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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 $$\mu(n) = \mu(n+1) = -1?$$

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Dylan
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    Related: http://math.stackexchange.com/questions/564581/m%C3%B6bius-function-of-consecutive-numbers?rq=1 – Winther Nov 15 '15 at 18:51
  • Surely this happens because square free numbers have asymptotic density.I don't have a proof right now but I can imagine it is possible to find one. – Konstantinos Gaitanas Nov 15 '15 at 18:55
  • @KonstantinosGaitanas The fact that some set has density does not necessarily mean that there are consecutive numbers from that set. (Consider even numbers.) However, we know that for the square-free numbers density is $6/pi^2$. – Martin Sleziak Nov 15 '15 at 18:59
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    This MO post seems related, too: Are there infinitely many triples of consecutive square-free integers? – Martin Sleziak Nov 15 '15 at 19:04
  • It is almost certain that the primorials $p$#$\pm 1$ are squarefree. If this is actually true, there are infinite many triples $(n,n+1,n+2)$, such that $n,n+1,n+2$ are all squarefree. It would be even enough to show that sufficiently large primorials are squarefree. – Peter Nov 18 '15 at 18:31
  • Numbers $n$ such that $\mu(n)=\mu(n+1)=\mu(n+2)=1$ are tabulated at https://oeis.org/A063838. If $p=2r+1$ and $q=3r+1$ are both prime, then $2q$ and $3p$ are consecutive numbers with $\mu=1$, and by standard conjectures there are infinitely many such $r$. – Gerry Myerson May 20 '16 at 09:53

2 Answers2

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

Gerry Myerson
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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

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