300+1 \lt b \lt a \lt 2b-300.$ The number $300$ could easily be made as large as you please (unless a counter-example to your conjecture turned up) because each $j$ only allows a few possible pairs $a=2b+j$ and these can be checked. That does not prove anything but the method is simple and might interest you. 
Pick a particular $j$ (and you have assumed $j \le 2$), set $a=2b+j$ and consider first the weaker condition that $a+b \mid 2b^3+6ab^2-1$ which becomes $3b+j \mid 14b^3+6b^2j-1$. Any common divisor of those two must also divide $14b^2(3b+j)-3(14b^3+6b^2j-1)=-4b^2j+3.$ Continuing in this fashion we see that any common divisor must divide $27-4j^3.$ So 


*

*find all the divisors of $27-4j^3$ of the form $3k+j$ 

*For each set $b=k,a=2k-j$ and, provided that $b \lt a$ check if indeed  $3b+j \mid 14b^3+6b^2j-1$.

*On the rare occasions that that happens check the actual desired condition that  $(3b+j)^2 \mid 14b^3+6b^2j-1$.
For $-30 \le j \le 2$ the pairs $[b,j]$ which survive until the last test are 
$[2671, -28], [5865, -28], [451, -26], [3254, -23], [625, -22],$$ [843, -22], [322, -19], [179, -18], [602, -17], [18, -13],$$ [76, -13], [592, -13], [11, -8], [141, -8], [12, -5], [1, 2]$
But only $[1,2]$ survives the last step. 
I checked as far as $j=-300.$
A: Here's a reformulation of the conjecture, and a heuristic which suggests that the conjecture is false.  Note that $a+b$ must be odd.  Put $\ell = a+b$  (and $\ell$ is odd) and $m=a-b$ which is also odd.  The condition that $b<a<2b$ now translates to $0<m < \ell/3$.  The divisibility condition of the question translates nicely to the condition $\ell^2$ divides $m^3-2$.  
So the question really is can $m^3-2$ have a square factor of size exceeding $(3m)^2$?   Here's a heuristic that indicates why this should be possible.  Given $\ell$ odd, the number of solutions $\pmod {\ell^2}$ to $x^3-2\equiv 0 \pmod{\ell^2}$ is given by a multiplicative function $f(\ell)$.  Here if $p\ge 5$ then $f(p^k)$ is simply the number of solutions to $x^3\equiv 2 \pmod p$ which is $0$, $1$ or $3$ (with asymptotic densities as in Chebotarev).  For $p=3$ one should work a tiny bit harder.  Then the question is whether one of these $f(\ell)$ solutions $\pmod{\ell^2}$ happens to lie below $\ell/3$.  We may expect that probability to be about $f(\ell)/(3\ell)$.  Thus the expected number of $\ell$ below $x$ for which $\ell^2$ divides $m^3-2$ (with $m< \ell/3$) should be about 
$$ 
\sum_{\ell < x; \ell \text{ odd}} f(\ell)/(3\ell) \sim C \log x,
$$ 
for some positive constant $C$ -- with a little calculation, using the Dedekind zeta function one should be able to calculate $C$.   This suggests that there are infinitely many counterexamples.  
I didn't find any with $m$ below $10000$ (I searched for large square factors of $m^3-2$), but my guess would be that there is a reasonable chance of finding counterexamples with $m$ in the millions, and a very good chance if $m$ is in the billions.  
Added:  A relevant paper for such problems is Granville's paper in IMRN.
Granville shows that on the abc conjecture the largest square factor of any cubic polynomial (without repeated roots) $f(m)$ is at most $m^{2+o(1)}$.  This generalizes the observation in GH from MO's answer (the general case requires more work than the easy $x^3-2$ example).  Granville also gives reasons to believe that this exponent is best possible, and in particular shows that such large square factors are attained for cubic polynomials having a rational root -- this uses the solutions to Pell's equation.  Of course, Granville's construction doesn't apply for $x^3-2$ which is irreducible, but the results may be suggestive.  Also note that the Pell type constructions produce a logarithmic number of examples with large square factors, which is again suggestive.
A: This is a supplement to Lucia's answer. His/her heuristic analysis suggests that there are infinitely many odd $m$'s such that the largest square dividing $m^3-2$ exceeds $9m^2$, contradicting the conjecture. 
On the other hand, the abc conjecture implies that the largest square dividing $m^3-2$ is $\ll_\epsilon m^{2+\epsilon}$ for any $\epsilon>0$, which indicates that a slightly weaker version of the conjecture (i.e. one with a more restrictive constraint on the variables) is probably true.
Some calculations with SAGE show that for $m<10^7$ there are only two instances where the largest square dividing $m^3-2$ exceeds $m^2$, namely $m=3$ and $m=100$. (Noam Elkies extended the range to $m<\ell<10^8$ with a more efficient gp-pari code.) I found five further values with a square divisor exceeding $m^2/2$, and these are $\{1244,11317,296428,714417,722428\}$.
Added. The constant in Lucia's heuristic analysis (with $m$ restricted to odd numbers) equals
$$ \frac{1}{6}\frac{1}{2}\frac{2}{3}\prod_{\substack{p>3\\f(p)=0}}\left(1-\frac{1}{p^3}\right)\prod_{\substack{p>3\\f(p)=1}}\left(1-\frac{1}{p^2}\right)\prod_{\substack{p>3\\f(p)=3}}\left(1-\frac{3}{p^2}+\frac{2}{p^3}\right)$$
times the residue of the Dedekind zeta function of $\mathbb{Q}(\sqrt[3]2)$. The initial factor $1/6$ accounts for Lucia's denominator $3$ and the fact that we only look for odd $m$'s. The other factors come from comparing the Euler factors of $\sum_{\ell=1}^\infty f(\ell)\ell^{-s}$ to those of the Dedekind zeta function. My calculation in SAGE shows that
$ C\approx 0.0423$, which suggests that the average ratio between the consecutive $\ell$'s yielding a counterexample $(\ell,m)$ is about $1.867\times 10^{10}$. 
Summary of ongoing calculation. Counterexamples to the OP's conjecture correspond bijectively to pairs of odd numbers $(\ell,m)$ such that $\ell^2\mid m^3-2$ and $m/\ell<1/3$.
I extended Noam Elkies' gp-pari calculation to the range $\ell<10^{12}$ and am checking higher ranges on several machines. The best pairs I found so far (allowing $m$ even) are 
$$(\ell,m)=(10444012561, 5062142741)\quad\text{with ratio}\quad m/\ell\approx 0.485$$
$$(\ell,m)=(22713683537, 7950843140)\quad\text{with ratio}\quad m/\ell\approx 0.350$$
In particular, these beat the basic ratio $m/\ell=0.600$ that corresponds to $(\ell,m)=(5,3)$.
