Find all the possible values of gcd$(n^2 +2 , n^4 +4)$ if $n$ is a natural number? Is this a complete “proof”? Im new to number theory so it might be wrong. So, $$n^4 +4 = (n^2 +2+2n)(n^2 +2 -2n)\ ,$$ so if $d$ divides $n^2 +2 +2n$ and $n^2 +2$, then $d | 2n$. , so $d|2$ or $d|n$. 
Suppose $d| n$. We know $d | (n^2 +2)$ so $d| 2$ too, therefore $d=$gcd$(n , 2)$ , so $d=1$ (if $n$ odd) or $2$(if $n$ even) .  (If we work with the $n^2 +2 -2n$ we get the same result).
If you have different solutions to this, please write them down! Thanks

We have $n^4+4 = (n^2+2)(n^2 - 2) + 8$. If $n$ is odd, then $gcd(n^2+2, 8) = 1$, and if $n$ is even then $n^2+2 = 4k^2 + 2$ is not divisible by $4$ or $8$ so $gcd(n^2+2,8) = 2$. This gives your original calculation (via a Euclidean Algorithm argument)
