Find all numbers $n \in \Bbb Z$ such that $2n^2+3n-2$ and $n^2+1$ are coprime.

Question

Find all numbers $n \in \Bbb Z$ such that $2n^2+3n-2$ and $n^2+1$ are coprime.

I used Euclidean algorithm and got $(2n^2+3n-2,n^2+1)=(3n-4,n^2+1)=(3n-4,4n+3)=(3n-4,12n+9)=(3n-4,25)$

then I checked it when

$n=1 \implies (3,2)=1$

$n=2 \implies (12,5)=1$

$n=3 \implies (25,10) \neq 1$

is it somehow useful? What should I do next?

Answer 1

Assuming you've done your algebra correctly, you have shown that if they have a nontrivial gcd then they're both divisible by five. Now $n^2+1$ is divisible by five precisely when $n$ is two more than, or two less than, a multiple of five. The other polynomial is divisible by five when $n$ is two less than a multiple of five, but not when $n$ is two more than a multiple of five. So the answer is, exactly when $n$ is two less than a multiple of five. (That is, those are the values of $n$ for which the two numbers are not coprime).