CRT + Fermat's Little Theorem

Write $a=p^{q-1}+q^{p-1}$. Immediately we have

$$\begin{cases} a\equiv 0^{\,q-1}+q^{p-1} \equiv 1 \mod p \\ a\equiv p^{q-1}+0^{\,p-1} \equiv 1 \mod q.\end{cases}$$

What does CRT tell us $a$ must be $\bmod pq$?

Hint $\rm\ mod\ (p,q)\!:\ q^{p-1}\equiv (1,0),\ \ p^{q-1}\equiv (0,1)\:$ so their sum $\rm\:x \equiv (1,1)$

So $\rm\:p,q\ |\ x\!-\!1\ \Rightarrow\ pq = lcm(p,q)\ |\ x\!-\!1.\:$ (CRT isn't needed for this simple constant case)

More generally, if $\rm\:b,c\:$ are coprime then $\rm\: e' = b^{\phi(c)}$ satisfies $\rm\: e'\equiv 0\pmod{b},\ e'\equiv 1\pmod{c}\:.\ $ Hence, $\ $ using $\rm\:e'\:$ and its complement $\rm\:\ e = 1-e'\:,\ $ where $\rm\ \ e\: \equiv 1\pmod{b},\ e\:\equiv 0\pmod{c}\:,\:$ yields the following closed-form for solutions of congruences by the Chinese Remainder Theorem

$$\begin{eqnarray}\rm x\ \equiv\ a\pmod{b} \\ \rm x\ \equiv\ d\pmod{c}\end{eqnarray}\rm\ \ \iff\ \ x\ \equiv\ a\ e + d\ e'\pmod{b\:c}$$

The relationship between CRT and the orthogonal idempotents $\rm\:(1,0),\ (0,1)\:$ will become clearer when you study the Peirce decomposition induced by such idempotents.

first using Fermat's theorem,

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

Also,$\ \ \ p^{q-1}\equiv0 \pmod p$

Second using Fermat's theorem,

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

Also,$q^{p-1}\equiv0 \pmod q$

Using first part we get,

$q^{p-1}+p^{q-1}\equiv1 \pmod p$

using second part we get,

$p^{q-1}+q^{p-1}\equiv1 \pmod q$

Since $p$ and $q$ are distinct prime, therefore $\gcd (p,q)=1$

$q^{p-1}+p^{q-1}\equiv1 \pmod {pq}$