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λ(n) Liouville function
Ω(n) oeis.org/A001222
大师赛(RMMS) 2011 Day2 Problem 4
Solution Notice that we have $Ω(mn) = Ω(m) + Ω(n)$ for all positive integers $m, n$ ($Ω$ is a completely additive arithmetic function), translating into $λ(mn) = λ(m) · λ(n)$ (so $λ$ is a completely multiplicative arithmetic function), hence $λ(p) = −1$ for any prime $p$, and $λ(k^2) = λ(k)^2 = +1$ for all positive integers $k$.
The start (first 100 terms) of the sequence $\mathfrak{S} = (λ(n))_{n≥1}$ 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, +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.
i) The Pell equation $x^2-6 y^2=1$ has infinitely many solutions in positive integers; all solutions are given by $\left(x_n, y_n\right)$, where $x_n+y_n \sqrt{6}=(5+2 \sqrt{6})^n$. Since $\lambda\left(6 y^2\right)=1$ and also $\lambda\left(6 y^2+1\right)=\lambda\left(x^2\right)=1$, the thesis is proven.
Alternative Solution. Take any existing pair with $\lambda(n)=$ $\lambda(n+1)=1$. Then $\lambda\left((2 n+1)^2-1\right)=\lambda\left(4 n^2+4 n\right)=\lambda(4) \cdot \lambda(n)$. $\lambda(n+1)=1$, and also $\lambda\left((2 n+1)^2\right)=\lambda(2 n+1)^2=1$, so we have built a larger $(1,1)$ pair.
ii) 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.
Alternative Solution. 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.
Alternative Solution. 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, ?, . . . , ?, −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, . . . , 1, −1, 1, 1, 1, −1),
and so the "separating packets" of 1’s contain either one or three terms. Now assume some far enough (1, 1, 1, 1) or (−1, 1, 1, −1) subsequence of $\frak S$ 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 $λ(p) = λ(2p^2) = −1$ for all odd primes $p$, and these terms have different parity of their positions. A contradiction has been reached.
Alternative Solution for both i) and ii). (I. Bogdanov)
Take $ε ∈ \{−1, 1\}$. There obviously exist infinitely many $n$ such that $λ(2n +1) = ε$ (just take $2n +1$ to be the product of an appropriate number of odd primes). Now, if either $λ(2n) = ε$ or $λ(2n + 2) = ε$, we are done; otherwise $λ(n) = −λ(2n) =−λ(2n + 2) = λ(n + 1) = ε$. Therefore, for such an $n$, one of the three pairs $(n, n +1), (2n, 2n +1)$ or $(2n +1, 2n +2)$ fits the bill.
We have thus proved the existence in $\frak S$ of infinitely many occurrences of all possible subsequences of length 1, viz. (+1) and (−1), and of length 2, viz. (+1, −1), (−1, +1), (+1, +1) and (−1, −1). |
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