# Sum of two squares and implication of Bunyakovsky conjecture

Bunyakovsky's conjecture states that a polynomial with integer coefficients takes infinitely many prime values at integers, unless this is impossible for trivial reasons.

Let $a_1(x), a_2(x), a_3(x), a_4(x) \in \mathbb{Z}[x]$ and $a_i \ne \pm a_j,i \ne j$ satisfy $$a_1^2+a_2^2=a_3^2+a_4^2.$$ Let $p(x)=a_1^2+a_2^2$.

For integer $n$, $p(n)$ can be written as sum of two squares in two ways, except possibly for a finite number of exceptions.

This means $p(n)$ is composite.

Bunyakovsky's conjecture implies that either $p(x)$ is reducible or the content is not one (i.e. all coefficients have a common prime divisor) or $p(x)$ has a fixed divisor for congruence reasons when the content is one (like $x^2+x+2$, which is always even as pointed out in the comments).

It is possible for $p(x)$ to be irreducible, e.g.

$a_1=16 x^{2},a_2=-37 x^{2} + 80 x + 20,a_3=-35 x^{2} + 48 x + 12, a_4=20 x^{2} - 64 x - 16$ and $p(x)=\left(5\right) \cdot (325 x^{4} - 1184 x^{3} + 984 x^{2} + 640 x + 80)$ , but the content is not one.

Must the content of $p(x)$ be greater than one when it is irreducible (no matter if there is a fixed divisor)?

I believe similar identities exist for arbitrary large degree.

• Have you tried taking Euler's argument proving that a positive integer that can be written as a sum of two squares in two different ways is composite, and running it for polynomials? – Jeremy Rouse Nov 16 '14 at 17:46
• I think you are misapplying Bouniakowsky's conjecture. It says a nonconstant polynomial in ${\mathbf Z}[x]$ has prime values infinitely often unless the polynomial is reducible in $\mathbf Z[x]$ or all of its values are divisible by a common prime number, and that second condition is weaker than saying the polynomial has content 1. For example, $x^2+x+2$ is irreducible with content 1 but all of its values on $\mathbf Z$ are even. – KConrad Nov 16 '14 at 20:42
• @KConrad, indeed, thanks. I missed this case, probably will edit. – joro Nov 17 '14 at 6:52
• @JeremyRouse I couldn't solve it this way. Note that the original question was wrong, Bunyakovsky conjecture doesn't imply what I claimed, there is third possibility. – joro Nov 17 '14 at 12:15
• What is the content of a polynomial (by definition)? – GH from MO Nov 17 '14 at 12:37

$(a1-\textit{i}*a2)*(-4+3\textit{i})/5 = a3-\textit{i}*a4$.
$\mathbb{Z}[\textit{i}]$ is UFD and so is $\mathbb{Z}[\textit{i}][x]$.
• Maybe I should add as clarification that the property of $n$ being composite if it sum of 2 squares in 2 different ways, is a consequence of $\mathbb{Z}[\textit{i}]$ being UFD. So the same applies for $p(x)$ in $\mathbb{Z}[x]$ since $\mathbb{Z}[\textit{i}][x]$ is also UFD. – Yaakov Baruch Nov 17 '14 at 12:44