Suppose that $p$ and $q$ are distinct odd primes and $a$ is an integer such that $\text{gcd}(a,pq)=1$. Prove that $a^{(p-1)(q-1)+1}\equiv a \pmod{pq}$.
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1Welcome to Math Stack Exchange. Can you use the Chinese Remainder Theorem? – J. W. Tanner Jul 02 '19 at 18:18
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We are not taught that theorem yet, so I'd really appreciate if there is any other way to prove this. Thank you! – Hana Jul 02 '19 at 19:19
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See my comment to my answer below – J. W. Tanner Jul 02 '19 at 19:35
2 Answers
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Use Fermat's Little Theorem to say $a^{p-1}\equiv1 \mod p$ and $a^{q-1}\equiv 1 \mod q.$
Therefore $a^{(p-1)(q-1)}\equiv 1 \mod p,q.$
Therefore $a^{(p-1)(q-1)}\equiv 1 \mod pq,$ using the Chinese Remainder Theorem.
Now multiply both sides by $a$.
J. W. Tanner
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If you don’t want to invoke CRT, will you accept primes $p,q|x-1\implies pq|x-1$ ? – J. W. Tanner Jul 02 '19 at 19:35
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Okay, but then I cannot figure out how I can apply this rule to prove it... – Hana Jul 02 '19 at 20:13
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which rule? Fermat's little theorem? that was used to prove $a^{p-1}\equiv1\mod p$ and same for $q$; then $x=a^{(p-1)(q-1)}\equiv1 \mod p,q$; the next line in my answer uses either CRT or the alternative in my comment – J. W. Tanner Jul 02 '19 at 20:21
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