# Averages of euler

Example : \$sumlimits_dphi(d) = phi(1) + phi(2) + phi(4) = 1 + 1 + 2 = 4\$

I was told this has a simple proof. Problem is, I can not think of a way to lớn show this in a very simple & straightforward way.

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Let \$G=langle x angle\$ be the cyclic group of order \$n\$ generated by \$x\$. Then,eginalignn &= |G| \&= sum_1leq dleq n |\textelements of order d| \&= sum_n|\textelements of order d|endalignby Lagrange"s Theorem. But \$G\$ has a quality cyclic subgroup of order \$d\$ for each \$d|n\$, namely \$langle x^n/d angle\$. Moreover, each such subgroup has \$varphi(d)\$ generators, so\$\$n = sum_dvarphi(d)\$\$

Hint: You can show that \$sumlimits_nphi(d) = sumlimits_dphi(fracnd) \$ but \$phi(fracnd)\$ is all numbers with GCD of \$d\$ with \$n\$. Therefore the last sum counts all numbers till \$n\$ & is equal lớn \$n\$.

The simplest proof I know is this:

Consider all proper fractions of the size \$a/n\$. There are \$n\$ of those. When you consider their reduced forms you get fractions of the khung \$b/d\$ with \$d|n\$ and \$(b,d)=1\$. By definition, there are \$phi(d)\$ of those. The result follows.

Here"s one approach:

Consider the polynomial \$x^n-1\$ as a polynomial over the complex numbers \$au-79.netbbC\$. You can easily see, almost tautologically, that

where \$Phi_d(x)\$ is the polynomial whose roots are the primitive \$d^ extth\$ roots of unity (i.e. solutions to \$x^d-1\$ that bởi vì not satisfy \$x^c-1\$ for \$0

27

0

3

3

-3

12

8

2

5

0

1

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