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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?

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    $\begingroup$ raise both sides of $ax+by=1$ to the $ab$ power, or something like that $\endgroup$ – mathworker21 Nov 24 '18 at 5:08
  • $\begingroup$ How do you mean, you say that I need to use $x+y|x^n+y^n$ or something else? $\endgroup$ – Marko Škorić Nov 24 '18 at 5:15
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    $\begingroup$ no, just use binomial theorem to expand $\endgroup$ – mathworker21 Nov 24 '18 at 5:22
  • $\begingroup$ @mathworker21 Hint: $\,\bmod ab\!:\ (a\!+\!b)^{\large n}\equiv a^{\large n} + b^{\large n}.\,$ See my answer for details. $\endgroup$ – Bill Dubuque Nov 24 '18 at 17:18
  • $\begingroup$ @MarkoŠkorić We can do it directly without CRT - see my answer. $\endgroup$ – Bill Dubuque Nov 24 '18 at 17:20
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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.

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$\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 $\,\bmod \color{#c00}{ab}\!: \overbrace{\exists\, n\!:\ 1 \equiv (a\!+\!b)^{\large n}}^{\Large{\rm e.g.}\,\ n\ =\ {\rm ord}(a+b)}\!\equiv \underbrace{a^{\large n} + b^{\large n}+\overbrace{\color{#c00}{(ab)}(\cdots)}^{\large \equiv\ 0\ }}_{\large \rm Binomial\ Theorem}$

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