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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.

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    $\begingroup$ 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? $\endgroup$
    – Jeremy Rouse
    Commented Nov 16, 2014 at 17:46
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    $\begingroup$ 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. $\endgroup$
    – KConrad
    Commented Nov 16, 2014 at 20:42
  • $\begingroup$ @KConrad, indeed, thanks. I missed this case, probably will edit. $\endgroup$
    – joro
    Commented Nov 17, 2014 at 6:52
  • $\begingroup$ @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. $\endgroup$
    – joro
    Commented Nov 17, 2014 at 12:15
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    $\begingroup$ @GHfromMO The gcd of all coefficients: proofwiki.org/wiki/Definition:Content_of_Polynomial $\endgroup$
    – joro
    Commented Nov 17, 2014 at 12:48

1 Answer 1

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Notice that in your example

$(a1-\textit{i}*a2)*(-4+3\textit{i})/5 = a3-\textit{i}*a4$.

$\mathbb{Z}[\textit{i}]$ is UFD and so is $\mathbb{Z}[\textit{i}][x]$.

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  • $\begingroup$ Thank you, but what about the general question? $\endgroup$
    – joro
    Commented Nov 17, 2014 at 12:30
  • $\begingroup$ This implies that the answer is yes. For the polynomial part (divided by content) to be irreducible, it means the content part be non trivial. $\endgroup$
    – Yaakov Baruch
    Commented Nov 17, 2014 at 12:37
  • $\begingroup$ 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. $\endgroup$
    – Yaakov Baruch
    Commented Nov 17, 2014 at 12:44

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