Let $a,b,m,n \in N$ with $\gcd(m,n)=1$ prove that the modular system $ \{ x=a \mod m ; x =b\mod n \}$ has absolution and is unique modulo $mn$}
Note that had asked a question I got why there it is a solution but not why it is unique mod $mn$
I recall it is $\operatorname{lcm}(m,n)=1$ but it is unclear why? (if it is)
hopefully this helps
proof since $\gcd (m,n)=1, \exists u,v s.t mu+nv =1$ leads to 4 consequences
- i) $mu =0 \mod m$
- ii) $nv=1 (\mod m)$ [b.c $1-nv=mu$]
- iii) $nv=0\mod n$
- iv) $mu =1 \mod n $ [b.c $1-mu=nv$]
let $t=bmu +an \therefore $ by i),ii) and thm [if $a=b \mod m \wedge c=d \mod n \Rightarrow a+c=b+d \mod n$ and $ac=bd\mod n$]] $$t=bmu +anv=b0+a1=a\mod m$$ so that $t=a \mod m $ similarly by iii iv and prev. thm $t=bmu+anv=b*1+a*0=b \mod n$. so that $t=b\mod n $ $\therefore$ t is a solution to system
(Read on Chinese remainder theorem)
Suppose $\exists x_1$ that is a solution so,
$$x_1 = x \equiv a \bmod m \iff m \mid (x_1 - x)$$ $$x_1 = x \equiv b \bmod n \iff m \mid (x_1 - x)$$ since $\gcd(m, n) = 1$, $mn \mid (x_1 - x)$ $\therefore$ $x = x_1 \mod mn$ so the solution is unique $\bmod mn$.