For polynomials $f, g$ and $\gcd(f(X),g(X))=d(X)$ then $\gcd(f(X+a),g(X+a))=d(X+a)$ Suppose for polynomials f, g in $\mathbb{Q}[X]$ it holds that
$$\gcd(f(X),g(X))=d(X)$$ 
What we also want to now prove is that for $a \in \mathbb{Q}$: $$\gcd(f(X+a),g(X+a))=d(X+a)$$ 
So the question is asking that if we replace $X$ by $X+a$, the new gcd is the old one evaluated at $X+a$. I was thinking that we can always write:
 $$ \alpha(X) f(X)+ \beta(X)  G(X) =d(X)$$ 
For polynomials $\alpha$, $\beta$, but I don't see how I could use this. If I would just treat $d(X)$ as a function now and shift by $a$ to the left, the proof would immediately follow, but this feels a bit like cheating, is this statement valid?
I would then finish like:
If we now treat $d(X)$ as a function, we shift the function $a$ to the left and get: 
$$d(X+a)=\alpha(X+a) f(X+a)+ \beta(X+a) g(X+a) .$$
Since we still have $\alpha(X+a), \beta(X+a) \in  \mathbb{Q}[X]$ this statement is equivalent to: 
$$\gcd(f(X+a),g(X+a))=d(X+a) .$$
as desired $\square$.
 A: I think one problem might be 
$$p(x+a)f(x+a) + q(x+a)g(x+a) = d(x+a)$$
does not imply 
$$d(x+a)=\gcd(f(x+a),g(x+a)),$$
only that their $\gcd$ divides $d(x+a)$.  
For example you can multiply arbitrarily large degree $k(x)$ to $p(x+a), q(x+a)$ and $d(x+a)$.  
Perhaps arguing along with degree of $d(x), d(x+a)$ can work. I guess this also means $a$ must be defined as not a root.    
But if $a$ is not a root perhaps another way is: we can find $p(x),q(x)$ such that
$$
p(x)(f(x)/d(x)) + q(x)(g(x)/d(x)) = 1
$$
Therefore
$$
p(x+a)(f(x+a)/d(x+a)) + q(x+a)(g(x+a)/d(x+a)) = 1
$$
This time round we can say something concrete about their $\gcd$:
$$
D(x) = \gcd(f(x+a)/d(x+a),g(x+a)/d(x+a)) = 1,
$$
as otherwise there is some non-constant polynomial $D(x)$ that divides $1$.  
Therefore
$$
\gcd(f(x+a),g(x+a)) = d(x+a)
$$
A: Hint $ $ shifts $\,f\mapsto \bar f$ are automorphisms so preserve divisibility $\ cd=a\iff \bar c \bar d = \bar a,\,$ so gcds
$$ c\mid (a,b)\iff c\mid a,b\iff \bar c\mid \bar a,\bar b\iff \bar c\mid (\bar a ,\bar b)$$
