, 2023

(1) When $m$ is large, we expect that the summation in (i) is very close to the following integral:
$$\int_{0}^{m} \left\{ \frac{ax}{m} \right\} \left\{ \frac{bx}{m} \right\} dx = m \cdot \int_{0}^{1} \{ax\}\{bx\}dx.$$
Hence, we will prove the inequality (i) for
$$\lambda = \int_{0}^{1} \{ax\}\{bx\}dx, \quad \beta = 2(a+b).$$
To do this, we divide the summation and integral into several segments, and it suffices to prove that
$$\begin{aligned} & \left| \sum_{k=1}^{m-1} \left\{ \frac{ak}{m} \right\} \left\{ \frac{bk}{m} \right\} - \int_{0}^{m} \left\{ \frac{ax}{m} \right\} \left\{ \frac{bx}{m} \right\} dx \right| \\ & = \left| \sum_{k=0}^{m-1} \frac{1}{2} \left( \left\{ \frac{ak}{m} \right\} \left\{ \frac{bk}{m} \right\} + \left\{ \frac{a(k+1)}{m} \right\} \left\{ \frac{b(k+1)}{m} \right\} \right) - \int_{0}^{m} \left\{ \frac{ax}{m} \right\} \left\{ \frac{bx}{m} \right\} dx \right| \\ & (ii) \quad \le \sum_{k=0}^{m-1} \left| \frac{1}{2} \left( \left\{ \frac{ak}{m} \right\} \left\{ \frac{bk}{m} \right\} + \left\{ \frac{a(k+1)}{m} \right\} \left\{ \frac{b(k+1)}{m} \right\} \right) - \int_{k}^{k+1} \left\{ \frac{ax}{m} \right\} \left\{ \frac{bx}{m} \right\} dx \right| \end{aligned}$$
is less than or equal to $2(a+b)$.
If $m \le a+b$, each term in (ii) is not greater than 1, so the total sum is not greater than $2a+2b$. We now assume $m > a+b$.
For each $k$, we divide the discussion into two cases:
(a) If there exists a $k$ such that $\{\frac{ak}{m}\}$ or $\{\frac{bk}{m}\}$ is discontinuous on $(k, k+1]$. There are at most $a+b-1$ such $k$. Moreover, since the product $\{\frac{ak}{m}\}\{\frac{bk}{m}\}$ always takes values in $[0, 1]$, the contribution of each corresponding term in (**) is at most $a+b-1$.
(b) If there exists a $k$ such that $\{\frac{ak}{m}\}$ or $\{\frac{bk}{m}\}$ is continuous on $(k, k+1]$. Let $x = k+\delta$ ($\delta \in [0, 1]$), then $\{\frac{ak}{m}\} = \{\frac{bk}{m}\} + \frac{a\delta}{m}$ or $\{\frac{bk}{m}\} = \{\frac{ak}{m}\} + \frac{b\delta}{m}$. Let $\alpha = \{\frac{ak}{m}\}$ and $\beta = \{\frac{bk}{m}\}$.

The corresponding term in (ii) is then
$$\begin{aligned} & \frac{1}{2} \left( \alpha\beta + (\alpha + \frac{a}{m})(\beta + \frac{b}{m}) \right) - \int_{0}^{1} (\alpha + \frac{a\delta}{m})(\beta + \frac{b\delta}{m}) d\delta \\ &= (\alpha\beta + \frac{a\beta}{2m} + \frac{b\alpha}{2m} + \frac{ab}{2m^2}) - (\alpha\beta + \frac{a\beta}{2m} + \frac{b\alpha}{2m} + \frac{ab}{3m^2}) = \frac{ab}{6m^2}. \end{aligned}$$
Combining the discussions in (a) and (b), we know that the summation in (ii) (when $m > a + b$) is less than or equal to
$$a + b - 1 + m \cdot \frac{ab}{6m^2} < 2(a + b) - 1.$$
This completes the proof of (1).

(2) Step 1: A refined estimation of (1).
We first calculate the value of $\lambda = \int_{0}^{1} \{ax\}\{bx\}dx$. Intuitively, this corresponds to integrating along the diagonal of the $a \times b$ grid, as shown in the left figure ($a = 4, b = 3$). Since the function inside the integral actually only depends on the small square where the diagonal lies, we can translate the diagonal to a small square, as shown in the middle figure (here, the small square is appropriately enlarged). However, this small square can also be divided into $ab$ rectangular blocks such that the path of integration exactly follows the diagonal, as shown in the right figure. This essentially corresponds to partitioning the integral $\int_{0}^{1}$ into integrals from $\frac{i}{ab}$ to $\frac{i+1}{ab}$ ($0 \le i \le ab-1$). Note that inside each small square, $\{ax\}$ and $\{bx\}$ are both continuous.

We calculate specifically:
$$\begin{aligned} \lambda &= \int_{0}^{1} \{ax\}\{bx\}dx = \frac{1}{ab} \sum_{i=0}^{ab-1} \int_{i}^{i+1} \left\{\frac{x}{b}\right\} \left\{\frac{x}{a}\right\} dx \\ (iii) \quad &= \frac{1}{ab} \sum_{i=0}^{ab-1} \left\{\frac{i+\frac{1}{2}}{b}\right\} \left\{\frac{i+\frac{1}{2}}{a}\right\} + \frac{1}{ab} \sum_{i=0}^{ab-1} \int_{i}^{i+1} \left( \left\{\frac{x}{b}\right\} \left\{\frac{x}{a}\right\} - \left\{\frac{i+\frac{1}{2}}{b}\right\} \left\{\frac{i+\frac{1}{2}}{a}\right\} \right) dx \end{aligned}$$
where each term on the right side of (iii) equals
$$\begin{align*} & \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( \left\{ \frac{i + \frac{1}{2} + x}{b} \right\} \left\{ \frac{i + \frac{1}{2} + x}{a} \right\} - \left\{ \frac{i + \frac{1}{2}}{b} \right\} \left\{ \frac{i + \frac{1}{2}}{a} \right\} \right) dx \\ &= \int_{-\frac{1}{2}}^{\frac{1}{2}} \left( \left\{ \frac{i + \frac{1}{2}}{b} \right\} \frac{x}{a} + \left\{ \frac{i + \frac{1}{2}}{a} \right\} \frac{x}{b} + \frac{x^2}{ab} \right) dx \\ &= 0 + 0 + \frac{1}{3ab} \left( \left( \frac{1}{2} \right)^3 - \left( -\frac{1}{2} \right)^3 \right) = \frac{1}{12ab}. \end{align*}$$
Regarding the preceding summation in (iii), note that when $i$ takes values from $\{0, 1, \dots, ab-1\}$, the pairs of numbers $\left(\left\{\frac{i}{b}\right\}, \left\{\frac{i}{a}\right\}\right)$ exactly cover all the points in the set
$$\left\{0, \frac{1}{a}, \dots, \frac{a-1}{a}\right\} \times \left\{0, \frac{1}{b}, \dots, \frac{b-1}{b}\right\}.$$
From this, we know that
$$\sum_{i=0}^{ab-1} \left\{ \frac{i + \frac{1}{2}}{b} \right\} \left\{ \frac{i + \frac{1}{2}}{a} \right\} = \sum_{j=0}^{a-1} \sum_{k=0}^{b-1} \left\{ \frac{k + \frac{1}{2}}{b} \right\} \left\{ \frac{j + \frac{1}{2}}{a} \right\} = \frac{a}{2} \cdot \frac{b}{2} = \frac{ab}{4}.$$
Combining the above results, we obtain
$$\lambda = \frac{1}{ab} \cdot \frac{ab}{4} + \frac{1}{ab} \cdot ab \cdot \frac{1}{12ab} = \frac{1}{4} + \frac{1}{12ab}.$$
That is,
$$\left| \sum_{k=1}^{m-1} \left\{ \frac{ak}{m} \right\} \left\{ \frac{bk}{m} \right\} - \left( \frac{m}{4} + \frac{m}{12ab} \right) \right| \le 2|a| + 2|b| - 1.$$
Note that when $m$ is coprime to both $a$ and $b$, since $\{\frac{ak}{m}\} + \{\frac{-ak}{m}\} = 1$, the sum corresponding to $(a, b)$ and the sum corresponding to $(-a, b)$ add up to $\frac{m-1}{2}$. Therefore, the inequality holds when $a, b$ are taken as negative integers coprime to $m$.
$$(*) \quad \left| \sum_{k=1}^{m-1} \left\{ \frac{ak}{m} \right\} \left\{ \frac{bk}{m} \right\} - \left( \frac{m}{4} + \frac{m}{12ab} \right) \right| \le 2|a| + 2|b|.$$

Step 2: Transform the problem into a summation similar to (1).
Let $p$ be a sufficiently large prime number (e.g., $p > 10^{15}$). The congruence symbol $\equiv$
used below denotes congruence modulo $p$.
Let $a, b$, and $c$ represent $a_1, a_2$, and $a_3$ respectively, and let $a_4$ be chosen such that
$a_1 + a_2 + a_3 + a_4 \equiv 0$. For $k \in \{1, 2, \dots, p-1\}$, let
$$h_k = \left\{ \frac{a_1 k}{p} \right\} + \left\{ \frac{a_2 k}{p} \right\} + \left\{ \frac{a_3 k}{p} \right\} + \left\{ \frac{a_4 k}{p} \right\},$$
It is easy to see that $h_k$ is an integer and $0 \le h_k < 4$, i.e., $h_k \in \{0, 1, 2, 3\}$. Moreover,
$\{\frac{a_1 k}{p}\} + \{\frac{a_2 k}{p}\} + \{\frac{a_3 k}{p}\} \le 1$ is equivalent to $h_k = 0$ or 1.
Let $L_0, L_1, L_2$, and $L_3$ be the number of occurrences of 0, 1, 2, and 3 respectively among $h_1, h_2, \dots, h_{p-1}$. We need to prove that $L_0 + L_1 \ge \lfloor \frac{p}{12} \rfloor$.

If exactly $t$ out of $a_1, a_2, a_3, a_4$ are not multiples of $p$, then $h_{p-k} = t - h_k$. We have the following results: If $t=0$, then $L_0 = p-1$. If $t=2$, then $L_1 = p-1$. If $t=3$, then $L_1 = L_2 = \frac{p-1}{2}$.
Without loss of generality, we assume that $a_1, a_2, a_3, a_4 \neq 0$. In this case, $L_1 = L_3$, and $L_1 + L_2 + L_3 = p-1$. We can consider $\{\frac{ak}{p}\}$ as a random variable, and the estimation of $L_1$ will be related to the variance of the sum $h_k$. Specifically, we consider the sum of squares:
$$\sum_{k=1}^{p-1} h_k^2 = L_1 + 4L_2 + 9L_3 = 4(p-1) + 2L_1,$$
On the other hand,
$$H = \sum_{k=1}^{p-1} h_k^2 = \sum_{k=1}^{p-1} \left( \sum_{i=1}^{4} \left\{ \frac{a_i k}{p} \right\} \right) \left( \sum_{j=1}^{4} \left\{ \frac{a_j k}{p} \right\} \right) = \sum_{i=1}^{4} \sum_{j=1}^{4} \left( \sum_{k=1}^{p-1} \left\{ \frac{a_i k}{p} \right\} \left\{ \frac{a_j k}{p} \right\} \right).$$
For any $a, b \neq 0$, let's define the function $f(a, b) = \sum_{k=1}^{p-1} \left\{ \frac{ak}{p} \right\} \left\{ \frac{bk}{p} \right\}$. In particular,
$$f(a_i, a_i) = f(1, 1) = \frac{1^2 + 2^2 + \dots + (p-1)^2}{p^2} = \frac{p}{3} - \frac{1}{2} + \frac{1}{6p}.$$
Now, let's consider the sum of six terms $F = \sum_{1 \le i < j \le 4} f(a_i, a_j)$. Since $H = 2F + 4f(1, 1) = 4(p-1) + 2L_1$, the desired inequality $L_1 \ge \lfloor \frac{p}{12} \rfloor$ holds if we have $F \ge \frac{17}{12}p$.

Step 3: Translating into estimates for $a_1, a_2, a_3, a_4$.
We define an equivalence relation $(a, b) \sim (u, v)$ if there exists $r \neq 0$ such that $u \equiv ra$ and $v \equiv rb$. In this case,
$$f(u, v) = \sum_{k=1}^{p-1} \left\{ \frac{uk}{p} \right\} \left\{ \frac{vk}{p} \right\} = \sum_{k=1}^{p-1} \left\{ \frac{ark}{p} \right\} \left\{ \frac{brk}{p} \right\} = \sum_{rk=1}^{p-1} \left\{ \frac{a \cdot rk}{p} \right\} \left\{ \frac{b \cdot rk}{p} \right\} = f(a, b).$$
Thus, the value of the function $f(a_i, a_j)$ depends only on the equivalence class of $(a_i, a_j)$. In particular, combining with the calculation in Step 1 (*), we have
$$(***) \qquad \left| f(a_i, a_j) - \left( \frac{p}{4} + p \frac{\gcd(a_i, a_j)^2}{12a_i a_j} \right) \right| \le 2 \frac{|a_i| + |a_j|}{\gcd(a_i, a_j)}.$$
To obtain a more accurate estimation of the values of $f(a_i, a_j)$, we introduce additional considerations.
For any $a, b \neq 0$, let $b^{-1}$ denote the modular inverse of $b$ modulo $p$. Consider the residues modulo $p$ of the set $0, ab^{-1}, 2 \times ab^{-1}, \dots, \lfloor \sqrt{p} \rfloor \times ab^{-1}$. By the pigeonhole principle, there exist two residues whose difference is less than or equal to $\lfloor \sqrt{p} \rfloor$. Therefore, there exist $u, v \in \pm 1, \pm 2, \dots, \pm \lfloor \sqrt{p} \rfloor$ such that $u \equiv v \times ab^{-1}$, i.e., $(u, v) \sim (a, b)$. (Here, we assume $u > 0$, and we also assume that $u$ and $v$ are coprime; otherwise, we can replace them with $\frac{u}{(u,v)}$ and $\frac{v}{(u,v)}$.)
For $(a_i, a_j)$, let $(u_{ij}, v_{ij})$ be one of the pairs that satisfy the above condition. If there are multiple choices, we can select any of them, and let $g_{ij} = \frac{1}{u_{ij}v_{ij}}$. In this case, we have
$$f(a_i, a_j) = f(u_{ij}, v_{ij}),$$
and according to Equation (***), we have:
$$\left| f(a_i, a_j) - \left( \frac{p}{4} + \frac{p}{12u_{ij}v_{ij}} \right) \right| < 4\sqrt{p}.$$
Let $G = \sum_{1 \le i < j \le 4} g_{ij} = \sum_{1 \le i < j \le 4} \frac{1}{u_{ij}v_{ij}}$, and let $\delta = 10^{-5}$. As long as (there exists a choice of $g_{ij}$) $G \ge -1 + \delta$, we have
$$F = \sum_{i<j} f(a_i, a_j) > \sum_{i<j} \left( \frac{p}{4} + \frac{p}{12u_{ij}v_{ij}} - 4\sqrt{p} \right) = \frac{18+G}{12} \times p - 24\sqrt{p} > \frac{17}{12}p.$$
The condition $p \nmid (a+b)(a+c)(b+c)$ ensures that in the set $a_1, a_2, a_3, a_4$,
For each pair $(a_i, a_j)$, since $a_i+a_j \ne 0$, we know that $(u_{ij}, v_{ij}) \ne (1, -1)$, which implies $g_{ij} \ge -\frac{1}{2}$. We call a pair $(a, b)$ “good” if there exist $u, v \in \pm 1, \pm 2, \dots, \pm 10$ such that $(a, b) \sim (u, v)$. In this case, if $(a_i, a_j)$ is not a good pair, then $g_{ij} = \frac{1}{u_{ij}v_{ij}} \ge -\frac{1}{11}$. If among the six pairs $(a_i, a_j)$, $1 \le i < j \le 4$, there are either 0 or 1 good pairs, then $G \ge -\frac{1}{2} + 5 \times (-\frac{1}{11}) = -\frac{21}{22} > -1 + \delta$.
Therefore, assuming there are at least two good pairs, we know that $a_1+a_2+a_3+a_4 \equiv 0$ implies that the four numbers are in a (simple) proportion. For example:
If $(a_1, a_2) \sim (u_1, v_1)$ and $(a_1, a_3) \sim (u_2, v_2)$, then the four numbers are proportional to $(u_1u_2, v_1u_2, v_2u_1, -u_1u_2 - v_1u_2 - v_2u_1)$.
If $(a_1, a_2) \sim (u_1, v_1)$ and $(a_3, a_4) \sim (u_2, v_2)$, then the four numbers are proportional to $(u_1(u_2+v_2), v_1(u_2+v_2), -u_2(u_1+v_1), -v_2(u_1+v_1))$.
In conclusion, there exists $z_1+z_2+z_3+z_4 = 0$ such that $(a_1, a_2, a_3, a_4) \sim (z_1, z_2, z_3, z_4)$ and $1 \le |z_i| \le 300$, $z_i + z_j \ne 0$. In this case, we have $(a_i, a_j) \sim \left( \frac{z_i}{z_i z_j}, \frac{z_j}{z_i z_j} \right)$, which allows us to choose $g_{i,j} = \frac{(z_i, z_j)^2}{z_i z_j}$. We still focus on the value of $G = \sum_{i<j} g_{i,j}$, and there are two simple proportions satisfying $G = 1$:
* The four numbers form the “Golden Ratio 1” of $(1 : -2 : -3 : 4)$. Let $a_1 = p-1$, $a_2 = 2$, $a_3 = 3$, $a_4 = p-4$. In this case, $h_k = 1$ is equivalent to $\frac{2k}{p} + \frac{3k}{p} \le \frac{k}{p}$. Among the values $k \in 1, 2, \dots, p-1$, such $k$ satisfying $\frac{2}{3}p < k < \frac{3}{4}p$ are counted as $L_1 = \lfloor \frac{3}{4}p \rfloor - \lfloor \frac{2}{3}p \rfloor \ge \lfloor \frac{p}{12} \rfloor$.
* The four numbers form the “Golden Ratio 2” of $(1 : -3 : -4 : 6)$. Let $a_1 = p-1$, $a_2 = 3$, $a_3 = 4$, $a_4 = p-6$. In this case, $h_k = 1$ is equivalent to $\frac{3k}{p} + \frac{4k}{p} \le \frac{k}{p}$. Among the values $k \in 1, 2, \dots, p-1$, such $k$ satisfying $\frac{3}{4}p < k < \frac{5}{6}p$ are counted as $L_1 = \lfloor \frac{5}{6}p \rfloor - \lfloor \frac{3}{4}p \rfloor \ge \lfloor \frac{p}{12} \rfloor$.
Assuming $(a_1, a_2, a_3, a_4) \sim (z_1, z_2, z_3, z_4)$ does not correspond to the two golden ratios mentioned above, we will prove that $G \ge -1 + \delta$. Let's consider the signs of $z_1, z_2, z_3, z_4$.
If there are three positive and one negative (or three negative and one positive) numbers, without loss of generality, let $z_1, z_2, z_3 > 0$ and $z_4 = -(z_1 + z_2 + z_3)$. In this case, among the six terms in $G = \sum_{1 \le i < j \le 4} g(z_i, z_j)$, there are three positive and three negative terms. Specifically, $g(z_i, z_4) = -\frac{(z_i, z_4)^2}{z_i |z_4|} \ge -\frac{z_i}{|z_4|}$ for $i = 1, 2, 3$, and $g(z_1, z_2) \ge \frac{1}{z_1 z_2} > \delta$. Therefore, $G > -1 + \delta$.
Assuming that $z_1, z_2, z_3, z_4$ consist of two positive and two negative numbers, let $z_1, z_4 > 0$ and $z_2, z_3 < 0$.
If there exists $\frac{z_j}{z_i} = -2$, without loss of generality, let $z_1 = u > 0$, $z_2 = -2u$. Let $z_3 = -v < 0$ and $z_4 = u + v$ (where $u$ and $v$ are coprime). In this case,
$$G = -\frac{1}{2} - \frac{1}{uv} + \frac{1}{u(u+v)} + \frac{(2, v)^2}{2uv} - \frac{(2, u+v)^2}{2u(u+v)} - \frac{1}{v(u+v)}$$
$$G \geq -\frac{1}{2} - \frac{1}{uv} + \frac{1}{u(u+v)} + \frac{1}{2uv} - \frac{4}{2u(u+v)} - \frac{1}{v(u+v)} = -\frac{1}{2} - \frac{3}{2uv} > -1 + \delta.$$
The last step of the above derivation assumes that $(u, v) \neq (1, 1), (2, 1), (1, 3), (3, 1)$, which would lead to negations or the four numbers forming a golden ratio. Additionally, when $(u, v) = (1, 2)$, we have $G = 0$. In any case, we have $G > -1 + \delta$.
If there exists $\frac{z_j}{z_i} = -3$, without loss of generality, let $z_1 = u > 0$, $z_2 = -3u$. Let $z_3 = -v < 0$ and $z_4 = 2u + v$ (where $u$ and $v$ are coprime). In this case,
$$G = -\frac{1}{3} - \frac{1}{uv} + \frac{1}{u(2u + v)} + \frac{(3, v)^2}{3uv} - \frac{(3, 2u + v)^2}{3u(2u + v)} - \frac{(2, v)^2}{v(2u + v)}$$
$$G \geq -\frac{1}{3} - \frac{1}{uv} + \frac{1}{u(2u + v)} + \frac{1}{3uv} - \frac{9}{3u(2u + v)} - \frac{4}{v(2u + v)} = -\frac{1}{3} - \frac{8}{3uv} > -1 + \delta.$$
The last step of the above derivation assumes that $(u, v) \neq (1, 1), (1, 2), (1, 4)$, which would lead to negations or the four numbers forming a golden ratio. Additionally, when $(u, v) = (2, 1), (1, 3), (3, 1), (4, 1)$, we have $G = -\frac{4}{5}, \frac{2}{5}, -\frac{2}{3}, -\frac{2}{3}$ respectively. In any case, we have $G > -1 + \delta$.
If each $\frac{z_j}{z_i} \neq -2, -3$, then the four negative terms in $G$ are all greater than or equal to $-\frac{1}{4}$, and the two positive terms are both greater than $\delta$. Therefore, we still have $G > -1 + \delta$.
In conclusion, we have $L_1 + L_0 \geq \lfloor \frac{p}{12} \rfloor$, and thus the conclusion of the problem is established. $\square$