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I'm doing the following exercise and I don't know how to prove it.

Suppose $a,b \in \mathbb{Z}$, and $d=\gcd(a,b)$. I have to prove that if $x\equiv y \bmod d$, then the system

$$X\equiv x \mod a$$

$$X\equiv y \mod b$$

has a solution.

I don't even how to start...

Thanks for your help :)

Michael Hardy
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Alopiso
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  • I explained further in the Remark in the linked dupe how to scale the Bezout gcd equation to get a solution of the associated linear Diophantine equation. If anything remains unclear please feel welcome to ask questions there. – Bill Dubuque Jan 26 '17 at 16:20

1 Answers1

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Bezout's Lemma $a,b$ are relatively prime if and only if there exists an $x,y$ such that $ax+by=1$.

Another way to phrase this is that $ax=1\pmod{b}$ and $by=1\pmod{a}$. See if you can use this statement to solve your problem.

Chris
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Stella Biderman
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    We have that $ax+by=d$, so $ax=d$ (mod $b$) and $by=d$ (mod $a$). On the ohter hand, there exist a $k$ such that $kd=x-y$. Then, $kby=x-y$ (mod $a$), so $kby+y=x$ (mod $a$) – Alopiso Jan 26 '17 at 16:04