if $\gcd(a,b)=1$
is it true that, $$a^{\phi(b)}+b^{\phi(a)}=1\mod ab$$
I think the answer is yes, but I am not exactly sure about my reasoning
$a^{\phi(b)}=1\mod b$
and $b^{\phi(a)}=1\mod a$
by Euler's theorem.
The question asks if
$$ab| a^{\phi(b)}+b^{\phi(a)}-1$$
So what would a better way of seeing this be?
You're nearly done. Note that by $\text{CRT}$, all you have to note now is the fact that $$a^{\phi(b)}+b^{\phi(a)} \equiv b^{\phi(a)}\equiv 1\pmod a$$$$a^{\phi(b)}+b^{\phi(a)} \equiv a^{\phi(b)}\equiv 1\pmod b$$ From Euler's Theorem. So we can conclude $$a^{\phi(b)}+b^{\phi(a)}\equiv 1\pmod {ab}$$ As $\gcd(a,b)=1$.
