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I'm not entirely sure of how I should prove this statement:

$$ \gcd(a,b) = ax+by \Rightarrow \gcd(x,y) = 1 $$

So I've tried

$$ \begin{align} &\gcd(x,y) = d \Rightarrow x =x'd, y=y'd\\ \Rightarrow &\gcd(a,b) = ax'd+by'd = d(ax'+by')\\ \Rightarrow & \gcd(a,b) = d(\gcd(a,b)) \\ \Rightarrow &1= d \end{align} $$

But I'm not sure if this is correct. Even so, is there perhaps a better way to solve this?

hola
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Frank Vel
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4 Answers4

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Denote $\gcd(a,b)=c$, then $a=ca'$ and $b=cb'$, $$\gcd(a,b)=ax+by\Leftrightarrow c=ca'x+cb'y\Leftrightarrow a'x+b'y=1$$ so $\gcd(x,y)=1$.

Alfred Chern
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\begin{align*} d \mid x \text{ and } d \mid y &\implies d \gcd(a,b) \mid ax \text{ and } d \gcd(a,b) \mid by \\ &\implies d \gcd(a,b) \mid ax + by = \gcd(a,b) \\ &\implies d = \pm 1. \end{align*} As others have mentioned, your own solution has an error: just because $ax + by = \gcd(a,b)$ for this specific value of $x,y$ certainly does not mean it will be equl to $\gcd(a,b)$ for any $x,y$. For instance this does not hold in the cases $x,y = 0$ and $x, y = 1$.

Caleb Stanford
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Why would $ax'+by' = \gcd(a,b)$ hold? Your solution is $\color{#c00}{\text{wrong}}$.

Please note that the statement $\gcd(a,b) = ax+by$, holds for some integers $x,y$, and not all integers $x,y$ satisfy this relation. perhaps, you are mistaken here.

For example, $\gcd(2,3)=1$, that is we take $a=2, b=3$; so you would get $2\cdot2+3\cdot(-1)=1$. So, it holds with $x=2,y=-1$. Note that there are infinitely many choices of $x,y$; such as $(-4, 3), (5, -3), (8, -5)$ etc. But not all integer $x,y$ are applicable. You may like to test with, say, $x=0, y=0$, or $x=-1, y=2$ etc.

hola
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  • OK @Goos, I apologize. Even if I am so, it was not intentional... :-(. – hola Oct 17 '14 at 10:54
  • No worries I probably misread your tone. – Caleb Stanford Oct 17 '14 at 10:58
  • yes but if x and y could be factored by a common divisor, $ax'+by'$ would be a solution. So if $t|-1$ and $t|2$, then we would have a smaller solution $-1/t$ and $2/t$. A contradiction? Although I'm not sure if that would prove anything... – Frank Vel Oct 17 '14 at 11:12
  • @pushpen.paul for any $a,b$ that $\gcd(a,b) = ax+by \Rightarrow \gcd(x,y) = 1$ – Frank Vel Oct 17 '14 at 11:24
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    @fvel Please mention that $x\neq0,y\neq0$. This is true (I checked), but what I was telling (in response to your comment "I'm not sure if this is correct"): Your solution is wrong. Kindly see other answers, if that helps, for correct solution of your question. I am not posting an answer to it, since it is already here. – hola Oct 17 '14 at 12:06
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    @fvel Your argument is valid if it is known that $gcd(a,b)=ax+by = \mathbf1$ (note that there is one 1). – hola Oct 17 '14 at 12:06
  • @pushpen.paul I want to prove the implication whenever $\gcd(a,b) = d$, for any $d$. Is that false? Anyways, people are more likely to correct something wrong than they're willing to prove something right. I was pretty sure my proof was flawed and quickly got a lot of responses right away. – Frank Vel Oct 17 '14 at 12:09
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    @fvel Summary: Your question is correct (for all $\gcd$'s, may not be $1$). Here are correct solutions to it. But, what you did was wrong (unless you know that the $\gcd$ is $1$). Hope, that makes clear the fact. :-) – hola Oct 17 '14 at 12:12
  • @pushpen.paul ah thanks now I understand what you meant... anyways thanks for your time and help! – Frank Vel Oct 17 '14 at 12:16
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    @fvel You are always welcome. – hola Oct 17 '14 at 12:17
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The statement is probably that there exists such a pair of values $x,y$ that $\gcd(a,b)=ax+by$. Therefore, you cannot say that $ax'+by' = \gcd(a,b)$, your mistake comes from this replacement.

5xum
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