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

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

## 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. $$\blacksquare$$