Given that $a$ is an integer, compute the possibilities for $\gcd (2a^4 + 2a^2+3,2a^5+2a^3+a^2+a+3).$
Here's my work (I'm not entirely sure if it's correct).
Just to make sure we're on the same page, we say that for integers $a,b,$ $d=\gcd(a,b)$ is the integer such that $d\mid a$ and $d\mid b$ and for all $c\in\mathbb{Z}$ such that $c\mid a$ and $c\mid b,c\mid d$ (so there can be multiple possible $gcd$'s).
We use the fact that $\gcd (a,b)=\gcd(b,a-bq),q\in\mathbb{Z}.$
Dividing using the Euclidean Algorithm, we have the following:
$2a^5+2a^3+a^2+a+3=(2a^4+2a^2+3)(a)+a^2-2a+3\\ 2a^4+2a^2+3=(a^2-2a+3)(2a^2+4a+4)-4a-9\\ 4(a^2-2a+3)=4a^2-8a+12=(-4a-9)(-a+4)-a+48\\ -4a-9=4(-a+48)-201.$
Hence $\gcd (2a^5+2a^3+a^2+a+3,2a^4+2a^2+3)=\gcd (-a+48,-201)$ for all $a\in\mathbb{Z}.$ Let $f(a)=48-a, f:\mathbb{Z}\to \mathbb{Z}.$ Then since $f$ is a surjective, monic linear polynomial, the possibilities for $\gcd (-a+48,-201)$ are precisely the integer divisors of $201$ (since $\gcd$'s can be negative according to the definition).
Edit: I've edited my answer to include the proper possibilities.
Trusting on your calculations, it is enough to find the possible values of $\gcd(48-a, -201)$. Now, assume $f(a):=48-a, f:\mathbb{Z}\to\mathbb{Z}$. Since, $f$ is monic linear polynomial it is surjective. Therefore, possibilities of $\gcd(48-a, -201)$ will be all positive divisors of $201$.