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 mobius 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 aproach 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 aproach 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 $\mu(n)=\mu(n+1)=\mu(n+2)=-1$ or simply $\mu(n)=\mu(n+1)=1$.

Thanks in advance!