Values of the Liouville function

Let $\Omega: \mathbb{N} \to \mathbb{N}\cup\{0\}$ be the function which counts how many prime factors a number has, with multiplicity. For example, $\Omega(380) = 4$, $\Omega(108)= 5$. More generally, for $p$ prime, $\Omega(p) = 1, \Omega(p^k) = k$ and it is clear that $\Omega(mn) = \Omega(m) + \Omega(n)$. Furthermore, $\Omega(n) = 0 \iff n =1$.

The Liouville lambda function is defined to be $\lambda:\mathbb{N} \to \{\pm1\}$,

$$\lambda(n) = (-1)^{\Omega(n)}$$

It is apparent that $\lambda(mn) = \lambda(m) \lambda(n)$.

Consider the sequence given by $\lambda(n)$. Do there exist arbitrarily long stretches in this sequence of either $+1$ or $-1$?

That is to say, $\forall N \in \mathbb{N}, \exists n: \lambda(n)=\lambda(n+1)=\cdots=\lambda(n+N)$?

This seems like quite a natural question to ask about this function, but I was unable to find an answer after searching. I verified this for $N\le4$ by hand, and I would presume that this is true. But this seems like a difficult conjecture to prove, and I know very little number theory in the first place.

It would follow from the logarithmically averaged Chowla conjecture, the conjecture being that for any $0\le a_1<\ldots < a_K $ as $x\to \infty$ $$\sum_{n\le x} \frac{\prod_{k=1}^K \lambda(n+a_k)}{n}= o(\log x)$$ Unforntunately it seems this conjecture has been proven only for $K=2$ and $K$ odd.

Once proven for all $K$ we would have $$\sum_{n\le x} \frac{ \prod_{m=0}^M (1+\lambda(n+m))}{n}= \sum_{n\le x} \frac{1}n+o(\log x)$$