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Show that $2019^{2n}$ can be expressed as a sum of ten different positive squares, for every positive integer $n$.

I have tried to solve this problem considering first particular cases and then expressing 2019 as the sum of two integers and trying to use and apply Newton's binomial, but I have not taken what is wanted. I would like to know if there is a less cumbersome test using perhaps a simpler method.

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Hint: if $2019^{2m} = a_1^2 +\ldots + a_{10}^2$, then $2019^{2m+2} = (2019 a_1)^2 + \ldots + (2019 a_{10})^2$. So you really only need to do it for $m=1$.

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  • $\begingroup$ I don't understand why it is enough to do it for $m = 1$ Could you explain this please. $\endgroup$
    – wessi
    Feb 28, 2020 at 13:15
  • $\begingroup$ As the hint shows, if it's true for $m$ then it's true for $m+1$. Use mathematical induction. $\endgroup$ Feb 28, 2020 at 13:29

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