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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 think about fundamental theorem of arithmetic...2012-08-16
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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

2 Answers 2

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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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    Are you saying $ijpq=jp=iq$??2012-08-16
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    Let me edit my post...2012-08-16
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    Now that is the same as my answer except that my $x$ is your $a + b$.2012-08-16
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    True, and you were first. My first answer was bad.2012-08-16
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    Guys, guys, it's all for the joy of helping number theory idiots like me! Can't we all just get along? :) Also, if I ever meet Tom Apostol, I'm going to push him over. Fair warning given.2012-08-16
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    I think we are all getting along :)2012-08-16
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    I think so too!2012-08-16