Verify my proof of divisibility of a rational fraction

I would suggest the following approach :

First consider $(2n-1)(2n+1)=4n^2-1$ hence we have $$4\cdot \frac{n^2+2}{2n-1}=\frac{4n^2+8}{2n-1}=\frac{4n^2-1}{2n-1}+\frac{9}{2n-1}=2n+1+\frac{9}{2n-1}$$

Since $2n-1$ is odd, it must divide $n^2+2$ , if it divides $4(n^2+2)$ (try to find out why), hence you only have to check for which $n$ we have $2n-1|9$. Try to solve this.

By long division,

$$\frac{n^2+2}{2n-1}=\frac n2+\frac14+\frac9{4(2n-1)}$$ or $$4\frac{n^2+2}{2n-1}=2n+1+\frac9{2n-1}$$ so that $2n-1$ must divide $9$ and $n$ could be $1,2$ or $5$, which turn out to all be valid solutions.

If negatives are allowed, consider $0,-1,-4$ which are also possible.