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If $a$ and $b$ are relatively primes, with any number of $x$ and $y$, you could always find a set of $x$ and $y$ which makes $ax-by=1$

How is it possible?

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Given $a,b$ with $\gcd(a,b)=1$, let $c$ be the smallest positive integer of the form $ax-by$. We claim $c$ divides every number of the form $ax-by$. For let $am-bn=c$ and $ax-by=d$, then by the Division Theorem $d=cq+r$ with $0\le r<c$, so $$r=d-cq=ax-by-(am-bn)q=au-bv$$ where $u=x-mq$ and $v=y-nq$; by the minimality of $c$, we must have $r=0$, so $c$ divides $d$. Then $c$ must divide $a$ (take $x=1$, $y=0$) and $c$ must divide $b$ (take $x=0$ and $y=-1$), so $c=1$, QED.

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  • $\begingroup$ Thank you, but I don't think I can follow you at all. Why does au-bv have anything to do with c dividing a or b? Like you said, take x=1, y=0, then we get u=1-mq and v=-nq. Since r=0, that makes au=bv, meaning a-amq = bnq. And I'm stuck here. It's getting me nowhere. $\endgroup$
    – user800956
    Jul 5 '20 at 13:56
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    $\begingroup$ The Division Theorem says that given $c,d$ with $c>0$ you can divide $d$ by $c$ and get an integer quotient $q$ and an integer remainder $r$ with $0\le r<c$. If we let $c$ and $d$ be integers of the form $ax-by$, then the algebra that leads to $r=au-bv$ shows that $r$ is also of the form $ax-by$. If $c$ is the smallest positive integer of that form, then since $r<c$ we can't have $r$ being positive, so $r=0$, so $d$ is a multiple of $c$. So every integer of the form $ax-by$ is a multiple of $c$. So in particular, $a$ and $b$ are both multiples of $c$. But $\gcd(a,b)=1$, so $c$ must be $1$.... $\endgroup$ Jul 5 '20 at 23:02
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    $\begingroup$ (continued) so $1=ax-by$ for some $x,y$. $\endgroup$ Jul 5 '20 at 23:03

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