Prove or disprive that $n^{2}-n+17$ is prime for all integers $n$

Question

I am looking to prove this function is always prime for all integers $n$: $$n^{2}-n+17$$ I have tested it for the first $10$ integers and it seems to work but I am not sure how to prove it form all $n$. Any ideas?

Answer 1

$n^2-n+17=n(n-1)+17$. Taking $n=17$, you will get $17\times16+17$, a proper multiple of $17$. Taking $n-1=17$, you will get $18\times17+17$, another one.

More generally, a polynomial expression $P(n)$ with integer coefficients will fail for $n=P(0)$ and multiples.

Answer 2

Let $f(x)=a_0+a_1x+...+a_nx^n \in \mathbb Z[X]$ be a polynomial. Now note that

$$f(a_0)=a_0+a_1a_0+...+a_na_0^n=a_0(1+a_1+...+a_na_0^{n-1})$$

In other words, $a_0$ divides $f(a_0)$. Therefore, if $1<a_0<f(a_0)$ like in your case, $f(a_0)$ is not prime.

Answer 3

The formula $\quad n^2-n+C\quad $ is composite for many values where $n>C$ and always composite for $n=C^p, p\in\mathbb{N}.$