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I have seen this question and its answers. There are some nice answers but I was wondering if my original approach could be solved.

My approach: Let's assume $m$ is $\operatorname{lcm}(a,b)$ and $d$ is $\gcd(a,b)$ and $m = aq$, $m = br$
We know, $d = ax + by$
So, $\gcd(a,b) \operatorname{lcm}(a,b) = md $
$= m(ax + by) $
$ = (br) ax + (aq) by = ab(rx + qy) $

Is there any way to prove $rx + qy = 1$? I know $x,y$ are relatively prime, so are $q,r$ but couldn't prove this using this information.

Bernard
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alu
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  • Just note that the fact that $(x,y)=1$ and $(q,r)=1$ is not enough by itself to show $rx + qy = 1$ (easy counterexamples $x=y=q=r=1$ or $(x,y)=(1,2)$, $(q,r)=(1,-2)$ to include negative numbers). – Sil Jun 05 '20 at 14:24

1 Answers1

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Short answer: by your definition of $m$, we immediately conclude that $\gcd(q,r)=1$, and then you stated that there exists integers $x,y$ such that $d=ax+by$, so when you divide by $\gcd(a,b)$ you get $rx+qy=1$ and we know there exists such $x,y$ because $\gcd(q,r)=1$

Long answer/explanation: let's just make our convensions clear and try to prove it your approach:

Let $a=da_1$ and $b=db_1$ where $\gcd(a_1,b_1)=1$ and $\gcd(a,b)=d$ for some integers $d,a_1,b_1$.

Now ${\rm lcm}(a,b)=da_1b_1$ by definition. So, we can say if $m={\rm lcm}(a,b)=aq=br$, we have $q=b_1$ and $r=a_1$, so we can exchange $q,r$ this way. We know that there exists integers $x,y$ such that $$d=ax+by$$ But we can choose initial solutions $x_1$ and $y_1$ such that $$d=ax_1+by_1=da_1x_1+db_1y_1 \iff 1=a_1x_1 +b_1y_1$$ which deduces what you desired.

Anas A. Ibrahim
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