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 Post subject: Euler $\phi$ functionPosted: Tue Jan 19, 2016 4:33 pm

Joined: Mon Nov 09, 2015 4:32 pm
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Prove that

$\forall n\geq1 : \displaystyle\mathop{\sum}\limits_{d|n}\phi(d)=n$ where: $\phi(n) = \rvert\bigl\{k\in \mathbb{N} \rvert 1 \leq k \leq n \wedge (k,n)=1 \bigl\}\rvert$

is the Euler phi function.

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 Post subject: Re: Euler $\phi$ functionPosted: Tue Jan 19, 2016 4:34 pm
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Joined: Mon Nov 09, 2015 1:36 am
Posts: 440
Location: Ioannina, Greece
tziaxri wrote:
Prove that

$\forall n\geq1 : \displaystyle\mathop{\sum}\limits_{d|n}\phi(d)=n$ where: $\phi(n) = \rvert\bigl\{k\in \mathbb{N} \rvert 1 \leq k \leq n \wedge (k,n)=1 \bigl\}\rvert$

is the Euler phi function.

We give the classical proof which can be found in almost every number theory book. For example: Hardy and Wright, Introduction to the Theory of Numbers.

For $n=1$ is trivial. If $n>1$ and $n=\mathop{\prod}\limits_{i=1}^{k}p_i^{r_i}$ is the unique-prime-factorization of $n$, then the divisors of $n$ are the numbers $d=\mathop{\prod}\limits_{i=1}^{k}p_i^{s_i}\,, \; s_i\leqslant r_i\,,$ for every prime $p_i$. The function $F(n):=\displaystyle\mathop{\sum}\limits_{d|n}\phi(d)\,,\quad n\in {\mathbb{N}}\,,$ is multiplicative. For every $i=1,2,\ldots, k\,,$ we have
\begin{align*}
F(p_i^{r_i})&=\displaystyle\mathop{\sum}\limits_{j=0}^{r_i}\phi(p_i^{j})\\
&=\phi(1)+\mathop{\sum}\limits_{j=1}^{r_i}\phi(p_i^{j})\\
&=1+\mathop{\sum}\limits_{j=1}^{r_i}(p_i^{j}-p_i^{j-1})\\
&=p_i^{r_i}\,.
\end{align*} And from the multiplicity of $F$ we have $F(n)=\displaystyle\mathop{\prod}\limits_{i=1}^{k}F(p_i^{r_i})=\mathop{\prod}\limits_{i=1}^{k}p_i^{r_i}=n\,.$

_________________
Grigorios Kostakos

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