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Suppose that, for given integers $a$ and $b$, there exist integers $x$ and $y$ satisfying $ax + by = 6$. What are all the possible values of $\gcd(a, b)$?

Progress, we know that since $ax + by = 6$, then $x = (6 - by)/a$. In order for $x$ to be an integer, $6 - by$ must be some multiple of a. Similarly, $y = (6 - ax)/b$, so $6 - ay$ must be some multiple of $b$. Then for some integer $q$, $aq = 6 - by$. For some integer $r$, $br = 6 - ay$. Then $b = (6 - ay)/r$ and back substituting gives us $y = (6 - ax)/[(6 - ay)/r] = r(6 - ax)/(6 - ay)$. Thus $6y - ay^2 = r(6 - ax)$ and so we have the quadratic... I'm in a mess, any help and/or advice would greatly be appreciated.

Alex Meiburg
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WhatsDUI
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Let $d=\gcd(a,b)$ and write $a=a'd$ and $b=b'd$, such that $\gcd(a',b')=1$. Now: $$ax+by = d(a'x+b'y)=6$$ and therefore $d\mid 6$, though this doesn't mean all divisors of $6$ can be the $\gcd$

However, there is also a lemma (Bezout's identity) says that for all $a,b\in\mathbb{Z}$, there exist $x,y\in\mathbb{Z}$ such that: $$ax+by=\gcd(a,b)$$ The values of $x$ and $y$ can be found using the so-called Extended Euclidean Algorithm.

In conclusion: The gcd can be any (positive) divisor of $6$, so $6$, $3$, $2$ or $1$.

Mastrem
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