TITLE: $a^m + b^n \equiv 1 \mod ab$ for some $m,n$
QUESTION [5 upvotes]: If $a$ and $b$ are relatively prime integers, then there exist integers $m$ and $n$ such that $a^m + b^n \equiv 1 \mod ab$ .  How do I show this to be true?  (Artin's Algebra problem 11.1.16)

REPLY [4 votes]: My answer is 
$$a^{\phi(b)}+b^{\phi(a)}\equiv1\mod ab$$
Because

$$a^m+b^n\equiv1\mod ab$$
$$\Longleftrightarrow a^m+b^n\equiv1\mod a \text{ and } a^m+b^n\equiv1\mod b$$
$$\Longleftrightarrow b^n\equiv1\mod a \text{ and } a^m\equiv1\mod b \text{ ,}$$
then from Euler's theorem, we just need $n=\phi(a)$ and $m=\phi(b)$.

When is your assumption right? Firstly, we denote $\sigma_m(x)=\min\{d:x^d\equiv1\mod m\}$. Then get the conclusion that your assumption is right if and only if $\sigma_a(b)|a-1$ and $\sigma_b(a)|b-1$.