# About integer polynomials which are sums of squares of rational polynomials…

I have the following question for which I haven't been able to find any reference or proof.

Suppose we know that a univariate polynomial $P(X)$ with integer coefficients is the sum of squares of two polynomials with rational coefficients.

Is it true that $P(X)$ must also be the sum of squares of two polynomials with integer coefficients?

For example, take $P(X)=50X^2+14X+1$, then we see that $P(X)=(5X+3/5)^2+(5X+4/5)^2$, but it is also $X^2+(7X+1)^2$.

I would greatly appreciate any help pointing me into the right direction.

Thanks in advance, and regards, Guillermo

• I wonder if it helps to interpret "sum of two squares of rational polynomials" as the equivalent "norm of an element of ${\mathbb Q}[i][x]$". – Greg Martin Nov 28 '11 at 5:40
• I think this is false. Consider (5x^2+3x/5+4)^2+(5x^2+4x/5-3)^2. – Peter McNamara Nov 28 '11 at 7:20
• @Hsueh-Yung Lin: 3 is not a sum of two squares of rational numbers. – Marc van Leeuwen Nov 28 '11 at 7:49
• @Peter: (7x^2+x)^2 + (x^2+5)^2 – Andrés E. Caicedo Nov 28 '11 at 8:01

Yes. Suppose $n\in \mathbb N$ is minimal so that $P(x)=f_1^2+f_2^2$, where $nf_1$ and $nf_2$ are in $\mathbb Z[x]$.
Let $p$ be a prime with $p^\alpha||n$. Since $P\in \mathbb Z[x]$ we have $p^{2\alpha}| (p^\alpha f_1)^2+(p^\alpha f_2)^2$. Denoting $p^\alpha f_i$ by $g_i$, and letting $\beta$ be square root of $-1\pmod{p^{2\alpha}}$ (it is not hard to show that this must exist by looking at the coefficients of $g_i$ with lowest $p$-valuation).
We have $g_1^2+g_2^2\equiv 0\pmod{p^{2\alpha}}$ so $g_2^2\equiv (\beta g_1)^2\pmod{p^{2\alpha}}$ so that $p^{2\alpha}| ag_1+bg_2$ for some integers $a,b$ with $a^2+b^2=p^{2\alpha}$ and $(ab,p)=1$.
Now we can take $P(x)^2=\left(\frac{af_1+bf_2}{p^{\alpha}}\right)^2+\left(\frac{af_2-bf_1}{p^\alpha}\right)^2$ and both polynomials have coefficients with $\nu_p\geq 0$. Now repeat the procedure with other prime divisors of $n$ until you have polynomials with integer coefficients.
In fact, if $P(x)$ is a polynomial with integer coefficients and if every arithmetic progression contains an integer $n$ for which $P(n)$ is a sum of two rational squares, then $P(x) = u_1(x)^2 + u_2(x)^2$ identically, where $u_1(x)$ and $u_2(x)$ are polynomials with integral coefficients. This follows from a theorem of Davenport, Lewis, and Schinzel; see the Corollary to Theorem 2 in Polynomials of certain special types (Acta Arith. IX, 1964, 107--116).