TITLE: Determine all $a\in\mathbb{Z}$ such that $a^3-8a^2-14a+144$ is prime
QUESTION [9 upvotes]: Here is a problem from my teacher.
I'm not sure there is no typo.

Determine all $a\in\mathbb{Z}$ such that $a^3-8a^2-14a+144$ is prime.

I found that $a$ must in form $6k+5$,but had nothing to do next.
I tried to factor it while failed.

REPLY [3 votes]: I strongly suspect a typo.
Let $f(x)=x^3-8x^2-14x+144$.

$f(x)$ has leading coefficient $1$, which is positive.
$f(x)$ has no integer roots, hence irreducible in $\mathbb{Z}[x]$.
$f(5)=-1$ so $\gcd(f(1), f(2), \ldots)=1$.

Now $f(x)$ satisfies the conditions of Bunyakovsky's conjecture. To my knowledge, it remains unproved for polynomials with degree $>1$. (The degree $1$ case corresponds to Dirichlet's theorem on primes in arithmetic progressions)