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Problem

If $(a,b)=1$ then $(a, b+1)=1$.

Progress

So far I have,

  • Let $d = (a,b)$
  • Which implies $d\mid a$ and $d\mid b$
  • Which implies there exist $x$ and $y$ such that $d\mid (a)(x) + (b)(y)$
  • So I want to find an $x$ and $y$ that can make the equation, $ax+by$ into $(a+1)(x)+(ab)(y)$
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    @Anna: As has been pointed out, there are $b$ for which the assertion is false, so one cannot hope to prove it in general. Perhaps you did not quote the question exactly as it was posed.2012-12-09

2 Answers 2

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if $b=a+1$ then the second gcd is $a+1$

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    @GerryMyerson, yes, she changed it not long after I wrote this answer. I'm pretty sure she had time to see this. And, she has gone on to ask a bunch of questions at this level. I don't think she pays much attention after she posts things...I get it, also unregistered.2012-12-11
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This answer gives $u,v$ with $abu+(a+b)v=1$, given $ax+by=1$.

Start with $ax=1-by$, subtract $ay$ from both sides, and get $a(x-y)=1-(a+b)y.$ Now start with $by=1-ax$, subtract $bx$ from both sides, and get $b(y-x)=1-(a+b)x.$ Now multiply these equations to get $ab(x-y)(y-x)=1-(a+b)x -(a+b)y + (a+b)^2 xy,$ which may be rearranged as $ab(x-y)(y-x)+(a+b)(x+y-(a+b)xy)=1.$ With $u=(x-y)(y-x)$ and $v=x+y-(a+b)xy$ this is $abu+(a+b)v=1.$