Prove that if $(a,b)=1$ then there exist some $m,n$ such that $a^m+b^n\equiv 1\pmod {ab}$. Number $a$, $b$ are nature and positive number.
Since $(a,b)=1$ then there some number $x$, $y\in \mathbb Z$ such that $ax+by=1$. Since $(a,b)=1$ then I can use Fermath's theorem $a^{b-1}\equiv 1 \pmod b$, and $b^{a-1}\equiv 1 \pmod a$ so then $a^b\equiv a \pmod {ab}$ and $b^a \equiv b \pmod {ab}$. So $a^b+b^a\equiv a+b \pmod {ab}$. But I am not so sure that I going in right direction. Can you help me?
On
$\overbrace{(a,b)=1\,\Rightarrow(\color{#c00}{ab},a\!+\!b)=1}^{\Large (\color{#c00}a,a+b)=(a,b)= (\color{#c00}b,a+b)},\ $ so $\,\bmod \color{#c00}{ab}\!: \overbrace{\exists\, n\!:\ 1 \equiv (a\!+\!b)^{\large n}}^{\Large{\rm e.g.}\,\ n\ =\ {\rm ord}(a+b)}\!\equiv \underbrace{a^{\large n} + b^{\large n}+\overbrace{\color{#c00}{(ab)}(\cdots)}^{\large \equiv\ 0\ }}_{\large \rm Binomial\ Theorem}$
By the Chinese remainder theorem, you need to find $m$, $n\in\Bbb N$ with $a^m+b^n\equiv1\pmod a$ and $a^m+b^n\equiv1\pmod b$. That is $b^n\equiv1\pmod a$ and $a^m\equiv1\pmod b$. By the Fermat-Euler theorem, you can take $n=\varphi(a)$ etc.