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With $\mu$ being the Möbius function, there exist infinite possibilities of square roots. For example, for each $n$ such that $\mu(n)\neq 0$ there is a choice: if $\mu(n)=-1$, we can choose to define $\sqrt\mu(n) = \pm i$, and if $\mu(n)=1$, we can choose to define $\sqrt\mu(n) = \pm 1$. And any of those specially chosen $\sqrt\mu$ will satisfy $\sqrt\mu^2 =\mu$.

  1. We can remark that there are infinitely many choices of square roots $\sqrt\mu$ such that $$ \forall x,\,\Bigl| \sum_{n\leq x} \sqrt\mu (n) \Bigr|\leq C, $$ where $C$ will be a constant $C\geq\sqrt 2$. Just choosing "online" a square root of $\sqrt\mu(n)$ to satisfy the bound.

  2. We can remark that there are infinitely many choices of square roots $\sqrt\mu$ that are multiplicative. One can choose $\sqrt\mu (1)=1$, then make an infinite number of choices for each prime $p$: $\sqrt\mu (p) = \pm i$. From those choices, we can impose the multiplicativity of the square root $\sqrt\mu$ infering all the other values of $\sqrt\mu$ from our "offline" choices on primes with the multiplicative rule.

Question: Is a multiplicative (2.) and bounded (1.) square root $\sqrt\mu$ of the Möbius function known to exist?

Bonus 1: If so, does multiplicativity (2.) imply the boundedness property (1.)?

Bonus 2: What would be the lowest possible value of $C$ ?

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    $\begingroup$ Please use a high-level tag like "nt.number-theory". I added this tag now. I also removed the tag "algebraic-number-theory" as irrelevant. Regarding high-level tags, see meta.mathoverflow.net/q/1075 $\endgroup$
    – GH from MO
    Commented Aug 20 at 8:32
  • $\begingroup$ Certainly multiplicativity does not imply boundedness. Given $x$, once you've chosen $\sqrt\mu(p)$ for each prime $p \leq x/2$, you have determined $\sum_{n\leq x} \sqrt\mu(n)$ except for the $\pi(x) - \pi(x/2) \sim x / (2 \log x)$ terms $\sqrt\mu(p)$ with $x/2 < p \leq x$, each of which you can choose to be either $+i$ or $-i$. So in the worst case the sum can grow at least as fast as a positive multiple of $x / \log x$. $\endgroup$
    – Noam D. Elkies
    Commented Aug 20 at 14:46

1 Answer 1

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EDIT: Corrected my previous answer, which contained an incorrect reduction of the problem to a result in the literature.

The answer is likely negative. By [1, Theorem 1.1], the partial sums of any multiplicative function with values $\pm 1$ supported on squarefree numbers are unbounded. As indicated by Will Sawin in the comments, it may be possible to adapt the proof to non-pretentious multiplicative functions with $|f(p)|=1$ ($p$ prime) supported on squarefrees, which applies to $\sqrt\mu$ in particular. However I would be cautious about this without carefully checking the details. The corresponding result for completely multiplicative functions (not supported on squarefrees) is known by Theorem 1.4 of [https://arxiv.org/pdf/1911.06265], so in particular if we replace $\sqrt\mu$ with $\sqrt\lambda$ (Liouville function) the answer to the question is negative.

1 Marco Aymone, The Erdős discrepancy problem over the squarefree and cubefree integers, Mathematika 68 (2022), no. 1, pp. 51–73, doi 10.1112/mtk.12117.

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    $\begingroup$ I don’t understand how the reduction works. If I replace each $\pm i$ with $\pm1$, and make it multiplicative, the sign of the function will be negated for square-free numbers with $2\pmod4$ prime factors, but stay the same for square-free numbers with $0\pmod4$ prime factors. Thus I do not see why this should preserve the property of bounded partial sums. $\endgroup$
    – Emil Jeřábek
    Commented Aug 20 at 9:44
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    $\begingroup$ Explicitly: let $f$ be a square root of $\mu$ as in the question, let $g$ be the multiplicative function such that $f(p)=\pm i\implies g(p)=\pm1$ for prime $p$, and for $j=0,1,2,3$, let $S_j(x)\in\mathbb Z$ be the sum of $g(n)$ over square-free $n\le x$ with $j\pmod4$ prime factors. Then the partial sums of $g$ are $\sum_jS_j(x)$, and the partial sums of $f$ are $\sum_ji^jS_j(x)=(S_0(x)-S_2(x))+i(S_1(x)-S_3(x))$. Thus, the assumption that $f$ has bounded sums means that $|S_0(x)-S_2(x)|$ and $|S_1(x)-S_3(x)|$ are bounded. How does this imply that $|(S_0(x)+S_2(x))+(S_1(x)+S_3(x))|$ is bounded? $\endgroup$
    – Emil Jeřábek
    Commented Aug 20 at 10:38
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    $\begingroup$ I agree with @EmilJeřábek that you can't easily reduce the problem this way. $\endgroup$
    – Virgile Dine
    Commented Aug 20 at 10:40
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    $\begingroup$ @VirgileDine The reduction does not work, but the method of proof easily transfers. One only needs the analogue of Proposition 2.1 for arbitrary $S^1$-valued multiplicative functions, because the squareroot of Möbius is not pretentious as that would force Möbius itself to be pretentious. But one just has to copy the proof of Proposition 2.1 and insert absolute values and complex conjugates in appropriate places to establish this. $\endgroup$
    – Will Sawin
    Commented Aug 20 at 10:59
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    $\begingroup$ I agree, my reduction doesn't work. I've updated my answer. $\endgroup$
    – Alexei Entin
    Commented Aug 20 at 13:53

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