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Edit: The question this is associated with has weaker hypothesis, so whilst technically the answers there also answer this question, some people may find it easier to follow the solution to this easier question. I have therefore voted to reopen and invite others to do the same.

I am trying to prove the following proposition. I am sure that it should be not so difficult but I failed to prove it.

Suppose that $\text{gcd}(s,t)=1$. Let $y$ such that $\text{gcd}(y,s)=1.$ Prove that $\exists x$ such that $\text{gcd}(x,st)=1$ and $x\equiv y\pmod s$.

I was trying to apply Bezout identity to coprime pairs but it did not work out. Can anyone show the proof please?

tkf
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RFZ
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  • By Stieltjes $,\exists, n\in\Bbb Z!:\ (\overbrace{\color{#c00}y+n,\color{#c00}s}^{\large x},,\color{#c00}{st}) = 1,$ by $,(\color{#c00}{y,s,st})=1\ $ (by $(\color{#c00}{y,s})=1$ by hypothesis) – Bill Dubuque Dec 02 '21 at 08:38
  • Or directly: $,(y!+!ns,s)=(y,s)=1$ and we can make $(y!+!ns,t)=1$ by choosing $,n,$ so $\bmod t!:\ y!+!ns\equiv 1,,$ i.e. $,n\equiv (1!-!y)/s\ $ $(s^{-1}$ exists by $(s,t)=1)\ \ $ – Bill Dubuque Dec 02 '21 at 08:56

1 Answers1

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As $\gcd(s,t)=1$, we have some integer $r$ such that $sr \equiv 1 \mod t$. Then set $$x=y-sr(y-1)$$

Then $x\equiv y \mod s$ and $x\equiv 1 \mod t$. In particular $\gcd(x,t)=1$. Also $\gcd (x,s)=1$ as any common factor of $x$ and $s$ is a common factor of $y$ and $s$.

Combining $\gcd(x,t)=1$ and $\gcd(x,s)=1$, we get $\gcd(x,st)=1$.

tkf
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