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Prove that gcd(m+1,n+1)|mn-1 ,where m and n are integers.

I have already done:

(m+1)(n+1)=mn+m+n+1

Clearly gcd(m+1, n+1)|(m+1)(n+1)

And obviously gcd(m+1),(n+1)|(m+1)(n+1)

Therefore gcd(m+1),(n+1)|mn-1. QED

Can someone give me an alternate proof?

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1  
What's the difference between your "clearly" and "obviously" statements? –  user2345215 Feb 27 at 20:33
3  
Your proof does not really make sense to me. –  drhab Feb 27 at 20:39

2 Answers 2

up vote 0 down vote accepted

${\rm mod}\ (m\!+\!1,n\!+\!1)\!:\ m\!+\!1\equiv0\equiv n\!+\!1,\,$ so $\,\color{#0a0}{m,n}\equiv \color{#c00}{-1}\,\Rightarrow\, \color{#0a0}{mn}-1\equiv (\color{#c00}{-1})(\color{#c00}{-1})-1\equiv 0.$

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1  
Wow, such color. Much insight. –  user2345215 Feb 27 at 20:50

$mn-1=(m+1)n-(n+1)$ so if $d|m+1$ and $d|n+1$ then also $d|mn-1$.

You can take $d=\gcd(m+1,n+1)$

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