Are there infinitely many natural numbers $n$ such that $\mu(n)=\mu(n+1)=\pm 1$? 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
$$\mu(n) = \mu(n+1) = -1?$$
 A: 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 $\mu(3p)=\mu(2q)=\mu(5r)=1$. 
A: This is not an answer, just a different perspective to the problem. 
Let's look at the following Diophantine equation: 
$$3\cdot x=2\cdot y + 1$$
It has one solution $(1,1)$ and as a result infinitely many $x=1+2\cdot t$, $y=1+3\cdot t$, $t \in \mathbb{N}$. 
Both these arithmetic progressions will generate infinitely many primes (https://en.wikipedia.org/wiki/Dirichlet%27s_theorem_on_arithmetic_progressions). Question is, will they both generate infinitely pairs of primes for the same $t$?
And, Green-Tao extension (https://en.wikipedia.org/wiki/Green%E2%80%93Tao_theorem#Extensions_and_generalizations) says that $k+2\cdot t$, $k+3\cdot t$ will be simultaneously primes for infinitely many $k$ and $t$, but we need $k=1$.
Obviously, any prime pair $(p_1, p_2)$ (e.g. $(5, 7)$), solution of this equation, satisfies:
$$1=\mu(2\cdot p_2)=\mu(3\cdot p_1)=\mu(2\cdot p_2+1)$$ 
Some related materials:
http://arxiv.org/pdf/1310.8140.pdf - Primes Solutions Of Linear Diophantine Equations
http://arxiv.org/pdf/math/0404188v6.pdf - The Primes Contain Arbitrarily Long Arithmetic Progressions
