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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 months, 1 week ago

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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. Jesse Nieminen · 2 months, 1 week ago

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@Jesse Nieminen Have you considered looking at prime factorisations? It might work in the \(k=3\) case. Sharky Kesa · 3 weeks, 4 days ago

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Could you post the case \(k \leq 2\)? Perhaps something in your solution will inspire Alex G · 2 months, 1 week ago

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