For a positive integer $n$, does there exist a permutation of all its positive integer divisors $(d_1 , d_2 , \ldots, d_k)$ such that the equation $d_kx^{k-1} + \ldots + d_2x + d_1 = 0$ has a rational root, if:
a) $n = 2024$;
b) $n = 2025$?
Proposed by Mykyta Kharin
The answer for a) is yes and for b) is no.
Let's first start with b) as it's easier; more generally, any odd perfect square doesn't work. Suppose that $x=\frac{p} {q}$ is a root for coprime $p, q$; then $\sum_{i=1}^{k} d_ip^{i-1}q^{k-i}=0$. Since all $d_i$ are odd, both $p, q$ are odd otherwise one of them is even and the other is odd and we obtain contradiction modulo $2$. But since $k$ is odd, the sum on the LHS consists of odd number of odd terms, hence it's also odd.
For a), we claim that any $n$, for which $\nu_p(n)=1$ for some prime $p$ works. Indeed, split the divisors of $n$ into pairs $(d, pd)$ for $p \nmid d$ and observe that $dx^i+dpx^{i+1}$ has root $x=\frac{-1}{p}$. Now, we can easily find a permutation for which $x=\frac{-1}{23}$ is a root of the polynomial.