Prove $a^{\frac{\phi(mn)}2} \equiv 1\pmod {mn}$

Question

I want to show that given $\gcd(m,n)=1$ with $m,n \gt2$ and $\gcd(a, mn)=1$, then $a^{\frac{\phi(mn)}2} \equiv 1\pmod {mn}$, where $\phi(mn)$ is the Euler Totient function.

My approach started using Euler's theorem, given $\gcd(a, n)= 1$ then $ a^{\phi(n)} \equiv 1\pmod {n}$. And since $(m,n)=1$ then: $$a^{\phi(mn)} \equiv 1\pmod {mn}$$

I just cannot seem to get the $\frac12$ factor to appear. What I tried doing was just raising both sides to the $\frac 12$

$$a^{\phi(mn)} \equiv 1\pmod {mn} \Rightarrow a^{\frac{\phi(mn)}2} \equiv 1^{\frac12}\pmod {mn} \equiv 1\pmod{mn}$$ But I somehow feel this is wrong. So another way I tried was again with Euler's theorem:

$$a^{\phi(n)} \equiv 1\pmod {n} \Rightarrow (a^{\phi(m)})^{\phi(n)} \equiv 1\pmod {n} \Rightarrow a^{\phi(m)\phi(n)} \equiv 1\pmod {n} $$

$$a^{\phi(m)} \equiv 1\pmod {m} \Rightarrow (a^{\phi(n)})^{\phi(m)} \equiv 1\pmod {m} \Rightarrow a^{\phi(n)\phi(m)} \equiv 1\pmod {m} $$

Multiplying one by the other yields:

$$a^{2\phi(m)\phi(n)} \equiv 1\pmod {mn} \Rightarrow a^{2\phi(mn)} \equiv 1\pmod {mn}$$

Or using the CRT:

$$a^{\phi(m)\phi(n)} \equiv 1\pmod {mn} $$

From here I have no idea on where to go or what to try, I was thinking perhaps something related to Wilson's Theorem to make the $\frac12$ factor appear, but since m and n may not be prime, they are just coprime, I don't know if I can use that. Am I going in the right direction?

Please search to avoid addinge dupe questions, Here they can be quickly located by approach0. — Bill Dubuque, Feb 24 '21 at 02:16

score 1 · Accepted Answer · answered Feb 23 '21 at 18:16

1

Hint: Prove that $\phi(N)$ is even for all $N\geq 3$. So both $\phi(m),\phi(n)$ are even, hence you have the spare factor of 2: $\operatorname{lcm}(\phi(m),\phi(n))\mid \frac12\phi(m)\phi(n)$.

answered Feb 23 '21 at 18:16

user10354138

33,239

So since $\phi(mn)$ is always even I can divide by $2$ without any restrictions just because I have the spare factor of $2$? – Rodrigoss Feb 23 '21 at 18:35
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No. $\phi(m)$ divides $\phi(m)\frac{\phi(n)}{2}$ and $\phi(n)$ divides $\frac{\phi(m)}{2}\phi(n)$. So $a^{\phi(m)\phi(n)/2}\equiv(a^{\phi(m)})^{\phi(n)/2}$ is $1$ mod $m$ and similarly mod $n$, and CRT gives the result. – user10354138 Feb 23 '21 at 18:38
Please strive not to add more dupe answers to FAQs. Here they can be quickly located by approach0. – Bill Dubuque Feb 24 '21 at 02:13
@BillDubuque $m,n$ not primes and all results seems to be asking for prime case (or question 3491175 with 3n and 9n a very specific version)? – user10354138 Feb 24 '21 at 02:26
See this answer in the linked dupe (there are also many other dupes - it's a FAQ). – Bill Dubuque Feb 24 '21 at 02:33

Prove $a^{\frac{\phi(mn)}2} \equiv 1\pmod {mn}$

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