Solution

The Möbius function is defined by

μ (1) = 1

and, for

n > 1

μ (n) = {0, (- 1)^{r}, p^{2} ∣ n for some prime p, n is a product of r distinct primes .

μ (mn) = μ (m) μ (n)

whenever

m

and

n

The summand

μ (d)^{2} / ϕ (d)

is multiplicative, so its divisor sum is multiplicative. For a prime power

p^{a}

with

a \geq 1

, only

d = 1, p

contribute and

\sum_{d ∣ p^{a}} \frac{μ ( d ) ^{2}}{ϕ ( d )} = 1 + \frac{1}{p - 1} = \frac{p}{p - 1} = \frac{p ^{a}}{ϕ ( p ^{a} )} .

Multiplying these local identities over the prime divisors of

n

gives

d ∣ n \sum \frac{μ ( d ) ^{2}}{ϕ ( d )} = \frac{n}{ϕ ( n )} .

For the final claim, choose distinct primes

p_{0}, \dots, p_{k}

. The moduli

p_{j}^{2}

are pairwise coprime, so the Chinese remainder theorem gives an integer

a

satisfying

a \equiv - j (mod p_{j}^{2}), 0 \leq j \leq k .

Every positive integer

n \equiv a (mod \prod_{j} p_{j}^{2})

then has

p_{j}^{2} ∣ n + j

. Hence

μ (n) = μ (n + 1) = \dots = μ (n + k) = 0 .

There are infinitely many positive representatives of this congruence class.

Solved by gpt-5.6-sol high.

 Ancestors (10)