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Consider this equation

\begin{equation} y^2 = x^3 + (36n + 27)^2 \cdot x^2 + (15552 n^3 + 34992 n^2 + 26244 n + 6561) \cdot x + (46656 n^4 + 139968 n^3 + 157464 n^2 + 78713 n + 14748) \end{equation}

where $(n, x, y)$ are integers.

For each integer $n$, this defines an elliptic curve $E_n$ over $\mathbb{Q}$. I am interested in all integer solutions $(n, x, y)$, i.e., all integer points on all curves $E_n$ as $n$ varies over $\mathbb{Z}$.

So far, I have found the following solutions by naive search (up to some bounds on $|n|, |x|$):

\begin{align} (n, x, y) &= (-1, 18, \pm 167), \\ &= (94, -562, \pm 17722), \\ &= (-110, 646, \pm 40812), \\ &= (-64, 144840, \pm 333523318), \\ &= (147498, -449511, \pm 2312387148693). \end{align}

These suggest that the problem is nontrivial and that there might be infinitely many solutions, or at least more than I have found.

Its Weierstrass equation was calculated on Magma:

\begin{equation} y^2 = x^3 + \left(-559872n^4 - 1664064n^3 - 1854576n^2 - 918540n - 170586\right) \cdot x + 161243136n^6 + 718875648n^5 + 1335341376n^4 + 1322837568n^3 + 737088984n^2 + 219032405n + 27118239. \end{equation}

My questions:

  1. How can I determine all integer solutions $(n, x, y)$ to this equation?
  2. If that is not feasible, what is known about the structure of integer points on this family? Could there be infinitely many $n$ for which $E_n$ has nontrivial integer points?

Any references, computational methods, or theoretical approaches would be greatly appreciated.


Note: I have performed a computational search up to $|n| \leq 1.5 \times 10^5$ and $|x| \leq 5 \times 10^6$, and found only the solutions listed above aside from small trivial ones. However, the rapid growth of the last two solutions suggests that more exist outside this range.

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    $\begingroup$ If you explained how those $E_n$ arise, it might be easier to identify if any integer points come up there. $\endgroup$ Commented 2 days ago
  • $\begingroup$ @Wojowu $E_n$ arose from this equation: $y^2 = x^3 + m^2 \cdot x^2 + \dfrac{1}{3} m^3 \cdot x + \dfrac{m^4 - 19m}{36} \quad \text{where $m \neq 0$}$ and if we were to let $t = \dfrac{m^4 - 19m}{36}$, a valid family of solutions would be $m(n) = 36n + 27$ and $t(n) = 46656n^4 + 139968n^3 + 157464n^2 + 78713n + 14748$ and making the necessary substitutions yield the initial problem I posed earlier. $\endgroup$ Commented 2 days ago

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