Suppose for polynomials f, g in $\mathbb{Q}[X]$ it holds that


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 $\square$.

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    $\begingroup$ Why do you think a valid inference is "cheating"? $\endgroup$ – Bill Dubuque Oct 11 '18 at 15:35
  • $\begingroup$ Because I didn't really know why I was allowed to do it ;) $\endgroup$ – 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 $$ \gcd(f(x+a),g(x+a)) = d(x+a) $$

  • $\begingroup$ 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$ ? $\endgroup$ – Wesley Strik Oct 11 '18 at 12:59
  • $\begingroup$ Since the relations holds for all X, it must also hold for $X+a$? $\endgroup$ – Wesley Strik Oct 11 '18 at 13:33
  • $\begingroup$ @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 $x$. $\endgroup$ – Yong Hao Ng Oct 11 '18 at 14:03

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