- Thread starter
- #1

- Mar 22, 2013

- 573

Now, being an analytic and algebraic number theorist, I cannot help state another proof at this point.

Let $C_k$ be the cyclic group of order $k$. All number that is coprime to $k$ are then order of the generators of the cyclic group. As for any $n$, $C_n$ is the union of the cyclic groups $C_k$ for some $k$ that divides $n$ (to prove this, consider each element of $C_n$ as generators of some cyclic group). Using the fact that $\varphi(k)$ is the number of generators of $C_k$, it is now easy to arrive at

$$\sum_{k | n} \varphi(k) = n$$

Now calculations afterwards is the tricky part. We need something far more strong and powerful than just elementary number theory. What we actually need is

$$g(n) = \sum_{d|n} f(d)$$

for $n \in \mathbb{N}$ then

$$f(n) = \sum_{d|n} \mu(d) g\left(\frac{n}{d}\right)$$

where $\mu$ is the Mobius function. Well, if we apply it to the Euler's formula,

$$\varphi(n) = n \sum_{d|n} \frac{\mu(d)}{d}$$

Now the left side looks familiar. This is essentially dual to the Mobius-dirchlet series for $\zeta$! Using this fact, one could actually derive an Euler-like product for $\zeta$, applied to $\varphi$ instead, which is essentially the same mentioned above

Let $C_k$ be the cyclic group of order $k$. All number that is coprime to $k$ are then order of the generators of the cyclic group. As for any $n$, $C_n$ is the union of the cyclic groups $C_k$ for some $k$ that divides $n$ (to prove this, consider each element of $C_n$ as generators of some cyclic group). Using the fact that $\varphi(k)$ is the number of generators of $C_k$, it is now easy to arrive at

*Euler's formula*$$\sum_{k | n} \varphi(k) = n$$

Now calculations afterwards is the tricky part. We need something far more strong and powerful than just elementary number theory. What we actually need is

*Mobius inversion theorem*. I think a proof would be off topic at this point, so I just jump in to the argument and results instead. The theorem states that if$$g(n) = \sum_{d|n} f(d)$$

for $n \in \mathbb{N}$ then

$$f(n) = \sum_{d|n} \mu(d) g\left(\frac{n}{d}\right)$$

where $\mu$ is the Mobius function. Well, if we apply it to the Euler's formula,

$$\varphi(n) = n \sum_{d|n} \frac{\mu(d)}{d}$$

Now the left side looks familiar. This is essentially dual to the Mobius-dirchlet series for $\zeta$! Using this fact, one could actually derive an Euler-like product for $\zeta$, applied to $\varphi$ instead, which is essentially the same mentioned above

Last edited: