Relatively Prime Relationship Equation Proof

Question

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.

Answer 1

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

Answer 2

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.