k-dimensional cube-tetrahedrons

I conjecture that $n^k = \dfrac{m(m+1)\cdots(m+k-1)}{k!}$ , where $n,m,k$ are positive integers, has an infinite amount of solutions for $n$ if $k \leq 2$ and that $n = 1$ is the only possible value of $n$ if $k > 2$ .

I am able to prove the case where $k \leq 2$ , but I am unable to prove the case where $k > 2$ .

If you can prove this or find a counter example, please post a comment!

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Comments

Proof for case $k \leq 2$ :

If $k = 1$ clearly $n = m$ which has an infinite amount of solutions.

If $k = 2$ ,

$\begin{aligned} &n^2 = \dfrac{m(m+1)}{2} \\ &2n^2 = m^2 + m \\ &m^2 + m - 2n^2 = 0 \\ &m = \dfrac{-1 \pm \sqrt{1 + 8n^2}}{2} \\ &1 + 8n^2 = x^2 \\ &x^2 - 8n^2 = 1\end{aligned}$

This is now a Pell's equation which is well know to have an infinite amount of solutions.

Jesse Nieminen - 4 years, 11 months ago

Have you considered looking at prime factorisations? It might work in the $k=3$ case.

Sharky Kesa - 4 years, 9 months ago

Could you post the case $k \leq 2$ ? Perhaps something in your solution will inspire

Alex G - 4 years, 11 months ago

Math	Appears as
Remember to wrap math in `\(` ... `\)` or `\[` ... `\]` to ensure proper formatting.
`2 \times 3`	$2 \times 3$
`2^{34}`	$2^{34}$
`a_{i-1}`	$a_{i-1}$
`\frac{2}{3}`	$\frac{2}{3}$
`\sqrt{2}`	$\sqrt{2}$
`\sum_{i=1}^3`	$\sum_{i=1}^3$
`\sin \theta`	$\sin \theta$
`\boxed{123}`	$\boxed{123}$