3.3. Goldbach 问题的上界

在前一节中, 我们通过 Dirichlet 级数展开法研究了被筛集合为

$$\mathcal{A}=\{n(n-h):n\le x\}$$

时的筛函数 $S(\mathcal{A},\mathcal{P},z)$ 从而改良了之前对 ${\pi }_h(x)$ 的上界估计. 但在由于当时的上界仅是对于固定 $h$ 成立, 所以对于 Goldbach 问题我们还是需要用初等方法得到对所有 $h$ 一致成立的 $G(x)$ 展开式.

由前一节的推导可知:

$$h(p)=\{\begin{array}{ll}1/(p-1)& p|h\\ 2/(p-2)& p\nmid h\end{array},$$

所以有:

$$\begin{array}{rl}G(x)& =\sum _{d<x}{\mu }^2(d)h(d)=\sum _{k|h}\sum _{\begin{array}{c}d<x\\ (d,h)=k\end{array}}{\mu }^2(d)h(d)\\ & =\sum _{k|h}{\mu }^2(k)h(k)\underset{S}{\underbrace{\sum _{\begin{array}{c}t<x/k\\ (t,h)=1\end{array}}{\mu }^2(t)h(t)}}.\end{array}$$

为了估计 $S$, 设积性函数 $u(m)$ 满足 $u(p)=p-2$, 则交换求和次序可得:

$$\begin{array}{rl}S& =\sum _{\begin{array}{c}t<x/k\\ (t,h)=1\end{array}}{\mu }^2(t)\prod _{p|t}\frac{2}{p-2}=\sum _{\begin{array}{c}t<x/k\\ (t,h)=1\end{array}}\frac{{\mu }^2(t)2^{\omega (t)}}{t}\prod _{p|t}\left(1+\frac{2}{p-2}\right)\\ & =\sum _{\begin{array}{c}t<x/k\\ (t,h)=1\end{array}}\frac{{\mu }^2(t)2^{\omega (t)}}{t}\sum _{m|t}\frac{{\mu }^2(m)2^{\omega (m)}}{u(m)}\\ & =\sum _{\begin{array}{c}m<x/k\\ (m,h)=1\end{array}}\frac{{\mu }^2(m)4^{\omega (t)}}{u(m)m}\sum _{\begin{array}{c}n<x/mk\\ (n,mh)=1\end{array}}\frac{{\mu }^2(n)2^{\omega (n)}}{n}.\end{array}$$

为了继续估计 $S$, 我们代入定理 7.5.3.1, 便有:

$$\begin{array}{rl}S& =\frac{1}{2}\prod _p{\left(1-\frac{1}{p}\right)}^2\left(1+\frac{2}{p}\right)\sum _{\begin{array}{c}m<x/k\\ (m,h)=1\end{array}}\frac{{\mu }^2(m)4^{\omega (m)}}{u(m)m}\prod _{p|mh}\frac{p}{p+2}{\log }^2\frac{x}{mk}\\ & +O(\log x\log \log 3hx)+O\{(\log \log xh{)}^2\}\\ & =\frac{1}{2}\prod _p{\left(1-\frac{1}{p}\right)}^2\left(1+\frac{2}{p}\right)\prod _{p|h}\frac{p}{p+2}\prod _{p\nmid h}\left(1+\frac{4}{p^2-4}\right){\log }^2\frac{x}{k}\\ & +O(\log x\log \log 3hx)+O\{(\log \log 3xh{)}^2\}.\end{array}$$

化简乘积可知:

$$\begin{array}{rl}& \prod _p{\left(1-\frac{1}{p}\right)}^2\left(1+\frac{2}{p}\right)\prod _{p|h}\frac{p}{p+2}\prod _{p\nmid h}\left(1+\frac{4}{p^2-4}\right)\\ & =\frac{1}{4}\prod _{p>2}\frac{(1-p{)}^2}{p^2}\frac{p+2}{p}\frac{p^2}{p^2-4}\prod _{\begin{array}{c}p|h\\ p>2\end{array}}\frac{p}{p+2}\frac{p^2-4}{p^2}\\ & =\frac{1}{4}\prod _{p>2}\frac{(p-1{)}^2}{p(p-2)}\prod _{\begin{array}{c}p|h\\ p>2\end{array}}\frac{p-2}{p},\end{array}$$

所以有:

$$\begin{array}{rl}S& =\frac{1}{8}\prod _{p>2}\frac{(p-1{)}^2}{p(p-2)}\prod _{\begin{array}{c}p|h\\ p>2\end{array}}\frac{p-2}{p}{\log }^2\frac{x}{k}\\ & +O(\log x\log \log 3hx)+O\{(\log \log 3xh{)}^2\}.\end{array}$$

为了继续将 $S$ 代入到 $G(x)$ 中进行计算, 我们需要先估计下面这种形式的和:

$$T_n(h)=\sum _{k|h}{\mu }^2(k)h(k){\log }^{n}k.$$

其中很明显 $T_0(h)=h/\phi (h)$, 而 $n\ge 1$ 时:

$$\begin{array}{rl}{T}_n(h)& =\sum _{k|h}{\mu }^2(k)h(k){\log }^{n-1}k\sum _{p|k}\log p\\ & =\sum _{p|h}\log p\sum _{\begin{array}{c}k\\ p|k|h\end{array}}{\mu }^2(k)h(k){\log }^{n-1}k\\ & =\sum _{p|h}h(p)\log p\sum _{t|h/(p,h)}{\mu }^2(t)h(t){\log }^{n-1}(pt)\\ & =\sum _{p|h}h(p)\log p\sum _{t|h/(p,h)}{\mu }^2(t)h(t)\sum _{0\le r\le n-1}\left(\genfrac{}{}{0ex}{}{n-1}{r}\right){\log }^{r}t{\log }^{n-1-r}p\\ & =\sum _{0\le r\le n-1}\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)T_r\left(\frac{h}{(p,h)}\right)\sum _{p|h}h(p){\log }^{n-r}p.\end{array}$$

由于 $p|h$ 时总有 $h(p)=1/(p-1)$, 所以当 $n$ 固定时总有:

$$\begin{array}{rl}{T}_n(h)& \ll \sum _{0\le r\le n-1}\left(\genfrac{}{}{0ex}{}{n-1}{r}\right)T_r(h)\sum _{p|h}\frac{{\log }^{n-r}p}{p}\\ & \ll \sum _{0\le r\le n-1}{T}_r(h)(\log \log 3h{)}^{n-r}.\end{array}$$

由此可知对于所有的 $n\ge 0$, 均存在 $C_n>0$ 使得:

$$T_n(h)<C_n\frac{h}{\phi (h)}(\log \log 3h{)}^n.$$

将这个结果与 $S$ 和 $G(x)$ 结合, 我们就得到了最终的展开式:

$$\begin{array}{rl}G(x)& =\frac{1}{4}\prod _{p>2}\frac{(p-1{)}^2}{p(p-2)}\prod _{\begin{array}{c}p|h\\ p>2\end{array}}\frac{p-2}{p-1}{\log }^{2}x\\ & +O\left(\frac{h}{\phi (h)}\log x\log \log 3hx\right)+O\left\{\frac{h}{\phi (h)}(\log \log 3hx{)}^2\right\}.\end{array}$$

现在用 $r_x(a,b)$ 表示将偶数 $x$ 分解成一个不超过 $a$ 个素数的乘积与一个不超过 $b$ 个素数的乘积之和的方法数时, 则根据 $S(\mathcal{A},\mathcal{P},z)$ 的定义可知:

$$r_x(1,1)\le S(\mathcal{A},\mathcal{P},x^{1/2})+2x^{1/2}.$$

再将其与定理 3.4.1 和 $G(x)$ 的展开式结合, 便有:

定理 3.3.0.1. 当 $x$ 为大偶数时, 总有: $$r_x(1,1)<\frac{16x}{{\log }^{2}x}\prod _{\begin{array}{c}p|x\\ p>2\end{array}}\frac{p-1}{p-2}\prod _{p>2}\left(1-\frac{1}{(p-1{)}^2}\right)+O\left\{\prod _{\begin{array}{c}p|x\\ p>2\end{array}}\frac{p-1}{p-2}\frac{\log \log x}{{\log }^{3}x}\right\}.$$