number theory - Find all intergers such that $2n^2+1$ divides $n^3+9n-17$

Find all intergers such that $2n^2+1$ divides $n^3+9n-17$.

Answer : $n=(2 \ and \ 5)$

I did it.

As $2n^2+1$ divides $n^3+9n-17$, then $2n^2+1 \leq n^3+9n-17 \implies n \geq 2$

So $n =2$ is solution and doens't exist solution when n<2. How can I do now to find 5 ? Or better, how can you solve this with another good method ?

Thanks

Answer

HINT:

If integer $d$ divides $n^3+9n-17,2n^2+1$

$d$ must divide $2(n^3+9n-17)-n(2n^2+1)=17n-34$

$d$ must divide $17(2n^2+1)-2n(17n-34)=68n+17$

$d$ must divide $68n+17-4(17n-34)=153$

So the necessary condition is $2n^2+1$ must divide $153$

$\implies2n^2+1\le153\iff n^2\le76\iff-9