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  • If $(a,b) = 1$ then $(a+b,ab) = 1$.
  • If $(a,b) = 1$ then $(a+b,a-b) = 1$ or $2$.
  • If $(a,b) = 1$ then $(a+b,a^2-ab+b^2) = 1$ or $3$.

Is there an algorithm to compute the gcd of two polynomials applied to coprime numbers like this?

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up vote 2 down vote accepted

For homogeneous polynomials, as in your examples, you can basically apply the Euclidean algorithm.

In the general case, you can still eliminate one variable from one of the polynomials, but this is rarely conclusive.

For a special case, where it is still possible to find the values of $a$ and $b$ where one of the expressions divides the other, you could look at the example of Vieta jumping.

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"Vieta jumping" is a strange name for what is essentially descent in a group of integer points on a conic (realized by reflection). This is a special case of results on Pell equations (or equivalent theories). – Bill Dubuque Apr 23 '11 at 15:22
I wanted to indicate why there is no hope for a general result. "Vieta jumping" is easily googlable and the resulting pages are easy to understand with little background. If you have a link to a similarly simple explanation in terms of integer points on a conic, I will be happy to know it and refer to it in the future. – Phira Apr 23 '11 at 18:00
Perhaps a more appropriate term would be "integral conic reflection". Alas, there is a shortage of good expositions on group laws on conics (a degenerate case of the group law on elliptic curves). There lies much beauty which deserves good exposition. – Bill Dubuque Apr 23 '11 at 18:05
@Bill Dubuque: Is it evident from the conic point of view when this procedure will be a descent? – Phira Apr 25 '11 at 20:34

In the case where one of the polynomials is $a+b$, as in all three of your instances above, you can do a simple substitution if $a\equiv -b \pmod d$ in the second polynomial. For example, in the case $gcd(a+b,a^2-ab+b^2)$, if $d$ is a common divisor, then we must have $0\equiv 3b^2 \pmod d$. Since $d$ must be relatively prime to $b$, this means that $3 \equiv 0 \pmod d$.

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