That is to say $\forall k \in \mathbb{N}$, there exist infinitely many $n\in \mathbb{N}$ s.t. $\mu \left ( n+1 \right )=\mu \left ( n+2 \right )=\cdots=\mu \left ( n+k \right )$, where $\mu$ is the Möbius function.
I saw this problem in a number theory textbook and it appears in the exercises for the section talking about the Möbius inversion formula. I cannot come up with an effective idea to tackle it.
This is not a homework problem.
Many thanks for your help!
Let $p_1,p_2,\cdots,p_k$ be any distinct prime numbers.
Then by the Chinese Remainder Theorem, the system of congruences $$\begin{align} n + 1 & \equiv 0 \mod p_1^2 \\ n + 2 & \equiv 0 \mod p_2^2 \\ n + 3 & \equiv 0 \mod p_3^2 \\ \vdots \\ n + k & \equiv 0 \mod p_k^2 \end{align}$$
has infinitely many solutions $n$. For any such solution $n$, we have that $n + m$ is divisible by $p_m^2$ for $m=1,2,3,\cdots,k$, and so we have that $\mu(n+m)=0$ for $m=1,2,3\cdots,k$
We see that there are infinitely many natural numbers $n$ such that $$ \mu(n+1)=\mu(n+2)=\mu(n+3)=\cdots=\mu(n+k)=0$$
edit
In fact, if $k \geq 4$ and for some $n$ we have that $$ \mu(n+1)=\mu(n+2)=\mu(n+3)=\cdots=\mu(n+k)$$ then it must in fact be the case that $$ \mu(n+1)=\mu(n+2)=\mu(n+3)=\cdots=\mu(n+k)=0$$
This is because amongst the set of numbers $$\lbrace n+1, n+2, n+3, n+4\rbrace$$ there will be a multiple of $4$. The Möbius function evaluated at this multiple of $4$ will be $0$.
