If $a \equiv b \mod m$, $a \equiv b \mod n$, and $\gcd(m,n)=1$, then $a \equiv b \mod mn$.

Question

So I know that $mk=a-b$ and $nj=a-b$ for some $k,j \in \mathbb{Z}$ and that $\gcd(m,n)=1$ can be represented as $mx+ny=1$ for some $x,y \in \mathbb{Z}$.

I know to prove this I must end with an equation looking like this: $(mn)f=a-b$ for some $f \in \mathbb{Z}$.

I'm just not connecting something in the middle. Any help is appreciated.

See this answer in the dupe for a handful of ways to prove it. — Bill Dubuque, Sep 16 '20 at 21:51

score 0 · Accepted Answer · answered Sep 16 '20 at 21:41

0

Note that\begin{align}a-b&=(a-b)(mx+ny)\\&=(a-b)mx+(a-b)ny\\&=njmx+mkny\\&=mn(jx+ky).\end{align}

answered Sep 16 '20 at 21:41

José Carlos Santos

427,504

i.e. $,m,n\mid c\Rightarrow mn\mid mc,nc\Rightarrow,mn\mid c(\color{#c00}{mx!+!ny})= c\color{#c00}{\gcd(m,n)},,$ so $,mn/\gcd(m,n)\mid c,,$ for $,c = a!-!b,$ here. This is the common Bezout-basied proof of ${\rm lcm}(m,n) = mn/\gcd(m,n),$ in the case $\gcd(m,n)=1,,$ already posted here hundreds of times, e.g. here. Please strive to not post yet more dupe answers. – Bill Dubuque Sep 16 '20 at 22:16

If $a \equiv b \mod m$, $a \equiv b \mod n$, and $\gcd(m,n)=1$, then $a \equiv b \mod mn$.

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