为何下述莫比乌斯函数求和等式成立？是否与欧拉函数相关？

Hey there! Let's break down this identity step by step, and then explore how it ties into Euler's totient function.

Step 1: Proving the Identity #

First, let's recall key properties that will simplify our work:

• The Möbius function $\mu(n)$ equals 0 if $n$ has any squared prime factors. This means we can restrict our sum to square-free $h$ and $k$—any non-square-free terms will vanish entirely.
• Both the Möbius function and gcd are multiplicative functions. This lets us analyze the contribution of each prime $p \leq X$ independently, then multiply the results (this is the classic Euler product trick).

Analyzing a Single Prime $p \leq X$ #

For a fixed prime $p \leq X$, we consider all valid square-free combinations of $p$ in $h$ and $k$ (exponents are either 0 or 1):

1. Neither $h$ nor $k$ contains $p$: $\mu(h)$ and $\mu(k)$ both include $\mu(1)=1$, $\gcd(h,k)$ has no factor of $p$, so the contribution is $\frac{1 \cdot 1 \cdot 1}{1 \cdot 1} = 1$.
2. Only $h$ contains $p$: $\mu(h)$ includes $\mu(p)=-1$, $\mu(k)=1$, $\gcd(h,k)$ has no factor of $p$, contribution is $\frac{(-1) \cdot 1 \cdot 1}{p \cdot 1} = -\frac{1}{p}$.
3. Only $k$ contains $p$: Symmetric to case 2, contribution is $\frac{1 \cdot (-1) \cdot 1}{1 \cdot p} = -\frac{1}{p}$.
4. Both $h$ and $k$ contain $p$: $\mu(h)=\mu(k)=-1$, $\gcd(h,k)$ has a factor of $p$, contribution is $\frac{(-1) \cdot (-1) \cdot p}{p \cdot p} = \frac{1}{p}$.

Adding these contributions for a single prime $p$ gives:
$$1 - \frac{1}{p} - \frac{1}{p} + \frac{1}{p} = 1 - \frac{1}{p}$$

Combining All Primes #

For primes $p > X$, $h$ and $k$ can't include $p$ (per the problem's condition), so their only contribution is 1 (case 1 above). Multiplying the contributions of all primes $p \leq X$ gives us:
$$\sum_{\substack{h,k \ p|hk \implies p \leq X}} \frac{\mu(h)\mu(k)}{h \cdot k} \gcd(h,k) = \prod_{p \leq X} \left(1 - \frac{1}{p}\right)$$
This confirms the identity we needed to prove!

Step 2: Connection to Euler's Totient Function #

Absolutely, this identity is directly linked to Euler's totient function $\phi(n)$, which counts the number of integers from 1 to $n$ that are coprime to $n$.

Key Link via Euler Product #

The Euler product formula for $\phi(n)$ states that for any positive integer $n$:
$$\phi(n) = n \prod_{p|n} \left(1 - \frac{1}{p}\right)$$
If we take $n$ to be the primorial of $X$ (denoted $X#$)—the product of all primes $\leq X$:
$$X# = \prod_{p \leq X} p$$
Substituting into the totient formula gives:
$$\phi(X#) = X# \prod_{p \leq X} \left(1 - \frac{1}{p}\right)$$
Rearranging this, we get:
$$\prod_{p \leq X} \left(1 - \frac{1}{p}\right) = \frac{\phi(X#)}{X#}$$

What This Means #

The right-hand side of our original identity is the proportion of integers from 1 to $X#$ that are coprime to $X#$—a quantity directly defined by Euler's totient function. So our combinatorial sum (involving Möbius functions and gcds) simplifies to this proportion, creating a concrete bridge between the sum's structure and the totient function's number-theoretic meaning.

内容的提问来源于stack exchange，提问作者user257