Exercise 4.3.15* ( $\mu(n+1) = \mu(n+2) = \mu(n+3) = \cdots = \mu(n+k).$)

Question

Given any positive integer $k$, prove that there exist infinitely many integers $n$ such that
$$\mu(n+1) = \mu(n+2) = \mu(n+3) = \cdots = \mu(n+k).$$

Richard Ganaye · Accepted Answer

\begin{proof} 
By Problem 2.3.18, for any positive integer $k$, there are $k$ consecutive integers which are not square-free, say $n+1, n+2,\ldots, n +k$, where 
\begin{align*}
&n \equiv -1\pmod {{p_1}^2},\
&n \equiv -2 \pmod {{p_2}^2},\
&\ldots\
&n \equiv -k \pmod {{p_k}^2}.\
\end{align*}
and $p_n$ is the $n$th prime number.

Then
$$\mu(n+1) = \mu(n+2) = \mu(n+3) = \cdots = \mu(n+k) = 0.$$
Moreover, if $m = n + \lambda p_1^2p_2^2\cdots p_k^2 \equiv n \pmod {p_i^2},\ i = 1,\ldots,k$, $\lambda \geq 0$, then 
$$\mu(m+1) = \mu(m+2) = \mu(m+3) = \cdots = \mu(m+k) = 0.$$
There exist infinitely many integers $n$ such that
$$\mu(n+1) = \mu(n+2) = \mu(n+3) = \cdots = \mu(n+k).$$
\end{proof}

{\bf Example}: We obtain $n$ such that $\mu(n+1) = \mu(n+2) = \mu(n+3) = \cdots = \mu(n+100) = 0$ with Sagemath.
\begin{verbatim}
sage: k = 100

sage: residues = [-i for i in range(1, k + 1)]

sage: moduli   = [nth_prime(i)^2 for i in range(1, k + 1)]

sage: n = crt(residues, moduli); n

2028519938484645749139755369116060455678748407443319354737460070073924492819785
4040382614695687706408740206868628498272793718588896202223910231463399215573873
9012381727960514850058588890854654083644895991303964587404471780374260311538323
6857566704853860107939976936867759161606950484378691662341830612904337714320818
8508900267290633971736704641930711015662384249431378593012253349604837712376159
999392740948376596612422028902589512942936547

sage: print([moebius(n + i) for i in range(1, k + 1)])

[0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0,
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
 0, 0, 0, 0]
\end{verbatim}
Here $n \simeq 2\cdot 10^{439}$.

Exercise 4.3.15* ( $\mu(n+1) = \mu(n+2) = \mu(n+3) = \cdots = \mu(n+k).$)

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Exercise 4.3.15* ( $\mu(n+1) = \mu(n+2) = \mu(n+3) = \cdots = \mu(n+k).$)

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