irreducibility of polynomials with integer coefficients

Question

Consider the polynomial $$p(x)=x^9+18x^8+132x^7+501x^6+1011x^5+933x^4+269x^3+906x^2+2529x+1733$$ Is there a way to prove irreducubility of $p(x)$ in $\mathbb{Q}[x]$ different from asking to PARI/GP?

Answer 1

A starting point: Modulo $3$ the polynomial $p$ takes the form $$ red_3(p) = x^9-x^3-1\in \mathbb{F}_3[x]. $$ Since raising to $3$ is the frobenius automorphism we have $$ red_3(p) = (x^3-x-1)^3. $$ The polynomial $x^3-x-1$ is irreducible modulo three.

From all of this we get that if $p$ factors to a product $$ p=p_1\cdots p_r, $$ with $p_1, \ldots, p_r\in \mathbb{Z}[x]$ monic and irreducible over $\mathbb{Q}$, then $r\leq 3$ and the degree of each factor is at least $3$.

Answer 2

This polynomial has the element $\alpha^2+\beta$ described in this question as a root. My answer to that question implies among other things that the minimal polynomial of that element is of degree 9, so this polynomial has to be irreducible.

Answer 3

If you don't mind evaluating $p(x)$ several times and factoring large numbers, you could find $10$ values for $x$ which are separated by more than $2$, for which $p(x)$ is prime (or a unit). This would imply that $p(x)$ (having degree $9$) is irreducible.

The logic being: If $p(\alpha)$ is prime and if $p(x) = q(x) r(x)$, then $q(\alpha)$ or $r(\alpha)$ must be a unit ($\pm1$). Given that $q(x), r(x)$ must also have degrees that sum to that of $p(x)$, the number of times they can take the value of a unit is restricted - maximum is two times the degree of $p(x)$, to account for two possible signs for the units. However, if the $x$ values are separated by more than $2$, then for any of $q(x)$ and $r(x)$ the sign of the unit must be the same, and then there cannot be then more prime values than the degree of $p(x)$.

Here, we have $p(x)$ prime for $x \in \mathrm{A} = \{-16, -10, -3, 0, 3, 6, 14, 17, 28, 39 \}$. (I have omitted values such as $\{-12, -4, -2, -1, 5\}$ which also give primes but are too close to elements of $\mathrm{A}$, to avoid having to find $19$ prime values). As these are $10$ values separated by more than $2$, $p(x)$ is irreducible.