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 Post subject: On Euler's $\phi$ functionPosted: Mon Jan 18, 2016 7:44 pm

Joined: Sat Nov 07, 2015 6:12 pm
Posts: 828
Location: Larisa
Let $\phi$ denote Euler's phi function. If

$$\phi(n)=\phi(2n)$$

prove that $n$ is odd.

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 Post subject: Re: On Euler's $\phi$ functionPosted: Tue Apr 12, 2016 9:47 pm

Joined: Sat Nov 14, 2015 6:32 am
Posts: 137
Location: Melbourne, Australia
If $\gcd(2,n)=1$ (that is $n$ is odd) then:

$$\phi(2n)=\phi(2)\phi(n)=\phi(n)$$

holds.

Now, suppose that $n$ is even, that is $n=2^r m$ and $\gcd(2,m)=1$. Then:

$$\phi(n)=\phi\left(2^r \right) \phi(m)=2^r \left (1 - \frac{1}{2} \right) \phi(m)=2^{r-1} \phi(m)$$

while

$$\phi(2n)=\phi \left (2^{r+1} m \right)= \phi \left(2^{r+1} \right) \phi(m)=2^r \phi(m)$$

Thus $\phi(n) \neq \phi(2n)$ if $n$ is even. As a conclusion we have:

$$\phi(n)=\phi(2n) \Leftrightarrow n \; \text{is odd}$$

Note: A similar identity if $n$ is even holds:

$$2\phi(n)=\phi(2n) \Leftrightarrow n \; \text{is even}$$

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