Suppose that $\displaystyle a,b$ are two integers with gcd(m, n) = 1. Prove that there exist integers $\displaystyle m, n$
such that $\displaystyle a^{m} + b^{n} \equiv 1 \: mod \: ab.$
As an illustration, here's what happens when a=5 and b=7. The equation $\displaystyle 5^m\equiv1\!\!\!\pmod7$ has the solution m=6, and the equation $\displaystyle 7^n\equiv1\!\!\!\pmod5$ has the solution n=4. Then the number $\displaystyle 5^6+7^4$ will be congruent to 1 (mod 5) and also (mod 7), and therefore also (mod 35).
[Check: $\displaystyle 5^6+7^4 = 18026 = 515\times35+1$.]