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Find all odd positive integers $n$ for which there exists odd positive integers $x_1,x_2,..,x_n$ such that $$x_1^2+x_2^2+\cdots+x_n^2=n^4\,.$$

My work so far:

  1. For $n=3$, the equation $$x_1^2+x_2^2+x_3^2=81$$ has no solutions because if a solution exists the, in modulo $4$, we have $$3\equiv 1\pmod{4}$$ which is a contradiction.
  1. $n\ge 5$?

I need help here.

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  • $\begingroup$ Only a remark: if you take $n=m^2$, $m$ odd, then $x_k=m^3$ for all $k $ is a solution. $\endgroup$ – Kelenner Jun 28 '16 at 14:12
  • $\begingroup$ Is this a contest problem? $\endgroup$ – almagest Jun 28 '16 at 16:48
  • $\begingroup$ @almagest: Yes. It is the problem of the book "Collection of contest problems 2007-2008" $\endgroup$ – Roman83 Jun 28 '16 at 16:55
  • $\begingroup$ @Roman83 I have added the tag. This tag is useful for highlighting that the problem is more likely to require thought than knowledge! As I have just discovered to my cost, having spent 5 mins failing to solve it :) $\endgroup$ – almagest Jun 28 '16 at 17:12
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all odd squares are $1 \pmod 8.$ This includes $n^4.$ Meanwhile, $$ x_1^2 +x_2^2+ \cdots+ x_n^2 \equiv n \pmod 8. $$ So it is necessary to have $n \equiv 1 \pmod 8.$

In the other direction, all numbers $k \equiv 3 \pmod 8$ are the sum of three odd squares. This is a result of Gauss, and equivalent to the fact that all positive integers ae the sum of three triangular numbers, including $0$ if needed.

As a result, take any $n \equiv 1 \pmod 8.$ Take $x_4, x_5, \ldots, x_n$ to be anything (odd) you like, as long as the sum of squares is below $n^4.$ The leftover requires $$ x_1^2 + x_2^2 + x_3^2 = n^4 - \left( x_4^2 + \cdots+ x_n^2 \right) $$ which can always be solved in (odd) integers.

After Greg's comment, found a fairly greedy solution that does not require quoting Gauss. We have $n \equiv 1 \pmod 8.$ Note $37 \equiv 5 \pmod 8.$

Let $$ K = \frac{3n - 3}{8}, $$ $$ W = \frac{5n - 37}{8}, $$ so that $$ K + W = n-5, $$ $$ 9K + W = 4n-8. $$ The solution will have $K$ of the $x_j$ equal to $3,$ so those squares are $9,$ and their sum is $9K.$ We will also have $W$ of the $x_j$ set to $1,$ so their sum is just $W.$ Then, with a total of $n$ (odd) squares, $$ \color{red}{ (n^2 - 2)^2 + n^2 + n^2 + n^2 + (n-2)^2 + 9K + W = n^4} $$ With $n=9$ we get $K=3, W=1, $ $ \; \; 9^4 = 6561,$ then $$ 79^2 + 9^2 + 9^2 + 9^2 + 7^2 + 9 \cdot 3 + 1 = 6241 + 81 + 81 + 81 + 49 + 27 + 1 = 6561. $$

With $n=17$ we get $K=6, W=6, $ $ \; \; 17^4 = 83521,$ then $$ 287^2 + 17^2 + 17^2 + 17^2 + 15^2 + 9 \cdot 6 + 6 = 82369 + 289 + 289 + 289 + 225 + 54 + 6 = 83521. $$

With $n=25$ we get $K=9, W=11, $ $ \; \; 25^4 = 390625,$ then $$ 623^2 + 25^2 + 25^2 + 25^2 + 23^2 + 9 \cdot 9 + 11 = 388129 + 625 + 625 + 625 + 529 + 81 + 11 = 390625. $$

With $n=33$ we get $K=12, W=16, $ $ \; \; 33^4 = 1185921,$ then $$ 1087^2 + 33^2 + 33^2 + 33^2 + 31^2 + 9 \cdot 12 + 16 = 1181569 + 1089 + 1089 + 1089 + 961 + 108 + 16 = 1185921. $$

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  • $\begingroup$ So for example, take $x_4=\cdots=x_n=1$ and represent $n^4-n+3$ as the sum of three odd squares. $\endgroup$ – Greg Martin Jun 28 '16 at 17:52
  • $\begingroup$ @GregMartin added in a (mostly) greedy solution at the end that could be found by a student who did not know Gauss's result. $\endgroup$ – Will Jagy Jun 28 '16 at 18:26

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