Möbius function of consecutive numbers

This question arose from a problem in Niven & Zuckerman's book "Introduction to the Theory of Numbers". In the chapter that the authors introduce the Möbius function, the first exercise is the following:

Find a positive integer $n$ such that $\mu(n)+\mu(n+1)+\mu(n+2)=3$, i.e, $\mu(n)=\mu(n+1)=\mu(n+2)=1$. A brute-force approach reveals the solution $n=33$.

My question is simply: Are there infinitely many $n$ in the previous conditions?

I really don't know how to approach this problem. I tried various things (factorials, Chinese remainder theorem, etc.) and i didn't come up with nothing. Also, for the first thousand numbers, there are the solutions $n=33,85,93,141,201,213,217,301,393,445,633,697,869,921$. One can also think in the variation of the problem with $\mu(n)=\mu(n+1)=\mu(n+2)=-1$.

Thanks in advance!

This is not an answer, I do some prediction from the probability. Interestingly, the predicted results and the actual results are very similar.

Since $\mu(n)=0$ if $n$ is not squarefree, we should exclude this possibility, $$n\not\equiv 0,-1,-2 \pmod {p^2}, \forall \in \mathbb P.$$ $$\prod_{p\in\mathbb P} (1-\frac{3}{p^2})\approx 0.125802 \approx \frac{1}8.$$

Hence we have about $\dfrac{1}8$ possibilities for $n$ such that $\mu(n),\mu(n+1),\mu(n+2)$ are not equal to $0$.

We assume that if $\mu(m)\neq 0$ then the probability of a coin thrown sides are equal, the probability of $\mu(n)=\mu(n+1)=\mu(n+2)=1$ is equal to $$0.125802 \times \frac{1}2\times \frac{1}2 \times \frac{1}2 \approx \frac{1}{64}.$$

Hence there are about $\dfrac{N}{64}$ such integers $n\leq N.$ A063838 says $a_{1000}=67105$, this is very close to our result.

Again, this isn't really an answer, but perhaps is an interesting interpretation of the problem that warrants some further discussion (Its utility is not known to me): Since the sum of the primitive $n$th roots of unity is $\mu(n)$, this is equivalent to finding $n$ such that the coefficients of the $x^{\varphi(n)-1}, x^{\varphi(n+1)-1}$, and $x^{\varphi(n+2)-1}$ terms of the cyclotomic polynomials $\Phi_n(x), \Phi_{n+1}(x)$, and $\Phi_{n+2}(x)$ respectively are all equal to $1$.