It's well known that $1$ is the sum of three cubes infinitely many different ways but is it true for perhaps the tetrahedral numbers as well? Let $T_n = (1/6)n(n+1)(n+2)$. Then the following are the first solutions less than 6000 of the form $T_a - T_b - T_c = 1$. $$ \begin{array}{c c c} a & b & c\\ 6 & 5 & 4\\ 8 & 7 & 5\\ 11 & 9 & 8\\ 23 & 19 & 17\\ 33 & 32 & 14\\ 45 & 43 & 22\\ 51 & 49 & 24\\ 76 & 75 & 25\\ 85 & 84 & 27\\ 209 & 207 & 63\\ 238 & 228 & 117\\ 323 & 304 & 177\\ 340 & 323 & 177\\ 369 & 318 & 262\\ 380 & 317 & 284\\ 449 & 422 & 248\\ 715 & 707 & 229\\ 1105 & 1022 & 655\\ 1493 & 1438 & 707\\ 2319 & 2173 & 1302\\ 2406 & 2405 & 258\\ 3183 & 2982 & 1789\\ 5950 & 5093 & 4282\\ 5985 & 5904 & 2047\\ \end{array} $$ Are there infinitely many solutions?
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1$\begingroup$ Is your large slash-separated list supposed to be a table? If so, could you organise it as such? $\endgroup$– LSpiceCommented Oct 7, 2021 at 4:00
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3$\begingroup$ You could create an OEIS entry for the $a$ such that $T_a$ is of the form $1+T_b+T_c$. The first five examples above are the not beginning of any sequence there yet: oeis.org/… $\endgroup$– user44143Commented Oct 7, 2021 at 12:48
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2$\begingroup$ 5 solutions from the list have a=b+1. This particular form corresponds to" Triangular nuber=1+Tetrahedral number", which is maybe easier? The next of this form seems to be [9878, 9877, 663] $\endgroup$– Pietro MajerCommented Oct 7, 2021 at 14:54
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8$\begingroup$ For any fixed $k$, the equation $T_{b+k}-T_b=T_c+1$ defines an elliptic curve, and as such it can only have finitely many integral points by Siegel's theorem. $\endgroup$– WojowuCommented Oct 7, 2021 at 16:26
1 Answer
There are infinitely many solutions. I'll show below that there are infinitely many positive integers $k$ for which $93k^{2} - 288k + 276 = z^{2}$ for some positive integer $z$. From such a $z$, we get a solution by setting $a = \frac{z+3k}{6} - 1$, $b = 2k-3$, and $c = \frac{z-3k}{6} - 1$. Here's a table of solutions in this family:
| $k$ | $z$ | $a$ | $b$ | $c$ |
|---|---|---|---|---|
| $1$ | $9$ | $1$ | $-1$ | $0$ |
| $23$ | $207$ | $45$ | $43$ | $22$ |
| $163$ | $1557$ | $340$ | $323$ | $177$ |
| $18005$ | $173619$ | $37938$ | $36007$ | $19933$ |
| $135460$ | $1306314$ | $285448$ | $270917$ | $149988$ |
| $15104876$ | $145666134$ | $31830126$ | $30209749$ | $16725250$ |
The choice of variables here is related to Wojowu's comment that setting $a = c+k$ yields an elliptic curve. This elliptic curve is isomorphic to $y^{2} = x^{3} - 144k^{2}x + (-432k^{6} + 1728k^{4} + 10368k^{3})$. This elliptic curve always has the point $(12k^{2} - 24k, 36(k-4)k^{2})$ on it, but this corresponds to the "trivial solution" $a = b = k-3$ and $c = -3$. I observed that setting $x = dk^{2} - 24k$ yields $y^{2} = k^{4} \cdot (\text{quadratic in } k)$. Setting $d = 24$ gives $y^{2} = 144k^{4} (93k^{2} - 288k + 276)$.
The equation $93k^{2} - 288k + 276 = z^{2}$ is equivalent after completing the square to $w^{2} - 93z^{2} = -4932$ with the restriction that $w \equiv 42 \pmod{93}$. (We have $w = 93k-144$.) We can rewrite $w^{2} - 93z^{2} = -4932$ as $N_{\mathbb{Q}(\sqrt{93})/\mathbb{Q}}(w + z \sqrt{93}) = -4932$. One solution is $-51 + 9 \sqrt{93}$. To find more solutions, note that a fundamental unit is $u = \frac{29 + 3 \sqrt{93}}{2}$ and $u^{6} = 295293601 + 30620520 \sqrt{93} \equiv 1 \pmod{93}$. So an infinite family of such $z$'s can be found by looking at the coefficient of $\sqrt{93}$ in $(-51+9\sqrt{93}) (295293601 + 30620520 \sqrt{93})^{r}$ for $r \geq 1$.
This family of solutions is quite sparse, but other families can be found by choosing a different value of $d$ in $x = dk^{2} - 24k$.
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$\begingroup$ This is fantastic thank you so much! $\endgroup$ Commented Oct 8, 2021 at 16:37