Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ Given that n is a positive integer show that $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$. 
I'm thinking that I should be using the property of gcd that says if a and b are integers then gcd(a,b) = gcd(a+cb,b). So I can do things like decide that $\gcd(n^3 + 1, n^2 + 2) = \gcd((n^3+1) - n(n^2+2),n^2+2) = \gcd(1-2n,n^2+2)$ and then using Bezout's theorem I can get $\gcd(1-2n,n^2+2)= r(1-2n) + s(n^2 +2)$ and I can expand this to $r(1-2n) + s(n^2 +2) = r - 2rn + sn^2 + 2s$ However after some time of chasing this path using various substitutions and factorings I've gotten nowhere. 
Can anybody provide a hint as to how I should be looking at this problem?

As you note, $\gcd(n^3+1,n^2+2) = \gcd(1-2n,n^2+2)$.
Now, continuing in that manner,
$$\begin{align*}
\gcd(1-2n, n^2+2) &= \gcd(2n-1,n^2+2)\\
 &= \gcd(2n-1, n^2+2+2n-1)\\
 &= \gcd(2n-1,n^2+2n+1)\\
 &= \gcd(2n-1,(n+1)^2).
\end{align*}$$
Consider now $\gcd(2n-1,n+1)$. We have:
$$\begin{align*}
\gcd(2n-1,n+1) &= \gcd(n-2,n+1) \\
&= \gcd(n-2,n+1-(n-2))\\
&=\gcd(n-2,3)\\
 &= 1\text{ or }3.
\end{align*}$$ 
Therefore, the gcd of $2n-1$ and $(n+1)^2$ is either $1$, $3$, or $9$. Hence the same is true of the original gcd.
