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I could use some help with this proof.

Let $n, m ∈ Z^+$ and $a, b ∈ Z$. Suppose that $ a ≡ b$(mod n) and $a ≡ b$(mod m) and $(m, n) = 1.$ Show that $a ≡ b$(mod mn).

From what I understand, it is obvious that $n|(a-b)$ and $m|(a-b)$. So since $(m, n) = 1$ does it mean that there is some prime factor that divides $mn$ like so: $p^a | mn$? Can someone help me figure this proof out, I'm not totally sure if I'm doing this right.

Any help is greatly appreciated! Thanks!

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$a-b=mp = nq$ for some $p,q$ integers. Now you need to show that $q$ is divisible by $m$ and $p$ is divisible by $n$. But that follows from the relatively prime condition.

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There is a property stating:
If $a|c$ and $b|c$ and $(a,b)=1$, then $ab|c$.
This can easily be proved using Bezout's Identity.


If you apply it to what you have already stated concerning: $m/(a-b)$ and $n/(a-b)$, you obtain:
$$mn|(a-b)$$
If we suppose that $mn=k$ this leads to the result:
$a \equiv b \pmod k$
N.B: I used $k$ because i couldn't fit the two letters $m$ and $n$ in the $mod$ for some reason.. Perhaps my relatively minimal knowledge of MathJaX.

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