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?



Prove that $1$ is the sum of three tetrahedral numbers infinitely many different ways
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 $x = dk^{2} - 24k$.


