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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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    what if you let $a = 5$ and $b = 6$?2012-12-09
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    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
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    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
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    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
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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
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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

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