Prove that there exists two integers $m, n$ such that $a^m + b^n \equiv 1 \mod{ab}$ with $(a,b) = 1$. Prove that there exists two integers $m, n$ such that $a^m + b^n \equiv 1 \pmod{ab}$ with $(a,b) = 1$.
Could you please give me some clues on how to solve this problem, not just show me the answer? 
 A: *

*Show that there is $m>0$ such that $a^m\equiv 1 [b]$. and $n>0$ such that $b^n\equiv 1 [a]$.

*Let $c=a^m+b^n$. Show that $c\equiv1[b]$ and $c\equiv1[a]$. Deduce that $c\equiv1[ab]$


You need to use the fact that $(a,b)=1$.
A: Observation: If $\gcd(a,b)=1$, then $\gcd(a+b,ab)=1$.
(Proof: If prime $p$ divides $\gcd(a+b,ab)$ then, without loss of generality, $p$ divides $a$.  Since $p$ divides both $a$ and $a+b$, we know $p$ divides $b$, contradicting that $\gcd(a,b)=1$.)
Hence $$(a+b)^{\varphi(ab)} \equiv 1 \pmod {ab}$$ by Euler's Theorem.  And $$(a+b)^{\varphi(ab)} \equiv \sum_{i=0}^{\varphi(ab)} \binom{\varphi(ab)}{i} a^i b^{\varphi(ab)-i} \equiv a^{\varphi(ab)}+b^{\varphi(ab)} \pmod {ab}$$ by the Binomial Theorem, since $\varphi(ab) \geq 1$.
A: Another solution involving modular arithmetic:
$a^m\equiv 1 \text{ mod b} $, $b^n\equiv 0 \text{ mod b}\rightarrow b^n+a^m\equiv 1 \text{ mod b} $
$b^n\equiv 1 \text{ mod a} $, $a^m\equiv 0 \text{ mod a}\rightarrow b^n+a^m\equiv 1 \text{ mod a} $
Using the Chinese remainder theorem we can conclude $a^m+b^n\equiv 1 \text{ mod ab}$
