TITLE: SOS polynomials with integer coefficients
QUESTION [27 upvotes]: A well known theorem of Polya and Szego says that every non-negative univariate polynomial $p(x)$ can be expressed as the sum of exactly two squares: $p(x) = (f(x))^2 + (g(x))^2$ for some $f, g$. Suppose $p$ has integer coefficients. In general, its is too much to hope that $f, g$ also have integer coefficients; consider, for example, $p(x) = x^2 + 5x + 10$. Are there simple conditions we can impose on $p$ that guarantee that $f, g$ have integer coefficients?

REPLY [48 votes]: There is the following result of Davenport, Lewis, and Schinzel [DLS64, Cor to Thm 2]:

Theorem. Let $p \in \mathbf Z[x]$. Then the following are equivalent:

$p$ is a sum of two squares in $\mathbf Z[x]$;
$p(n)$ is a sum of two squares in $\mathbf Z$ for all $n \in \mathbf Z$;
Every arithmetic progression contains an $n$ such that $p(n)$ is a sum of two squares in $\mathbf Z$.


Criterion 3 is really weak! For example, it shows that in 2, we may replace $\mathbf Z$ by $\mathbf N$. Because it's short but takes some time to extract from [DLS64], here is their proof, simplified to this special case.
Proof. Implications 1 $\Rightarrow$ 2 $\Rightarrow$ 3 are obvious. For 3 $\Rightarrow$ 1, factor $p$ as
$$p = c \cdot p_1^{e_1} \cdots p_r^{e_r}$$
with $p_j \in \mathbf Z[x]$ pairwise coprime primitive irreducible and $c \in \mathbf Q$. We only need to treat the odd $e_j$ (and the constant $c$). Let $P = p_1 \cdots p_r$ be the radical of $p/c$, and choose $d \in \mathbf N$ such that $P$ is separable modulo every prime $q \not\mid d$. Suppose $P$ has a root modulo $q > 2d\operatorname{height}(c)$; say
$$P(n) \equiv 0 \pmod q$$
for some $n$. Then $P'(n) \not\equiv 0 \pmod q$, hence $P(n+q) \not\equiv P(n) \pmod{q^2}$. Replacing $n$ by $n+q$ if necessary, we see that $v_q(P(n)) = 1$; i.e. there is a $j$ such that
$$v_q\big(p_i(n)\big) = \begin{cases}1, & i = j, \\ 0, & i \neq j.\end{cases}.$$
If $e_j$ is odd, then so is $v_q(p(n))$, which equals $v_q(p(n'))$ for all $n' \equiv n \pmod{q^2}$. By assumption 3 we can choose $n' \equiv n \pmod{q^2}$ such that $p(n')$ is a sum of squares, so we conclude that $q \equiv 1 \pmod 4$. If $L = \mathbf Q[x]/(p_j)$, then we conclude that all primes $q > 2d\operatorname{height}(c)$ that have a factor $\mathfrak q \subseteq \mathcal O_L$ with $e(\mathfrak q) = f(\mathfrak q) = 1$ (i.e. $p_j$ has a root modulo $q$) are $1$ mod $4$. By Bauer's theorem (see e.g. [Neu99, Prop. VII.13.9]), this forces $\mathbf Q(i) \subseteq L$.
Thus we can write $i = f(\theta_j)$ for some $f \in \mathbf Q[x]$, where $\theta_j$ is a root of $p_j$. Then $p_j$ divides
$$N_{\mathbf Q(i)[x]/\mathbf Q[x]}\big(f(x)-i\big) = \big(f(x)-i\big)\big(f(x)+i\big),$$
since $p_j$ is irreducible and $\theta_j$ is a zero of both. Since $f(x)-i$ and $f(x)+i$ are coprime and $p_j$ is irreducible, there is a factor $g \in \mathbf Q(i)[x]$ of $f(x)+i$ such that
$$p_j = u \cdot N_{\mathbf Q(i)[x]/\mathbf Q[x]}(g) = u \cdot g \cdot \bar g$$
for some $u \in \mathbf Q[x]^\times = \mathbf Q^\times$. Applying this to all $p_j$ for which $e_j$ is odd, we get
$$p = a \cdot N_{\mathbf Q(i)[x]/\mathbf Q[x]}(h)$$
for some $h \in \mathbf Q(i)[x]$ and some $a \in \mathbf Q^\times$. By assumption 3, this forces $a$ to be a norm as well, so we may assume $a = 1$. Write $h = \alpha H$ for $\alpha \in \mathbf Q(i)$ and $H \in \mathbf Z[i][x]$ primitive. Then
$$p(x) = |\alpha|^2 H \bar H,$$
so Gauss's lemma gives $|\alpha|^2 \in \mathbf Z$. Since $|\alpha|^2$ is a sum of rational squares, it is a sum of integer squares; say $|\alpha|^2 = |\beta|^2$ for somce $\beta \in \mathbf Z[i]$. Finally, setting
$$F + iG = \beta H,$$
we get $p = F^2 + G^2$ with $F, G \in \mathbf Z[x]$. $\square$

Footnote: I am certainly surprised by this, given that the version for four squares is clearly false. Indeed, the condition just reads $p(n) \geq 0$ for all $n \in \mathbf Z$. But the OP's example cannot be written as any finite sum of squares in $\mathbf Z[x]$, because exactly one of the terms can have positive degree. (However, it might be different in $\mathbf Q[x]$.)

References.
[DLS64] H. Davenport, D. J. Lewis, and A. Schinzel, Polynomials of certain special types. Acta Arith. 9 (1964). ZBL0126.27801.
[Neu99] J. Neukirch, Algebraic number theory. Grundlehren der Mathematischen Wissenschaften 322 (1999). ZBL0956.11021.