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I have this math question that I am stuck on. This is the question:

Suppose that $a$ and $b$ are positive integers, with $d = \gcd(a, b)$. Suppose that there exists integers $r$ and $s$ so that $ar + bs= d$. Show that $\gcd(r, s) = 1$ using the relatively prime equation.

This is the relatively prime relationship equation: Let $a,b$ be non-zero integers. $a$ and $b$ are relatively prime iff there exists integers $x$ and $y$ such that $ax+by=1$

I know that the relatively prime equation that I have to solve is $ar + bs = 1$. However, I'm not sure how to start this. Thanks.

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  • $\begingroup$ What is this "relatively prime equation"? As stated, you can't use Bezout's identity because that's what you are trying to prove. $\endgroup$ – A.P. Oct 8 '15 at 19:24
  • $\begingroup$ @A.P. Let $a,b$ be non-zero integers. $a$ and $b$ are relatively prime iff there exists integers $x$ and $y$ such that $ax+by=1$ $\endgroup$ – KFC Oct 8 '15 at 19:26
  • $\begingroup$ That's a special case of what you are trying to prove. $\endgroup$ – A.P. Oct 8 '15 at 19:27
  • $\begingroup$ So, how do I show it using that equation? I don't start off with the equation? $\endgroup$ – KFC Oct 8 '15 at 19:29
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Dividing $a$ and $b$ by $d$, we have two relatively prime numbers. By the relatively prime numbers equation, it exist $r$ and $s$ such that

$r\frac{a}{d} + s\frac{b}{d} = 1.$

Those numbers are relatively prime. If not, the left part could be factorise giving an integer factorisation of 1...

Multiplying everything by $d$ again to obtain what you need

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  • $\begingroup$ How is that using the Relatively Prime Relationship Equation? $\endgroup$ – KFC Oct 8 '15 at 19:25
  • $\begingroup$ Since $\frac{a}{d}$ and $\frac{b}{d}$, there is $r$ and $s$ such that $r\frac{a}{d} = s\frac{b}{d}=1$. $r$ and $s$ are relatively prime. $\endgroup$ – Alain Remillard Oct 8 '15 at 19:30
  • $\begingroup$ +1: The idea's there, but you should try to flesh-out your answers a bit more. Right now this is more a comment than an answer... $\endgroup$ – A.P. Oct 8 '15 at 19:34
  • $\begingroup$ My goal was to give you a hint, I will flesh out my solution. $\endgroup$ – Alain Remillard Oct 8 '15 at 19:35
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We know that $a = d\bar{a}$ and $b = d\bar{b}$, so dividing $$ ar + bs = d $$ by $d$ we get $$ \bar{a}r + \bar{b}s = 1 $$ which is what you call the "relatively prime equation", hence $r$ and $s$ must be coprime.

To conclude without assuming this, just suppose that there is an $f > 1$ that divides both $r$ and $s$. Then the above equation implies $f \mid 1$, which is absurd.

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