Find all $x\in \mathbb Z$ for which $y=x^3-8x^2+20x-13$ is a prime number

Question

Question : Find all $x\in \mathbb Z$ for which $y=x^3-8x^2+20x-13$ is a prime number

---

I don't know how to start here. By Brute-Forcing, I could start plugging $x=1,2,3,...$ but that's very inefficient. If $y$ was a quadratic expression, I would have easily applied the Sridharacharya Formula, but that's not the case here.

Is there any other method or trick for this question, other than hit and trial?

Answer 1

For $x=1$, we get $y=0$.

So, we see that $y$ is divisible by $x-1$ to have $$y=(x-1)(x^2-7x+13)$$

In order for this to be a prime, we have to have $$x-1=\pm 1\qquad\text{or}\qquad x^2-7x+13=\pm 1$$ which implies $$x=0,2,3,4$$ giving $$y=-13,3,2,3$$ respectively.

So, the answer is $x=2,3,4$.

Answer 2

Hint:

Note that $$f(x) = (x-1)(x^2-7x+13)$$ If $x$ is an integer, then both factors of $f(x) \in \mathbb{Z}$. Then, if $f(x)$ is prime, the two factors should be $\pm 1$ and $\pm p$, where $p$ is a prime.

Equate both factors to $\pm 1$ and take the intersection of the set of solutions.