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.
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.