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Prove that:$$\sum_{d\mid n} \varphi(d)=n$$

Where $\varphi(n)$ denotes the number of positive integers $m$ less than or equal to $n$ such that $\gcd(m,n)=1$

I am lost here, any help would be appreciated.

Martin Sleziak
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Guy
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2 Answers2

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Here is my proof, based on counting arguments.

Consider the fractions $$\frac1{n},\frac{2}{n},\frac{3}{n},\cdots,\frac{n}{n}$$

Obviously there are $n$ such fractions.

Now consider these fractions, simplified to their lowest terms.

In each of these fractions, the denominator has to be a divisor of $n$. The number of fractions, in which the denominator is still equal to $n$, is the number of fractions whose numerator was originally relatively prime to $n$, i.e, $\varphi(n)$.

Similarly, for any given $d$, where $d$ is a divisor of $n$, there will be $\varphi(d)$ such fractions where the denominator is equal to $d$. Adding all these $\varphi(d)$ thus returns the total number of fractions, $n$.

So we arrive at the equality,

$$\sum_{d\mid n}\varphi(d)=n$$

Guy
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  • The last parraph's argument isn't clear to me: there could possibly be fractions for which $;\frac kd=\frac{k'}{d'};$ , for two different divisors $;d,d';$ of $;n;$ , and thus apparently your method would count these ones twice... – DonAntonio Mar 17 '14 at 13:16
  • @DonAntonio lowest terms – Guy Mar 17 '14 at 13:19
  • What sticks in my craw about this argument is that fractions are really inessential to the problem; they get brought in only because they come with a built-in, widely recognized "simplified to lowest terms" operation. IOW, fractions show up in this theorem only as, basically, a cute hack, and IMO they obscure what's really happening in the theorem, namely that (1) every $m \in \def\nset{{1,\dots,n}}\nset$ can be written uniquely as the product $du$ where $d\mid n$ and $\gcd(u, n/d) = 1$, and (2) every such product $du$ belongs to $\nset$. $;;\dots$ – kjo Mar 17 '14 at 13:50
  • @kjo i never said my solution was elegant it just worked for me. – Guy Mar 17 '14 at 13:52
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    $\dots;;$ These means that $\nset$ can be partitioned according to the $d$ factor of these products, and the size of each partition is $\varphi(n/d)$. But don't get me wrong: such proofs are OK, as quick ways to convince oneself that something is true, but one needs to go beyond them to get at the essence of why something is true. – kjo Mar 17 '14 at 13:52
  • @Sabyasachi: sorry, I had not finished with my comment. Also, on the contrary, I think your proof is extremely elegant. I'm making the case for inelegance here. – kjo Mar 17 '14 at 13:53
  • @kjo really? thank you then. – Guy Mar 17 '14 at 13:57
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    FWIW I think this proof is very elegant. – Caleb Stanford Mar 17 '14 at 14:08
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Consider the cyclic group $C_n$. Then, for every $g\in C_n$, $o(g)$ divides $|C_n|=n$. Moreover, for any $d|C_n$, $\exists g\in C_n:o(g)=d$. Thereofore, if $A_k$ is the set of all the elements of $C_n$ with order $k$, $A_k \neq \emptyset \Longleftrightarrow k | n$. Therefore, $\{A_k : k | n\}$ is a partition of $C_n$. So:

$$|C_n|= n = \displaystyle{\sum _{g \in C_n}} 1 = \displaystyle{\sum _{d|n} |A_d|} = \displaystyle{\sum _{d|n} \varphi(d)}$$

Since the number of elements of order $d$ in $C_n$ is $\varphi(d)$.

Imanol Pérez Arribas
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