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 $f_1^2+f_2^2+\cdots+f_k^2=f_{k+1}^2$?

  • 2
    $\begingroup$ 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. $\endgroup$ Mar 12 '14 at 15:34

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.


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 $f_1 = x$, $f_2 = \sqrt{-1} x$, $f_3 = 0$, $f_4 = f_5 = 1$.

  • $\begingroup$ 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 $\endgroup$
    – joro
    Mar 12 '14 at 16:35
  • $\begingroup$ 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. $\endgroup$ Mar 12 '14 at 16:46
  • $\begingroup$ You are right, sorry. $\endgroup$
    – Sasha
    Mar 12 '14 at 17:03

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