$x^6 + 69x^5 − 511x + 363$ is irreducible over $\mathbb Z$?

Question

As mentioned, I am trying to show that $x^6 + 69x^5 − 511x + 363$ is irreducible over $\mathbb Z$. To see that it has no roots and no cubic factors, I send the polynomial to $\mathbb F_7$ and $\mathbb F_2$ respectively. The thing I have not figured out is how to show it has no quadratic factors. Is there any way to do this other than brute force?

Answer 1

To see that it has no roots and no cubic factors, I send the polynomial to $\mathbb F_7$ and $\mathbb F_2$

Adding $\mathbb F_3$ and $\mathbb F_{11}$ to the mix, the constant term $c$ of a quadratic factor of $p(x)$ must satisfy:

• $p(x) = (x + 1)^2 (x^4 + x^3 + x^2 + x + 1)$ over $\mathbb F_2$ implies $c \equiv 1 \pmod{2}$

• $p(x) = x (x - 1) (x^4 + x^3 + x^2 + x + 1)$ over $\mathbb F_3$ implies $c \equiv 0 \pmod{3}$

• $p(x) = (x^2 - x + 4) (x^4 + 3 x^2 + 3 x + 5)$ over $\mathbb F_7$ implies $c \equiv 4 \pmod{7}$

• $p(x) = x (x + 2) (x^4 + x^3 + 9 x^2 + 4 x + 3)$ over $\mathbb F_{11}$ implies $c \equiv 0 \pmod{11}$

The general solution to the above is $c \equiv 165 \pmod{462}$, but there is no such $c$ that divides $363$, the constant term of $p(x)$, so no quadratic factor exists.

Answer 2

You can first search for some $\pm$ prime values and you find one at $P(-4)$ ($= -64153$). Assume $P(x)$ were reducible, that is $P(x) = P_1(x) \cdot P_2(x)$. Then one of the $P_i(-4)$ is $\pm 1$.

Now, consider $P(x)$ modulo $7$, factors as a product

$$P(x) = (x^2 + 6 x + 4)\cdot (x^4 + 3 x^2 + 3 x + 5) \mod 7$$

so in some order $\{P_1(x), P_2(x) \} \equiv \{(x^2 + 6 x + 4), (x^4 + 3 x^2 + 3 x + 5)\} \mod 7$ and now plugging in $x=-4$ we get

$$\{P_1(-4), P_2(-4)\} \equiv \{(x^2 + 6 x + 4), (x^4 + 3 x^2 + 3 x + 5)\}_{x=-4} \mod 7$$

However the RHS is $\{3,3\}$, so both $\not \equiv \pm 1 \mod 7$ , contradiction.