Suppose that $x \equiv 1 \pmod n$ and $x \equiv 1\pmod m$. Prove that $x \equiv 1 \pmod{ mn}$.

Question

I know the Chinese Remainder Theorem implies that there exists a unique solution modulo $mn$, but I can't spot how to show directly that you can deduce that $ x \equiv 1 \pmod {mn}$.

Thanks in advance!

Answer 1

If $m|x-1$ and $n|x-1$ and $\gcd(n,m)=1$, then $nm|x-1$.

(If $x-1=mk=nj$ and $\gcd(m,n)=1$, then $n|k$ by the generalized Euclid's lemma,

so $nm|x-1$.)

Answer 2

From $x\equiv1 \mod(m)$ it follows that $m \mid x-1$, hence we can write $x-1=m*p$, for some $p\in\mathbb{N}$. Similarly from $x\equiv1 \mod(n)$ we have $n \mid x-1$ or $n \mid m*p$. Since $n\nmid m$ because of $gcd(m,n)=1$ it must follow that $n\mid p$, hence $p=q*n$ for some $q\in\mathbb{N}$. So it follows that $x-1=q*m*n \Rightarrow x-1\equiv 0 \mod(mn) \Leftrightarrow x\equiv1\mod(mn)$