Consider the following conjecture.

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.

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.

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.

Consider the following proposition (well, conjecture currently!).

Theorem. Let $\alpha$ and $\beta$ be in $\mathbb{Z}_{>0}$, with $\beta < \alpha \le 2(\beta+1)$. If $(\alpha+\beta)^2$ divides $2\beta^3+6\alpha\beta^2-1$, then the quotient is $1$.

I believe this to be true, and furthermore believe that the only solution is $(\alpha,\beta)=(4,1)$. Inspired by a comment in another thread* suggesting Vieta jumping as a possible approach, I've developed a partial proof.

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 $\alpha < 2(\beta+1)$. Therefore $k=1$, as claimed.

Now assume $\xi_2$ is not an integer. Multiplying (2) by $k$ yields \begin{align} k\xi_2 &= 2\beta(3\beta-k) - k\alpha = \frac{\beta^2(k-2\beta)+1}{\alpha}. \tag{3} \end{align} By the first relation, $k\xi_2$ is an integer.

END OF PARTIAL PROOF

I don't know how to complete this proof. I do know in addition to what's there that 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 $k$, and in the case of the one known solution $(4,1)$, my $k=1$, which yields the degenerate Vieta root $(1,0)$. But I don't think that is enough to claim that I've found all solutions.

Finally, note that removing the restriction $\alpha > \beta$ reveals another solution, namely $(\alpha,\beta)=(11364,46061)$, where \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*} Notably, in this situation 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*} is not an integer. It would be enlightening to see if there are any other ``unrestricted'' integral or rational solutions to (1), or perhaps determine bounds on $\max(\alpha,\beta)$ or $\lvert\alpha \pm \beta\rvert$ in any hypothetical solution.

[*] Note: I moved this question here because I believe the proof has gone into territory more appropriate to MO than MSE.

Consider the following conjecture.

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.

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.

Consider the following proposition (well, conjecture currently!).

Theorem. Let $\alpha$ and $\beta$ be in $\mathbb{Z}_{>0}$, with $\beta < \alpha < 2\beta$. If $(\alpha+\beta)^2$ divides $2\beta^3+6\alpha\beta^2-1$, then the quotient is $1$.

I believe this to be true, and furthermore believe that the only solution is $(\alpha,\beta)=(4,1)$. Inspired by a comment in another thread* suggesting Vieta jumping as a possible approach, I've developed a partial proof.

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 $\alpha < 2\beta$. Therefore $k=1$, as claimed.

Now assume $\xi_2$ is not an integer. Multiplying (2) by $k$ yields \begin{align} k\xi_2 &= 2\beta(3\beta-k) - k\alpha = \frac{\beta^2(k-2\beta)+1}{\alpha}. \tag{3} \end{align} By the first relation, $k\xi_2$ is an integer.

END OF PARTIAL PROOF

I don't know how to complete this proof. I do know in addition to what's there that 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 $k$, and in the case of the one known solution $(4,1)$, my $k=1$, which yields the degenerate Vieta root $(1,0)$. But I don't think that is enough to claim that I've found all solutions.

Finally, note that removing the restriction $\alpha > \beta$ reveals another solution, namely $(\alpha,\beta)=(11364,46061)$, where \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*} Notably, in this situation 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*} is not an integer. It would be enlightening to see if there are any other ``unrestricted'' integral or rational solutions to (1), or perhaps determine bounds on $\max(\alpha,\beta)$ or $\lvert\alpha \pm \beta\rvert$ in any hypothetical solution.

[*] Note: I moved this question here because I believe the proof has gone into territory more appropriate to MO than MSE.

Consider the following proposition (well, conjecture currently!).

Theorem. Let $\alpha$ and $\beta$ be in $\mathbb{Z}_{>0}$, with $\beta < \alpha \le 2(\beta+1)$. If $(\alpha+\beta)^2$ divides $2\beta^3+6\alpha\beta^2-1$, then the quotient is $1$.

I believe this to be true, and furthermore believe that the only solution is $(\alpha,\beta)=(4,1)$. Inspired by a comment in another thread* suggesting Vieta jumping as a possible approach, I've developed a partial proof.

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 $\alpha < 2(\beta+1)$. Therefore $k=1$, as claimed.

Now assume $\xi_2$ is not an integer. Multiplying (2) by $k$ yields \begin{align} k\xi_2 &= 2\beta(3\beta-k) - k\alpha = \frac{\beta^2(k-2\beta)+1}{\alpha}. \tag{3} \end{align} By the first relation, $k\xi_2$ is an integer.

END OF PARTIAL PROOF

I don't know how to complete this proof. I do know in addition to what's there that 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 $k$, and in the case of the one known solution $(4,1)$, my $k=1$, which yields the degenerate Vieta root $(1,0)$. But I don't think that is enough to claim that I've found all solutions.

Finally, note that removing the restriction $\alpha > \beta$ reveals another solution, namely $(\alpha,\beta)=(11364,46061)$, where \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*} Notably, in this situation 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*} is not an integer. It would be enlightening to see if there are any other ``unrestricted'' integral or rational solutions to (1), or perhaps determine bounds on $\max(\alpha,\beta)$ or $\lvert\alpha \pm \beta\rvert$ in any hypothetical solution.

[*] Note: I moved this question here because I believe the proof has gone into territory more appropriate to MO than MSE.