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Show that for all $n\in\mathbb{N}$ there exist $a_i\in\mathbb{Z}, i=1,2,\dots,n$ distinct numbers so that: $$\sum_{i=1}^{n}a_i^2=\sum_{i=1}^{n}a_i^3$$

Using normal induction wont bring me anything, so maybe using an induction step on would be nice, like if it holds for n then it will also hold for n+2, then we need only to show that there exist two numbers so that $a^2+b^2=a^3+b^3$.

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    $\begingroup$ What are your own thoughts on this problem? $\endgroup$
    – Hirshy
    Jan 23, 2016 at 15:10
  • $\begingroup$ So the $a_i$'s are different in each sum, or not? $\endgroup$
    – Jimmy R.
    Jan 23, 2016 at 15:10
  • $\begingroup$ Both $a_i$ are the same $\endgroup$
    – CryoDrakon
    Jan 23, 2016 at 15:12
  • $\begingroup$ For $n=2$ we have $a=1,b=0$ as a solution $\endgroup$
    – Ross Millikan
    Jan 23, 2016 at 15:23
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    $\begingroup$ For $n=3$ I find $4,-3,-2$ and $13,-11,-8$ $\endgroup$
    – Ross Millikan
    Jan 23, 2016 at 15:35

1 Answer 1

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The statement is true. If you have two disjoint solutions, the union of the two is also a solution. If you have a solution that does not use $0$ or $1$, you can create new solutions with one or two more numbers by adding them in. We therefore have a proof if we can show there are arbitrarily large solutions for $n=3$.

For any positive $k$, we can let $$a_3=-k(k+2)\\ a_2=-k(k+1)(k+2)/2+1\\ a_1=k(k+1)(k+2)/2+1$$ and the equation is satisfied. The first few solutions are $$\begin {array} {r|r|r|r}k&a_3&a_2&a_1\\ \hline 1&-3&-2&4\\2&-8&-11&13\\3&-15&-29&31\\4&-24&-59&61\\5&-35&-104&106\\6&-48&-167&169\\7&-63&-251&253\\8&-80&-359&361\\9&-99&-494&496\\10&-120&-659&661 \end {array}$$ There are other three term solutions, but this is a nicely organized family.

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