# 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 $$\square$$.

• Why do you think a valid inference is "cheating"? – Bill Dubuque Oct 11 '18 at 15:35
• Because I didn't really know why I was allowed to do it ;) – Wesley Strik Oct 11 '18 at 16:29

Hint  shifts $$\,f\mapsto \bar f$$ are automorphisms so preserve divisibility $$\ cd=a\iff \bar c \bar d = \bar a,\,$$ so gcds

$$c\mid (a,b)\iff c\mid a,b\iff \bar c\mid \bar a,\bar b\iff \bar c\mid (\bar a ,\bar b)$$

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 $$\gcd(f(x+a),g(x+a)) = d(x+a)$$

• Why does $p(x)(f(x)/d(x)) + q(x)(g(x)/d(x)) = 1$ imply that $p(x+a)(f(x+a)/d(x+a)) + q(x+a)(g(x+a)/d(x+a)) = 1$ ? – Wesley Strik Oct 11 '18 at 12:59
• Since the relations holds for all X, it must also hold for $X+a$? – Wesley Strik Oct 11 '18 at 13:33
• @WesleyGroupshaveFeelingsToo Right, it really should have been written as $$f(x)/d(x) = u(x) \in \mathbb Q[x]$$ and similarly $v(x)$ for $g(x)/d(x)$. Then the change of variables hold and we can work with $u(x+a), v(x+a)$ first until $\gcd(u(x+a),v(x+a))=1$. Then $$\gcd(u(x+a)d(x+a),v(x+a)d(x+a)) = d(x+a)$$ and we can plug back $f,g$. Then translate $x+a$ back to $x$. – Yong Hao Ng Oct 11 '18 at 14:03