Subfield: Algebra


Question: Given a positive integer $n=\prod_{i=1}^{s} p_{i}^{\alpha_{i}}$, we write $\Omega(n)$ for the total number $\sum_{i=1}^{s} \alpha_{i}$ of prime factors of $n$, counted with multiplicity. Let $\lambda(n)=(-1)^{\Omega(n)}$ (so, for example, $\left.\lambda(12)=\lambda\left(2^{2} \cdot 3^{1}\right)=(-1)^{2+1}=-1\right)$.
Prove that there are infinitely many positive integers $n$ such that $\lambda(n)=\lambda(n+1)=-1$.

Solution:
Notice that we have $\Omega(m n)=\Omega(m)+\Omega(n)$ for all positive integers $m, n$ ( $\Omega$ is a completely additive arithmetic function), translating into $\lambda(m n)=\lambda(m) \cdot \lambda(n)$ (so $\lambda$ is a completely multiplicative arithmetic function), hence $\lambda(p)=-1$ for any prime $p$, and $\lambda\left(k^{2}\right)=\lambda(k)^{2}=+1$ for all positive integers $k$.



The start (first 100 terms) of the sequence $\mathfrak{S}=(\lambda(n))_{n \geq 1}$ is
+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1

The equation $3 x^{2}-2 y^{2}=1$ (again Pell theory) has also infinitely many solutions in positive integers, given by $\left(x_{n}, y_{n}\right)$, where $x_{n} \sqrt{3}+y_{n} \sqrt{2}=(\sqrt{3}+\sqrt{2})^{2 n+1}$. Since $\lambda\left(2 y^{2}\right)=$ -1 and $\lambda\left(2 y^{2}+1\right)=\lambda\left(3 x^{2}\right)=-1$, the thesis is proven.
Notice that we have $\Omega(m n)=\Omega(m)+\Omega(n)$ for all positive integers $m, n$ ( $\Omega$ is a completely additive arithmetic function), translating into $\lambda(m n)=\lambda(m) \cdot \lambda(n)$ (so $\lambda$ is a completely multiplicative arithmetic function), hence $\lambda(p)=-1$ for any prime $p$, and $\lambda\left(k^{2}\right)=\lambda(k)^{2}=+1$ for all positive integers $k$.



The start (first 100 terms) of the sequence $\mathfrak{S}=(\lambda(n))_{n \geq 1}$ is
+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1

Assume $(\lambda(n-1), \lambda(n))$ is the largest $(-1,-1)$ pair, therefore $\lambda(n+1)=1$ and $\lambda\left(n^{2}+n\right)=\lambda(n)$. $\lambda(n+1)=-1$, therefore again $\lambda\left(n^{2}+n+1\right)=1$. But then $\lambda\left(n^{3}-1\right)=\lambda(n-1) \cdot \lambda\left(n^{2}+n+1\right)=-1$, and also $\lambda\left(n^{3}\right)=$ $\lambda(n)^{3}=-1$, so we found yet a larger such pair than the one we started with, contradiction.
Notice that we have $\Omega(m n)=\Omega(m)+\Omega(n)$ for all positive integers $m, n$ ( $\Omega$ is a completely additive arithmetic function), translating into $\lambda(m n)=\lambda(m) \cdot \lambda(n)$ (so $\lambda$ is a completely multiplicative arithmetic function), hence $\lambda(p)=-1$ for any prime $p$, and $\lambda\left(k^{2}\right)=\lambda(k)^{2}=+1$ for all positive integers $k$.



The start (first 100 terms) of the sequence $\mathfrak{S}=(\lambda(n))_{n \geq 1}$ is
+1,-1,-1,+1,-1,+1,-1,-1,+1,+1,-1,-1,-1,+1,+1,+1,-1,-1,-1,-1+1,+1,-1,+1,+1,+1,-1,-1,-1,-1,-1,-1,+1,+1,+1,+1,-1,+1,+1,+1-1,-1,-1,-1,-1,+1,-1,-1,+1,-1,+l,-1,-1,+1,+1,+1,+1,+1,-1,+1-1,+1,-1,+1,+1,-1,-1,-1,+1,-1,-1,-1,-1,+1,-1,-1,+1,-1,-1,-1+1,+1,-1,+1,+1,+1,+1,+1,-1,+1,+1,-1,+1,+1,+1,+1,-1,-1,-1,+1

Assume the pairs of consecutive terms $(-1,-1)$ in $\mathfrak{S}$ are finitely many. Then from some rank on we only have subsequences $(1,-1,1,1, \ldots, 1,-1,1)$. By "doubling" such a subsequence (like at point ii)), we will produce



$$

(-1, ?, 1, ?,-1, ?,-1, ?, \ldots, ?,-1, ?, 1, ?,-1)

$$



According with our assumption, all ?-terms ought to be 1, hence the produced subsequence is



$$

(-1,1,1,1,-1,1,-1,1, \ldots, 1,-1,1,1,1,-1)

$$



and so the "separating packets" of l's contain either one or three terms. Now assume some far enough $(1,1,1,1)$ or $(-1,1,1,-1)$ subsequence of $\mathfrak{S}$ were to exist. Since it lies within some "doubled" subsequence, it contradicts the structure described above, which thus is the only prevalent from some rank on. But then all the positions of the $(-1)$ terms will have the same parity. However though, we have $\lambda(p)=\lambda\left(2 p^{2}\right)=-1$ for all odd primes $p$, and these terms have different parity of their positions. A contradiction has been reached.