There are several ways to prove this.
In my last article, I introduced you the proof as a corollary, but this only works if $m$ has a primitive root. https://wangdo.tistory.com/85
Let m > 2 be an integer with a primitive root r. Then, r^(ϕ(m)/2) ≡ −1 (mod m)
Let $m>2$ be an integer with a primitive root $r$. Then, $r^{\phi(m)/2}\equiv-1 \ (\text{mod} \ m)$.Proof.$\text{gcd}(-1,m)=1$. Therefore, there exists an integer $02$. Therefore, $2i=\phi(m)$. * Here we can obtain the corollary that $\phi(m)$ is an even n
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First way:
For an $m>2$, $\text{gcd}(m-1,m)=1$. So, $(m-1)^{\phi(m)}\equiv(-1)^{\phi(m)}\equiv1\ (\text{mod}\ m)$.
If $\phi(m)$ is odd, it becomes $-1\equiv1 (\text{mod}\ m)$, which is false because $m>2$.
So, $\phi(m)$ is even.
Second way: this proof is the most common, and easy-to-think-of way.
I will leave the proof made by google's AI.
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