# 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!

Note by Jesse Nieminen
2 years, 1 month ago

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## Comments

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Proof for case $$k \leq 2$$:

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

If $$k = 2$$,

\begin{align} &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{align}

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

- 2 years, 1 month ago

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

- 1 year, 11 months ago

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

- 2 years, 1 month ago

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