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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 $(a+1)(x)+(ab)(y)$
  • 2
    what if you let $a = 5$ and $b = 6$?2012-12-09
  • 4
    I don't think that the statement is true. Try $a =2$ and $b = 3$. Then $(2,3) = 1$ but $(2+1, 2\times3) = (3,6) = 3$. Am I missing something?2012-12-09
  • 1
    What about $(6,7)=1$ while $(7,42)=7$? (well, I was too late...)2012-12-09
  • 5
    Maybe you mean $(a+b,ab)=1$ in the conclusion?2012-12-09
  • 0
    Anna: I edited your question. I tried to make the formatting more like what I think you had intended it to be.2012-12-09
  • 0
    what i mean is you square it and then you get a^2*x^2 + 2abxy +b^2y^2 and continue to manipulate it2012-12-09
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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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    what i mean is you square it and then you get a^2*x^2 + 2abxy +b^2y^2 and continue to manipulate it2012-12-09
  • 3
    @Anna, the statement you are trying to prove is false. At this point, you should go back and check the source of the problem very carefully.2012-12-09
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    Anna = Cindy?${}$2012-12-11
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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 $abu+(a+b)v=1.$