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Let $p,q\in \mathbb{N}$ primes and $a,b\in \mathbb{Z}$ such that $a^e \equiv a \mod p$ and $b^e \equiv b \mod p$. By Chinese remainder theorem we know that existis a unique $x\in \mathbb{Z}$ (mod pq) which satisfies $$ x \equiv a \mod p \qquad x \equiv b \mod q$$ Calculate $x^e \mod pq$.

Sol. $$ x^e \equiv a^e \equiv a \equiv x\mod p \qquad x^e \equiv b^e \equiv b \equiv x \mod q $$ $$ p \mid x^e - x\qquad q \mid x^e - x$$ Using $p,q$ are primes, $pq \mid x^e-x$ so $x^e \equiv x \mod pq$.

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  • $\begingroup$ Are we supposed to calculate its residue class mod $pq$? $\endgroup$ – G Tony Jacobs May 19 '17 at 17:14
  • $\begingroup$ @GTonyJacobs Sorry, my fault. $\endgroup$ – Rafael Gonzalez Lopez May 19 '17 at 17:19
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Since we have $x\equiv a\pmod{p}$, then we can raise both sides to the power $e$, and say $x^e\equiv a^e\equiv a\pmod{p}$. We can do the same thing with our given fact modulo $q$.

Does that give you enough of a push in the right direction?

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  • $\begingroup$ I knew that, but let me think. Thank you. $\endgroup$ – Rafael Gonzalez Lopez May 19 '17 at 17:24
  • $\begingroup$ Sorry I delete some comments. It was wrong ideas. I can't do it :< @GTonyJacobs $\endgroup$ – Rafael Gonzalez Lopez May 20 '17 at 11:46
  • $\begingroup$ I've got it. Ty $\endgroup$ – Rafael Gonzalez Lopez May 20 '17 at 12:54

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