# Do monic irreducible polynomials assume at least one prime value?

We are given a monic irreducible polynomial with integer coefficients, $$f(x) = x^n + a_{n-1} x^{n-1} + \cdots + a_1 x + a_0.$$ Must there be an integer $n \geq 0$ such that $f(n)$ is prime?

In case it helps, it is fine to demand all $a_j \geq 0.$ It is not known, for example, whether $x^2 + 1$ assumes infinitely many prime values, but it does assume at least one, such as $2,5,17.$

In case it matters, the polynomial that caused this is $x^4 + m$ with $m > 0,$ from the question I answered just before asking this. Find all positive integers $m$, such that $n^4+m$ is not prime for any positive integer $n$

• I kept going "Wait, there's an obvious counterex-- oh blast that's reducible." – Dan Uznanski May 29 '17 at 5:20

## 1 Answer

$x^2 + x + 8 = (x+\frac{1}{2})^2 + \frac{31}{4}$ has no roots so is irreducible; and $x^2 + x + 8 = x(x+1) +8$ is always even and bigger than 2 hence cannot be prime ever!

• Well, that was disappointing. Haha, good job. – Fimpellizieri May 30 '17 at 1:12
• Sorry to disappoint! – SEWillB May 30 '17 at 1:12