Prove that there exists a positive integer that can be written as the sum of $2015$ $2014$th powers of distinct positive integers $x_1 <x_2 <\ldots <x_{2015}$ in at least two ways. How can I demonstrate it?

  • $\begingroup$ What is the source of this problem? $\endgroup$ – punctured dusk Apr 2 '15 at 8:38
  • 1
    $\begingroup$ Cesenatico 2014 $\endgroup$ – Domenico Vuono Apr 2 '15 at 8:41

Consider an integer $n> 2015$ and let us select $2015$ $x_i \in [1, N]$. Now there are $\displaystyle \binom{n}{2015}$ ways of selecting the $x_i$.

Assume there is no positive number as desired by the question. Then, in each case, $\sum x_i^{2014}$ must be distinct. Each sum is not greater than $2015 n^{2014}$, so by Pigeonhole principle, for all $n> 2015$ we must have, $$\binom n {2015} \le 2015 n^{2014}$$ But the LHS is a polynomial of degree $2015$, while the RHS has lesser degree, so for large enough $n$, this inequality will be violated. Hence there must exist such a positive number as desired.


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