If $p$ and $q$ are distinct primes and $a$ be any integer then $a^{pq} -a^q -a^p +a$ is divisible by $pq$. If $p$ and $q$ are distinct primes and $a$ be any integer then $a^{pq} -a^q -a^p +a$ is divisible by $pq$.
Factorising we get $a^{pq} -a^q -a^p +a =a^p(a^q -1) - a(a^{q-1}-1)$ and we know $p \mid a^{p-1}-1$ and $q \mid a^{q-1}-1$.
I can't proceed further with the proof. Please Help!

Hint $\,\ \{a^{pq},a\} \equiv \{a^p,a^q\}$ mod $ p,q,\,$ via $\,a^p\equiv a\pmod p,\ a^q\equiv a\pmod q\ $ [little Fermat]
Hence $\, a^{pq}\!+\!a \ \equiv\ a^p\! + a^q\,$ mod $\,p,q,\,$ so also mod $\,pq = {\rm lcm}(p,q)$
since addition $\,f(x,y)\, =\, x + y\,$ is $\rm\color{#c00}{symmetric}$  $\,f(x,y)= f(y,x),\,$  therefore its value depends only upon the (multi-)set $\,\{x,\,y\}.\,$  
Generally $ $ if a polynomial $\,f\in\Bbb Z[x,y]\,$ is $\rm\color{#c00}{symmetric}$ then
$\qquad\qquad\quad \{A, B\}\, \equiv\, \{a,b\}\,\ {\rm mod}\,\ m,\, n\,\ \Rightarrow\,\  f(A,B)\equiv f(a,b)\, \pmod{\!{\rm lcm}(m,n)}\qquad\quad$
a generalization of the constant-case optimization  of CRT = Chinese Remainder,  combined with a generalization of the Polynomial Congruence Rule to (symmetric) bivariate polynomials.
