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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