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Assume $\text{gcd}(m,n) = {mx + ny} = d$

Prove that $\text{gcd}(x,y) = 1 = d'$

Now for the solution, we know that $\text d'|x$ , $\text d'|y$

so $\text d'|mx + ny$

so $\text d'|d$

After this? Or is this the wrong direction?

Mike B
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  • hypotheses: $\text{lcm}(m,n) = {mx + ny} = d$ and $\text{lcm}(x,y) = 1 = d'$ \ you have to prove that $\text d'|x$ , $\text d'|y$ so $\text d'|d$, is it? – Henfe May 06 '13 at 21:30
  • http://math.stackexchange.com/questions/230997/prove-that-if-a-b-x-y-are-integers-with-ax-by-gcda-b-then-gcdx-y-1 – manthanomen May 06 '13 at 22:28

1 Answers1

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Let $r$ be a (positive) common divisor of $x$ and $y$, with $x=rx'$ and $y=ry'$: then

$$ d=mx+ny=mrx'+nry'=r(mx'+ry') $$

Then $r$ divides $d$ and so

$$\frac{d}{r}=mx'+ny'$$

which means that $d=\gcd(m,n)$ divides $d/r$; hence $r=1$.

We use the fact that $d$ divides any "linear combination" of the form $mh+nk$ for integer $h$ and $k$.

egreg
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