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?


1 Answer 1


Yes, this is a well-known. In fact one can prove the following:

Let $\{\alpha_n\}$ be an arbitrary complex sequence with $|\alpha_n| \leq 1$ for all $n \geq 1$ and finite $\ell^2$-norm. Then for any positive numbers $D, N$ one has

$$\displaystyle \sum_{d \leq D} \sum_{v^2 + 1 \equiv 0 \pmod{d}} \left \lvert \sum_{n \leq N} \alpha_n e \left(\frac{vn}{d} \right)\right \rvert^2 \ll (D + N) \lVert \alpha \rVert_2^2,$$

where $\alpha = \{\alpha_n\}$ and $\lVert \cdot \rVert_2$ is the $\ell^2$-norm.

The proof goes as follows. It clearly suffices to consider dyadic intervals of $D$, so we may assume that $D < d \leq 2D$ say. We then consider the equidistribution of the numbers $v/d$, with $v^2 + 1 \equiv 0 \pmod{d}, 0 \leq v < d$. For each such $v$ there exist positive integers $r,s$ such that $r^2 + s^2 = d$ and $r \equiv vs \pmod{d}$. This implies that we have an equation of the form

$$\displaystyle r - vs = \ell d,$$

which we rearrange as

$$\displaystyle \frac{r}{sd} - \frac{\ell}{s} = \frac{v}{d}.$$

By exploiting symmetry we can always assume $|s| \geq |r|$, and that $s > 0$. Since $r^2 + s^2 = d$ this implies

$$\displaystyle \frac{|r|}{sd} = \frac{|r|}{s(r^2 + s^2)} \leq \frac{|r|}{s(2|r|s)} \leq \frac{1}{2s^2},$$ with equality only when $r = s = 1$. Hence if we have two pairs $(v_1, d_1), (v_2, d_2)$ say we get the equality

$$\displaystyle \frac{v_1}{d_1} - \frac{v_2}{d_2} = \frac{r_1}{s_1 d_1} - \frac{r_2}{s_2 d_2} - \frac{\ell_1}{s_1} + \frac{\ell_2}{s_2},$$

which implies that the mod 1 distance between $v_1/d_1, v_2/d_2$ is $\gg 1/(s_1 s_2) \gg 1/D$ (this is because $s_i \gg d_i^{1/2}$, and $d_1, d_2 \in [D,2D)$). The large sieve inequality then gives the conclusion above.

With (a lot) more work one can even show that the fractions $v/p$ equi-distribute, where we take prime moduli only. See the following paper by Duke, Friedlander, and Iwaniec.

  • $\begingroup$ This is very helpful regarding equidistribution on average, thank you. Yet I don’t see how these kind of results imply that one has equidistribution for sequences of $d$s with $\omega(d)$ tending to infinity as in the question. Or is this something I am missing? $\endgroup$ May 31, 2020 at 21:17

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.