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I am trying to solve a very interesting problem about the ring $\mathbb{Z}/n\mathbb{Z}$ and Euler function $\phi (n)$, but i am not sure how to start, i have a few ideas, but none of them leads me to the end of the proof. So, here is the problem.

Let $n$ be a squarefree integer( integer is one divisible by no perfect square, except 1 ). Let $k\in \mathbb{Z}/n\mathbb{Z}$ and $e=1+j\phi (n)$, where $\phi (n)$ is the Euler function and $j\in \mathbb{N}$. Show that $k^{e}=k$.

My first thought, when i saw what i have to prove, was that i have to show the idempotence of the element $k$. I tried to show it, but i couldn't... Then i recalled that $\phi (n)$ is the order of the unit group of $\mathbb{Z}/n\mathbb{Z}$, but i don't know how should i use it here...or i also know that $\phi(n)$ is always positive and even number, so $j\phi(n)$ must be also even, so the number $e$ is odd...

Can anybody help me with this problem? I have the feeling the things must be easy. I would be glad to read your hints, ideas or remarks. Thank you in advance!

Lullaby
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    Strongly related: http://math.stackexchange.com/questions/518845/to-what-divisors-a-of-n-can-eulers-theorem-multiplied-by-a-be-generalized – Daniel Fischer Oct 11 '13 at 22:07

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Let $n$ be the product $p_1p_2\cdots p_m$ of distinct primes. Let $p_i$ be one of these primes.

If $p_i$ does not divide $k$, then $k^{1+j\varphi(n)}\equiv k\pmod{p_i}$. This follows from Fermat's Theorem, since $p_i-1$ divides $\varphi(n)$.

If on the other hand $p_i$ divides $k$, then trivially $k^{1+j\varphi(n)}\equiv k\pmod{p_i}$.

It follows that $k^{1+j\varphi(n)}\equiv k\pmod{n}$.

André Nicolas
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