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This question already has an answer here:

Show that $\text{gcd}(m+n,m) = \text{gcd}(m,n)$

Here is what I have so far, using the definition of $\text{GCD}$ for some integer d we have

$d = x(m+n) + y(m)=(x)(n) + (x+y) m$ Now let $x' = x+y$ and $y'=y$,

$d'= x'(m) + y'(n)=(x+y)m+y(n)$ where $\text{GCD}(m,n)=d'=d$

Is this reasoning sufficient?

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marked as duplicate by JMoravitz, Mikhail Katz, user228113, Shailesh, user296602 Jan 18 '16 at 3:19

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  • $\begingroup$ Here is meta.math.stackexchange.com/questions/5020/… for displaying math functions. $\endgroup$ – Arbuja Jan 18 '16 at 2:06
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    $\begingroup$ I don't see any reasoning, just equations. $\endgroup$ – pjs36 Jan 18 '16 at 2:16
  • $\begingroup$ Agreed. Explain what you're doing more carefully. $\endgroup$ – D_S Jan 18 '16 at 3:16

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