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Let $f_1,f_2,f_3,f_4,f_5 \in \mathbb{Q}[x]$ be linear and coprime and not all constant.

Is it possible $ f_1^2+f_2^2+f_3^2+f_4^2=f_5^2$?

I suppose the answer is negative.

If this is possible, solving $f_5(x)=N$ would give deterministic representation of $N$ as sum of four squares (probabilistic algorithms exist).

Couldn't solve this by equating coefficients.

What is the smallest natural $k$ such that subject to the same constraints $f_1^2+f_2^2+\cdots+f_k^2=f_{k+1}^2$?

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Equating coefficients (in either the case of $4$ or $k$) gives you two sums of squares equations, and one sum of products. Compare these with the Cauchy-Schwarz inequality to confirm that they must all be scalar multiples. – Zack Wolske Mar 12 '14 at 15:34
up vote 18 down vote accepted

To see that $f_{1}$, $f_{2}$, $\ldots$, $f_{k}$ must be scalar multiples of $f_{k+1}$, plug in the root of $f_{k+1}$ into both sides of the equation.

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Many thanks Jeremy Rouse. – joro Mar 12 '14 at 17:00

The answer would be positive if there would be no condition on rationality of the coefficients. With this condition the answer is negative. Indeed, assume first that $f_5$ is not constant. Then $f_5(t) = 0$ for some $t \in \mathbb{Q}$. Substituting it into the equation you get $\sum f_i^2(t) = 0$, hence each $f_i(t) = 0$ as we are over rationals. It follows that all $f_i$ are proportional to $f_5$. On the other hand, if $f_5$ is constant then by looking at the leading coefficients you see that the sum of their squares is zero, so each coefficient is zero, and so all $f_i$ are constant too.

Note that the same argument works for arbitrary number of squares and for any degree of polynomials.

Over $\mathbb{C}$ it is easy to give an example. Take $f_1 = x$, $f_2 = \sqrt{-1} x$, $f_3 = 0$, $f_4 = f_5 = 1$.

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Thanks what do you mean by any degree? I get (3-x^2)^2+(2*x)^2+(2*x)^2+(2*x)^2=(x^2+3)^2 – joro Mar 12 '14 at 16:35
You somehow assume that $f_5$ has a rational root, whereas in general a nonconstant polynomial can very well be irreducible over Q. Joro's example works perfectly well, since $x^2+3$ only has complex roots. – Achim Krause Mar 12 '14 at 16:46
You are right, sorry. – Sasha Mar 12 '14 at 17:03

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