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Given a positive integer $\displaystyle n = \prod_{i=1}^s p_i^{\alpha_i}$, we write $\Omega(n)$ for the total number $\displaystyle \sum_{i=1}^s \alpha_i$ of prime factors of $n$, counted with multiplicity. Let $\lambda(n) = (-1)^{\Omega(n)}$ (so, for example, $\lambda(12)=\lambda(2^2\cdot3^1)=(-1)^{2+1}=-1$). Prove the following

There are infinitely many positive integers $n$ such that $\lambda(n) = \lambda(n+1)=\lambda(n+2) = +1$

If this problem is following: (I can do it) There are infinitely many positive integers $n$ such that $\lambda(n) = \lambda(n+1) = +1$

it is clear $\lambda(mn)=\lambda(m)\lambda(n)\Longrightarrow \lambda(n^2)=1$, and if the postive integer $n$ such $\lambda(n) = \lambda(n+1) = +1$$(n=9)$,then we have $$\lambda((2n+1)^2-1)=\lambda(4n(n+1))=1=\lambda((2n+1)^2)=1$$

But I can't prove that there exist infinitely many Three Consecutive positive integers $n$ such $\lambda(n)=\lambda(n+1)=\lambda(n+2)=1$.Thanks

Vepir
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1 Answers1

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TL;DR

There is similar question on MathOverflow that I'm going to reference:

Your question was answered by Hildebrand (On consecutive values of the Liouville function, Enseign. Math. (2) 32 (1986), 219–226) when they proved that all 8 combinations $\pm 1,\pm 1,\pm 1$ occur infinitely often in the Liouville sequence. This is the $k=3$ case of the general problem.

Your question is one of those 8 combinations, namely, just $+1,+1,+1$.

If you want to learn more about the progress on consecutive values of the Liouville function $\lambda(n)$, you can watch Terence Tao's lecture at the Building Bridges II. conference from 2018. (Here is the direct video link.) $-$ The mentioned reference (Hildebrand, 1986.) for the solution of your problem is given at 18:20 mark of the linked video.


The Proof of $(+1,+1,+1)$

I have tracked down the result (Hildebrand, 1986.) which you should be able to read here: (On consecutive values of the Liouville function, Enseign. Math. (2) 32 (1986), 219–226).

I will copy the proof for the $\lambda(n)=\lambda(n+1)=\lambda(n+2)=1$ case, exactly as it is given in the linked paper, down below. (Via the magic of OCR.)

We first need to prove the following lemma.

LEMMA. Each of the equations $$ \lambda(15 n-1)=\lambda(15 n+1)=1 $$ and $$ \lambda(15 n-1)=\lambda(15 n+1)=-1 $$ holds for infinitely many positive integers $n$.

Proof. Given a positive integer $n_{0} \geqslant 2,$ define $n_{i}, i \geqslant 1,$ inductively by $$ n_{i+1}=n_{i}\left(4 n_{i}^{2}-3\right) \quad(i \geqslant 0) $$ It is easily checked that $$ n_{i+1} \pm 1=\left(n_{i} \pm 1\right)\left(2 n_{i}+1\right)^{2} \quad(i \geqslant 0) $$ so that $$ \lambda\left(n_{i+1} \pm 1\right)=\lambda\left(n_{i} \pm 1\right)-\ldots-\lambda\left(n_{0} \pm 1\right) \quad(i \geqslant 0) $$ Also, it follows by induction that $n_{0} | n_{i}$ for all $i \geqslant 0 .$

Therefore, taking in turn $n_{0}=15$ and $n_{0}=30$ and noting that $$ \lambda(14)=\lambda(16)=1, \quad \lambda(29)=\lambda(31)=-1 $$ we obtain two infinite sequences $\left(n_{i}^{(+)}\right)$ and $\left(n_{i}^{(-)}\right)$ with the required properties $$ n_{i}^{(\pm)} \equiv 0(\bmod 15), \quad \lambda\left(n_{i}^{(+)} \pm 1\right)=1, \quad \lambda\left(n_{i}^{(-)} \pm 1\right)=-1 $$ $$ \square $$

Now we are ready to prove the claim.

We shall show here that (2) $$ \lambda(n)=\lambda(n+1)=\lambda(n-1)=1 $$

has infinitely many solutions.

Call an integer $n \geqslant 2$ "good", if (2) holds for this $n$. We have to show that there are infinitely many good integers. To this end we shall show that for any positive integer $n$ satisfying (3) $$ n \equiv 0(\bmod 15), \quad \lambda(n+1)=\lambda(n-1)=1 $$ the interval (4) $$ I_{n}=\left[\frac{4 n}{5}, 4 n+5\right] $$ contains a good integer. Since by the LEMMA, (3) holds for infinitely many positive integers $n$, the desired result follows.

To prove our assertion we fix a positive integer $n$, for which (3) holds. We may suppose $\lambda(n)=-1,$ since otherwise $n \in I_{n}$ is good, and we are done. Put $N=4 n,$ and note that, by construction, $N$ is divisible by $3$, $4$ and $5$.

From (3) we get, using the multiplicativity of the function $\lambda$, $$ \lambda(N \pm 4)=\lambda(4(n \pm 1))=\lambda(4) \lambda(n \pm 1)=1 $$ and our assumption $\lambda(n)=-1$ implies $$ \lambda(N)=\lambda(4 n)=\lambda(4) \lambda(n)=-1 $$ If now $$ \lambda(N+5)=\lambda(N-5)=-1 $$ then $$ \lambda\left(\frac{N}{5} \pm 1\right)=\frac{\lambda(N \pm 5)}{\lambda(5)}=1=-\lambda(N)=\lambda\left(\frac{N}{5}\right) $$ and $N / 5=4 n / 5 \in I_{n}$ is good.

We may therefore suppose that at least one of values $\lambda(N+5)$ and $\lambda(\mathrm{N}-5)$ equals $1$.

For definiteness we shall assume $\lambda(N+5)=1 ;$ the other case is treated in exactly the same way.

If $\lambda(N+3)=1$ or $\lambda(N+6)=1,$ then $N+4 \in I_{n}$ or $N+5 \in I_{n}$ is good.

But in the remaining case $$ \lambda(N+3)=\lambda(N+6)=-1 $$ we have $$ \lambda\left(\frac{N}{3}\right)=\lambda\left(\frac{N}{3}+1\right)=\lambda\left(\frac{N}{3}+2\right)=1 $$ so that $(N+3) / 3 \in I_{n}$ is good.

Thus (3) implies the existence of a good integer in the interval as we had to show. $$ \blacksquare $$

Vepir
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