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Prove that for any positive integer $k\geq 3$ there exists an integer $n$ which is simultaneously a sum of $2, 3, \ldots, k$ distinct fourth powers.

I can prove this if we use distinct cubes instead. Note that $7^3+8^3=1^3+5^3+9^3$. Then if $$N=\sum_{j=1}^m a_{m_j}^3\quad\text{for}\quad m=2,3,\ldots,k$$ we will have $216N=\displaystyle\sum_{j=1}^m (6a_{m_j})^3$, so recalling that $3^3 + 4^3 + 5^3 = 6^3 = 216$, $$216N=\sum_{j=1}^k(6a_{k_j})^3=\left(\sum_{j=1}^{k-1}(6a_{k_j})^3\right)+(3a_{k_k})^3+(4a_{k_k})^3+(5a_{k_k})^3,$$ going from $k$ distinct cubes to $k+2$ distinct cubes.

But how can we extend this to the case of fourth powers?

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  • $\begingroup$ Is it known that it works for fourth powers, or is it a conjecture you're trying to settle? $\endgroup$ – coffeemath Jan 24 '16 at 16:23
  • $\begingroup$ It came from a test, so I'm pretty sure it's true. $\endgroup$ – jlammy Jan 24 '16 at 16:34
  • $\begingroup$ Not sure if this helps, but with a Python program, I found $7^4+28^4=3^4+20^4+26^4$. $\endgroup$ – Noble Mushtak Jan 24 '16 at 16:35
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    $\begingroup$ Also, according to Wolfram Alpha, $95800^4+217519^4+414560^4=422481^4$. I think we can use both of these facts to prove the theorem similarly to how you did for cubic powers above. $\endgroup$ – Noble Mushtak Jan 24 '16 at 16:41
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In order to make this easy, I'm literally just going to use your proof, except with fourth powers instead of third powers.

Note that $7^4+28^4=3^4+20^4+26^4$. Then if $$N=\sum_{j=1}^m a_{m_j}^4\quad\text{for}\quad m=2,3,\ldots,k$$ we will have $422481^4N=\displaystyle\sum_{j=1}^m (422481a_{m_j})^4$, so recalling that $95800^4 + 217519^4 + 414560^4 =422481^4$, $$422481^4N=\sum_{j=1}^k(422481a_{k_j})^4=\left(\sum_{j=1}^{k-1}(422481a_{k_j})^4\right)+(95800a_{k_k})^4+(217581a_{k_k})^4+(414560a_{k_k})^4,$$ going from $k$ distinct fourth powers to $k+2$ distinct fourth powers.

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