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Put a bit more coherently, given $p$ and $q$ as distinct prime numbers, and thus $(p,q)=1$, if

$p^{(q-1)} + q^{(p-1)} \equiv 1 \pmod p$ and $p^{(q-1)} + q^{(p-1)} \equiv 1 \pmod q,$

why does that lead to

$p^{(q-1)} + q^{(p-1)} \equiv 1 \pmod {pq}?$

The textbook I'm working with jumps to that conclusion as if it were obvious, but it's not. Not to me, at least.

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    I've had lots of hints. I can manipulate the CRT over and over mechanically but have no understanding of it. When it comes to number theory, I just need someone to tell me, and then I understand, much like C.Williamson has just done. Now that he's told me, I understand.2012-08-16

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Your question should be rephrased as:

Let $p$ and $q$ be distinct prime numbers. If $p | x$ and $q | x$, is it true that $pq | x$?

The fundamental theorem of arithmetic should answer this immediately.

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    It does. Many thanks, now perfectly clear.2012-08-16
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Since $p$ and $q$ are distinct prime numbers, if they both separately divide $a+b$, then they both occur in the prime factorization of $a+b$. Therefore, $pq$ would divide $a+b$.

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    I think so too!2012-08-16