Show that $a^{\phi(b)}+b^{\phi(a)} \equiv 1 (\text{mod }ab)$ , if a and b are relatively prime positive integers. Show that 
$a^{\phi(b)}+b^{\phi(a)} \equiv 1 (\text{mod} ab)$,
if a and b are relatively prime positive integers.
Note that $\phi(n)$ counts the number of positive integers not exceeding n which are relatively prime with n.

Use Euler's Theorem,
$$a^{\phi (b)}+b^{\phi (a)} \equiv a^{\phi (b)}\equiv 1 (mod \space b) $$
$$a^{\phi (b)}+b^{\phi (a)} \equiv b^{\phi (a)}\equiv 1 (mod \space a) $$
So, $a^{\phi (b)}+b^{\phi (a)} \equiv 1 (mod \space ab) $
