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$p$ and $q$ are distinct primes.

Where can I start with this proof?
It looks similar to the Chinese Remainder Theorem, but that deals with two different a values.

Hint $ $ Below are few proofs of this constant-case CCRT of Chinese Remainder Theorem (CRT). The first three proofs work for for arbitrary coprime naturals $rm,p,q.$

$rm(1) x equiv apmod {pq}:$ is clearly a solution, and the solution is $color{#C00}{unique}$ mod $rm,pq,$ by CRT.

$rm(2) p,q:|:x!-!aiff lcm(p,q):|:x!-!a.:$ Further $rm:(p,q)=1iff:lcm(p,q) = pq.$

$(3) , $ By Euclid's Lemma: $rm:(p,q)=1, p:|:qn =:x!-!a:Rightarrow:p:|:n:Rightarrow:pq:|:nq = x!-!a.$

$rm(4) , $ Since the prime factorization of $rm,x!-!a,$ is $color{#C00}{unique}$, and the prime $rm,p,$ occurs in one factorization, and the distinct prime $rm,q,$ occurs in another, both factorizations are the same up to order, hence contain both $rm,p,$ and $rm,q,:$ hence have $rm,pq,$ as a divisor.

Remark $ $ This constant-case optimization of CRT arises frequently in practice so is well-worth memorizing, e.g. see some prior posts for many examples.

Quite frequently, $color{#C00}{uniqueness} theorems$ provide powerful tools for proving equalities.

Let $[A,B]=lcm(A,B)$ and $(A,B)=gcd(A,B)$

If $p,q$ are different integers, $pmid(x-a)$ and $qmid(x-a)implies [p,q]mid(x-a)$

We know $[p,q]cdot (p,q)=pcdot q$

If $(p,q)=1, [p,q]=pcdot q$

If $p,q$ are distinct primes, $(p,q)=1$

Let $y=x-a$. We want to show that if $p$ divides $y$ and $q$ divides $y$ then $pq$ divides $y$.

Since $p$ divides $y$, we have $y=pz$ for some $z$. Thus $q$ divides $pz$. Since $q$ is prime, this implies $q$ divides $p$ or $q$ divides $z$. But $q$ cannot divide $p$, so $q$ divides $z$. Suppose that $z=qw$. Then $y=pqw$.