This is a question that arose during a conversation with a colleague regarding Landau's fourth problem, which asks whether there are infinitely many primes of the form $n^2+1$. The Conjectured asymptotic for

$$\sum_{n\leq X}\Lambda(n^2+1)$$ is $$-\sum_{n\leq x^2+1}\frac{\mu(n)\rho(n)\log n}{n}\sim \frac{CX}{\log X}$$

where $\rho(n)$ counts the number of solutions to the congruence $a^2+1\equiv 0 \pmod n$ and $C$ is specific arithmetic constant. Although Landau's fourth problem is considered to be out of reach, the question I want to ask seems to be a natural prerequisite. I have not proved this implication yet, but it seems to be a nice problem anyway.

**Question**
Let $n\in\mathbb{N}$ be squarefree and such that $p|n\Rightarrow p=4m+1$ for some $m\in\mathbb{N}$ so $-1$ is a square $\pmod n$ with $2^{\omega(n)}$ square roots $\pmod n$. Do these roots equidistribute $\pmod n$ as $\omega(n)\rightarrow\infty$?

Weyl's criterion asserts that equidistribution is equivalent to having

$$\sum_{a^2+1\equiv 0\pmod n}e\left(\frac{ak}{n}\right)=o\left(2^{\omega(n)}\right)$$

as $\omega(n)\rightarrow\infty$ for each $0<k<n$. By the chinese remainder theorem, the above statement equates to having

$$\prod_{p|n}\cos\left(\frac{2\pi k \overline{(n/p)}\rho_p}{p}\right)=o(1)$$

where $\rho^2_p+1\equiv 0\pmod p$ and $\overline{x}$ denotes the multiplicative inverse of $x \pmod p$.

Since $0$ and $p/2$ cannot be roots $\pmod p$, the limits of such trigonometric products will be zero if unboundedly many of the arguments do not converge to $0$ or $\pi$ as $\omega(n)\rightarrow\infty$, which amounts to having unboundedly many of the sequences

$$\frac{n}{p}\pmod p\hspace{1cm}p|n$$

diverge. Is it possible that all but finitely many of them could converge?