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Find the smallest number $n$ such that there exist polynomials $f_{1}, f_{2},...,f_{n}$ with rational coefficients satisfying $$x^{2}+7=f_{1}(x)^{2}+f_{2}(x)^{2}+...+f_{n}(x)^{2}.$$

It's Olympiad question.

My try is.. The equality $x^{2}+7=x^{2}+2^{2}+1^{2}+1^{2}+1^{2}$, shows that $n\leq 5$ It remains to show that $x^{2}+7$ is not a sum of four (or less) squares of polynomials with rational coefficients.

is there any one can show me a full solution?

Padrt
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  • Well you're on the right track in that you only have to figure out the smallest way to write $7$ as a sum of squares. The hitch is you have to deal with it over $\Bbb Q$ and not just $\Bbb Z$. – Gregory Grant Aug 13 '15 at 23:54
  • @GregoryGrant ok, but I should write a full solution, and I don't know how – Padrt Aug 13 '15 at 23:57
  • http://math.stackexchange.com/questions/582188/is-an-integer-a-sum-of-two-rational-squares-iff-it-is-a-sum-of-two-integer-squar and https://en.wikipedia.org/wiki/Legendre%27s_three-square_theorem – Asinomás Aug 14 '15 at 00:00
  • @Padrt There are quite a few different solutions here: http://www.artofproblemsolving.com/community/c6h418641p2362006 –  Aug 14 '15 at 00:01
  • You should be able to mix these two to prove you need at least four integers to make their squares add $7$ – Asinomás Aug 14 '15 at 00:01
  • after that you should prove having more than one polynomial of the form $(ax-b)^2$ won't help. – Asinomás Aug 14 '15 at 00:02
  • @dREaM Does that hold also for rational coefficients? – Gregory Grant Aug 14 '15 at 00:02
  • he would have to use the idea from the first link to extend to rational coefficients – Asinomás Aug 14 '15 at 00:03

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