how to start a proof of a system of congruences

Question

Let $a$, $b$, $m$, and $n$ be integers with $m > 0$, $n >0$, and $gcd(m,n) = 1$. Then the system $x\equiv a$ (mod n) and $x\equiv b$ (mod m) has a unique solution modulo mn.

This is not the Chinese Remainder Theorem just yet. That is the next proof. This is a proof leading up to it.

Help please!

Do you want to show existence too, or only uniqueness. If both then it is already CRT. — Bill Dubuque, Feb 26 '14 at 14:30
For some people CRT is for a general system, and a $2\times 2$ system is the lemma that makes the proof simple. — vadim123, Feb 26 '14 at 14:33
@vadim123 Right, the many variations on what CRT denotes are almost too many to count. — Bill Dubuque, Feb 26 '14 at 14:35
The previous theorem states, the system $x\equiv a$ (mod n) and $x \equiv b$ (mod m) has a solution if and only if $gcd(n,m) | a-b$. It may be useful to use? — Franky, Feb 26 '14 at 14:39
@Franky That yields the existence of solutions. This exercise yields the uniqueness. Presumably the next exercise combines them into CRT. Do you already know that $, m,n\mid k,\Rightarrow,{\rm lcm}(m,n)\mid k, ?\ \ $ — Bill Dubuque, Feb 26 '14 at 14:49
@Franky From that you have $\ m,n\mid k,\Rightarrow,mn\mid mk,nk,\Rightarrow,mn\mid \gcd(mk,nk)=\gcd(m,n)k,$ therefore $,{\rm lcm}(m,n) = mn/\gcd(m,n)\mid k.$ But there are more direct, conceptual ways — Bill Dubuque, Feb 26 '14 at 15:30

score 3 · Accepted Answer · answered Feb 26 '14 at 14:28

3

Hint $\ $ If $\,x',x\,$ are two solutions then $\,m,n\mid x'-x\,$ so $\ \ldots\mid x'-x$

answered Feb 26 '14 at 14:28

Bill Dubuque

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how to start a proof of a system of congruences

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