Are there infinitely many natural numbers $n$ such that $\mu(n)=\mu(n+1)=\pm 1$?

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 $\mu(3p)=\mu(2q)=\mu(5r)=1$.

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