Find p and q s.t. $({p-1})^{q-1}+({q-1})^{p-1} \equiv 101 \pmod {pq}$

The question is:

Find two distinct odd primes p and q such that $({p-1})^{q-1}+({q-1})^{p-1} \equiv 101 \pmod {pq}$ Well... I only have the idea to write down

${p-1}^{q-1} \equiv -1^{q-1} \equiv 1 \pmod {p}$

Actually, I solve this question by brute force and get p=11 q=3; however, I wanna know some wise solutions...

Mine is .... a bad one

Please help me to solve this question :(

Taking your equation mod $p$,

$$(-1)^{q-1} + (q-1)^{p-1} \equiv 101 \ (\text{mod }p)$$ But $q$ is an odd prime, so $(-1)^{q-1} = 1$, and thus $(q-1)^{p-1} \equiv 100 \ (\text{mod }p)$. Similarly $(p-1)^{q-1} \equiv 100 \ (\text{mod }q)$. Now use Fermat's "little" theorem (but don't forget the case where $p$ or $q$ is $5$).