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Given that $m,n > 2$ are relatively prime integers and that $a$ is an integer relatively prime to $mn$, prove that $$ a^{[\phi(m), \phi(n)]}\equiv 1 \pmod{mn} $$

I started by using the fact that

$$ [\phi(m), \phi(n)] = \alpha\phi(m)=\beta\phi(n) $$ for some positive integers $ \alpha , \beta $ to then write

$$ a^{\alpha\phi(m)}=(a^{\phi(m)})^\alpha\equiv(1)^\alpha\equiv1 \pmod{m} $$ and $$ a^{\beta\phi(n)}=(a^{\phi(m)})^\alpha\equiv(1)^\beta\equiv1 \pmod{n} $$

I'm wondering if what I did was correct and how to apply the Chinese Remainder Theorem to show that this congruence is true mod mn.

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Yes, it is correct. To finish, by CRT: $\,A\equiv 1\bmod m\ \&\ n\iff A\equiv 1\pmod{\!mn}.\,$ Or, w/o CRT, we have $\,m,n\mid A-1\iff {\rm lcm}(m,n)\mid A-1,\,$ and $\,{\rm lcm}(m,n) = mn\,$ by $\,\gcd(m,n)=1$

Remark $ $ This simple special case is known as CCRT = Constant-case Chinese Remainder Theorem

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  • $\begingroup$ what is A? and how did you get it using the CRT? $\endgroup$ – joseph Nov 19 '18 at 18:06
  • $\begingroup$ the CCRT case link attached above is for primes p and q.. does it matter that m and n are just relatively prime in this case? $\endgroup$ – joseph Nov 19 '18 at 18:07
  • $\begingroup$ @mjoseph $A := a^{\large [\phi(m),\phi(n)]}$. You proved $\,A\equiv 1\pmod m$ and $\,A\equiv 1\pmod n$ $\endgroup$ – Bill Dubuque Nov 19 '18 at 18:10
  • $\begingroup$ @mjoseph The first $3$ proofs in the link work for arbitrary coprime moduli (I added a remark emphasizing that). $\endgroup$ – Bill Dubuque Nov 19 '18 at 18:14
  • $\begingroup$ Just looked at the link a little more and understand it now. Thank you! $\endgroup$ – joseph Nov 19 '18 at 19:26

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