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?

  • 11
    $\begingroup$ Clearly a necessary condition is that $p(n)$ is a sum of two squares for every integer $n$. Could this condition be sufficient? Is there a nice characterization of polynomials with this property? $\endgroup$ Commented May 29, 2020 at 3:03
  • 4
    $\begingroup$ Not an answer, just a comment: The result that a nonnegative univariate polynomial is a sum of two squares is a direct consequence of the fundamental theorem of algebra. The result by Polya and Szegö is more subtle, it deals with polynomials on $[0,\infty[$. $\endgroup$ Commented May 29, 2020 at 4:47
  • $\begingroup$ @KasperAndersen Thanks for the correction. The result I found states that if $p(x) \geq 0$ for all $x \geq 0$ then there exist $f, g, h, k$ so that $p(x) = f(x)^2 + g(x)^2 + x(h(x)^2 + k(x)^2)$. Is that the result you were referring to? If so, I'd be interested in understanding when $f, g, h, k$ have integer coefficients as well. $\endgroup$
    – Gautam
    Commented May 29, 2020 at 18:12

1 Answer 1


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:

  1. $p$ is a sum of two squares in $\mathbf Z[x]$;
  2. $p(n)$ is a sum of two squares in $\mathbf Z$ for all $n \in \mathbf Z$;
  3. 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 $\mathbf Q[x]$.)


[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.

  • $\begingroup$ Thanks for this thoughtful answer. Do you have any thoughts about the situation in Q[x]? $\endgroup$
    – Gautam
    Commented Jul 9, 2020 at 0:59
  • 1
    $\begingroup$ Can't you just clear denominators? Like, multiply with a large square to assume $p \in \mathbf Z[x]$, and apply the criterion of DLS? $\endgroup$ Commented Jul 9, 2020 at 15:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.