A while ago, I answered this question here on StackExchange which asks if for any given integer $k$, whether there exists infinitely many natural numbers $n$ such that
$$ \mu(n+1)=\mu(n+2)=\mu(n+3)=\cdots=\mu(n+k) $$
This is indeed the case, and can be shown using the Chinese Remainder Theorem, as in my answer to the linked question. We find infinitely many $n$ such that the common value of $\mu$ evaluated at these points is $0$.
If $k \geq 4$, then one of the values $n+1, n+2, n+3, n+4$ is a multiple of $4$, and hence the common value must, in fact, actually be $0$. For $n<4$, this argument doesn't work, and we can't rule out the possibility that
$$ \mu(n+1)=\cdots=\mu(n+k)=\pm 1 $$
My question concerns the case of $k=2$. Are there infinitely many natural numbers $n$ such that either
$$\mu(n) = \mu(n+1) = 1$$
or
$$\mu(n) = \mu(n+1) = -1?$$
Best Answer
It is a standard conjecture in Number Theory that there exist infinitely many $s$ such that $p=10s+1$, $q=15s+2$, and $r=6s+1$ are all prime. Then $3p=30s+3$, $2q=30s+4$, and $5r=30s+5$ are consecutive integers, each a product of two primes, so $\mu(3p)=\mu(2q)=\mu(5r)=1$.