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.

Since $gcd(a,b)=1$, by Fermat's little theorem, $a^{\phi(b)}\equiv1 (\mod{b})$.
Now, $b^{\phi(a)}\equiv 0(\mod{b})(\because b\mid b^{\phi(a)}).$
So now we have, $a^{\phi(b)}+b^{\phi(a)}\equiv 1(\mod{b}).\tag{1}$
Again by Fermat little theorem,
$b^{\phi(a)}\equiv 1(\mod{a}).$
And $a^{\phi(b)}\equiv 0(\mod{a})(\because a\mid a^{\phi(b)})$
From this we have,
$a^{\phi(b)}+b^{\phi(a)}\equiv 1(\mod{a})\tag{2}$
From $(1)$ and $(2)$, we have,
$a^{\phi(b)}+b^{\phi(a)}\equiv 1(\mod{ab}), (\because (a,b)=1)$
