Does the equation $ax+by=\gcd(a,b)=\gcd(ax+by,b)=\gcd(a,ax+by)$ hold for $a, b, x, y \in \mathbb{Z}$?

Question

Does the equation $ax + by = \gcd(a, b) = \gcd(ax+by,b) = \gcd(a,ax+by)$ hold for $a, b, x, y \in \mathbb{Z}$? (Note that $x$ and $y$ are not uniquely determined by the equation $\gcd(a,b)=ax+by$.)

From this MSE question, I have the result $\gcd(a,b)=ax'+by'=c$. So I guess my main inquiry would then be: Can I push this finding further into $$\gcd(a,c)=\gcd(b,c) = \gcd(a,b) = c?$$

If not, under what conditions will such a system of equations hold?

By Euclid $,(ax+by,b) = (ax,b).,$ Why do you think $,(ax,b) = (a,b)?\ $ — Bill Dubuque, Sep 19 '17 at 14:57
@BillDubuque, I am looking for conditions under which the following equation holds: $$\gcd(\gcd(a,b),a)=\gcd(\gcd(a,b),b)=\gcd(a,b).$$ — Jose Arnaldo Bebita Dris, Sep 19 '17 at 15:05
@lulu, your counterexample is an actually an instance where my equation holds. — Jose Arnaldo Bebita Dris, Sep 19 '17 at 15:05
Recall that in the equation $\gcd(a,b) = ax' + by'$, $x', y' \in \mathbb{Z}$ are not unique. — Jose Arnaldo Bebita Dris, Sep 19 '17 at 15:06
@Jose Ah, your question was not clear. Yes, you could use Bezout to prove that, but it is easier to use the GCD Distributive Law, viz. $, d=(a,b)\mid a,$ so $,(d,a) = d(1,a/d) = d.,$ See this answer for a handful of proofs of the distributive law (including one by Bezout). — Bill Dubuque, Sep 19 '17 at 15:14
Thanks for illuminating me, @BillDubuque. I was thinking of a proof along exactly the same lines as you did, but could not get to articulate it properly. Your advice to use the GCD Distributive Law is very apt for this case. — Jose Arnaldo Bebita Dris, Sep 19 '17 at 15:21
@Jose Alternatively: if $d\mid a,$ then $,c\mid d,a\iff c\mid d,,$ so $,d,a,$ and $,d,$ have the same set $,C,$ of common divisors $,c,$ so they have the same greatest common divisor $(= \max C)$ $\ \ $ — Bill Dubuque, Sep 19 '17 at 15:23
@Jose Or, by Euclid $, (d,a) = (d,,a\bmod d) = (d,0) = d.\ $ There are many ways to prove it. — Bill Dubuque, Sep 19 '17 at 15:27
Or $\ ((a,b),b) = (a,(b,b)) = (a,b)\ $ by the GCD associative law. — Bill Dubuque, Sep 20 '17 at 13:06

score 1 · Accepted Answer · answered Sep 19 '17 at 15:29

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As hinted by Bill Dubuque in the comments, the equation $$\gcd(a,b)=\gcd(a,\gcd(a,b))=\gcd(b,\gcd(a,b))$$ can be proved using the GCD Distributive Law.

answered Sep 19 '17 at 15:29

Jose Arnaldo Bebita Dris

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Does the equation $ax+by=\gcd(a,b)=\gcd(ax+by,b)=\gcd(a,ax+by)$ hold for $a, b, x, y \in \mathbb{Z}$?

1 Answers1