For $n \geq 2$, denote the primes less than or equal to $n$ by $p_1 = 2, p_2 = 3, \ldots, p_{k(n)}$, and define a polynomial $P_n \in \mathbb{C}[z_1, \ldots, z_{k(n)}]$ by $$P_n(z_1, \ldots, z_{k(n)}) = \sum_{m = 1}^{n} m^{-1/2} \prod_{p_j^{\ell} \parallel m} z_j^{\ell}.$$ Here $p_j^{\ell} \parallel m$ denotes that $p_j^{\ell}$ is the highest power of $p_j$ dividing $m$.
To give concrete examples for small $n$, we have:
- $P_2(z_1) = 1 + 2^{-1/2} z_1$
- $P_3(z_1, z_2) = 1 + 2^{-1/2} z_1 + 3^{-1/2} z_2$;
- $P_4(z_1, z_2) = 1 + 2^{-1/2} z_1 + 3^{-1/2} z_2 + 2^{-1} z_1^2$;
- $P_5(z_1, z_2, z_3) = 1 + 2^{-1/2} z_1 + 3^{-1/2} z_2 + 2^{-1} z_1^2 + 5^{-1/2} z_3$.
The question: For which $n$ does $P_n$ have a zero in the closed unit polydisc, i.e. the set $$\{ (w_1, \ldots, w_{k(n)}) : | w_1 |, \ldots, | w_{k(n)} | \leq 1 \}?$$
Certainly there's no such zero for $n = 2$. For many $n \geq 3$, I can show that there's a zero by restricting the $z_j$ to real values and observing that $P_n$ is $1$ at the origin but negative either when every $z_j = -1$ or when $z_1 = 0$ and all other $z_j = -1$.
An interesting connection I've come across is that $$P_n(-1, -1, \ldots, -1) = \sum_{m = 1}^{n} \frac{\lambda(m)}{m^{1/2}} = L_{1/2}(n).$$ Here $\lambda$ is the Liouville function, and $L_{1/2}$ comes from the family of weighted summatory functions $$L_{\alpha}(x) = \sum_{m \leq x} \frac{\lambda(m)}{m^{\alpha}}.$$
An article I found mentions an unresolved conjecture of Mossinghoff and Trudgian that $L_{1/2}(x) \leq 0$ for all $x \geq 17$. This would settle the question, and apparently it's been verified that $L_{1/2}(x) \leq 0$ for $17 \leq x \leq 10^{12}$. But it would be great to find a simpler resolution that doesn't depend on solving Riemann Hypothesis-adjacent questions. Any thoughts are much appreciated.