| 1. | Lecture 1 |
| 2. | Lecture 2 |
| 3. | Lecture 3 |
| 4. | Lecture 4 |
| 5. | Lecture 5 |
| 6. | Lecture 6 |
| 7. | Lecture 7 |
| 8. | Lecture 8 |
| 9. | Lecture 9 |
| 10. | Lecture 10 |
| 11. | Lecture 11 |
| 12. | Lecture 12 |
| 13. | Lecture 13 |
The counting function for the number of representations of an odd integer \(N\) as the sum of three primes is \[r(N)=\sum_{p_{1}+p_{2}+p_{3}=N} 1.\]
The following is Vinogradov’s asymptotic formula for \(r(N)\).
There exists an arithmetic function \(\mathfrak{S}(N)\) and \(c_{1}, c_{2}\in\mathbb R_+\) such that \[c_{1}<\mathfrak{S}(N)<c_{2}\] for all sufficiently large odd integers \(N\), and \[r(N)=\mathfrak{G}(N) \frac{N^{2}}{2(\log N)^{3}}\left(1+O\left(\frac{\log \log N}{\log N}\right)\right).\]
The arithmetic function \(\mathfrak{S}(N)\) is called the singular series for the ternary Goldbach problem.
We begin by studying the arithmetic function
\[\mathfrak{S}(N)=\sum_{q=1}^{\infty} \frac{\mu(q) c_{q}(N)}{\varphi(q)^{3}},\] where \[c_{q}(N)=\sum_{\substack{a=1 \\(q, a)=1}}^{q} e(a N / q)\] is Ramanujan’s sum.
The function \(\mathfrak{S}(N)\) is called the singular series for the ternary Goldbach problem.
The Ramanujan sum \(c_q(n)\) is a multiplicative function of \(q\); that is, if \((q, q') = 1\), then \[c_{qq'}(n) = c_q(n) \cdot c_{q'}(n).\]
The Ramanujan sum can be expressed in the form
\[c_q(n) = \sum_{d \mid (q, n)} \mu\left(\frac{q}{d}\right) d,\]
In particular, if \((q, n) = 1\), then \(c_q(n) =\mu(q)\).
The singular series \(\mathfrak{S}(N)\) converges absolutely and uniformly in \(N\) and has the Euler product representation \[\mathfrak{S}(N)=\prod_{p}\left(1+\frac{1}{(p-1)^{3}}\right) \prod_{p \mid N}\left(1-\frac{1}{p^{2}-3 p+3}\right).\]
There exist positive constants \(c_{1}\) and \(c_{2}\) such that \[c_{1}<\mathfrak{S}(N)<c_{2},\] for all positive integers \(N\). Moreover, for any \(\varepsilon>0\),
\[\mathfrak{S}(N, Q)=\sum_{q \leq Q} \frac{\mu(q) c_{q}(N)}{\varphi(q)^{3}}=\mathfrak{S}(N)+O\left(Q^{-(1-\varepsilon)}\right),\] where the implied constant depends only on \(\varepsilon\in(0, 1)\).
We decompose the unit interval \([0,1]\) into two disjoint sets: the major arcs \(\mathfrak{M}\) and the minor arcs \(\mathfrak{m}\).
Let \(B>0\) and set \[Q=(\log N)^{B}\]
For \(1 \leq q \leq Q\) and \(0 \leq a \leq q\) satisfying \((a, q)=1\), the major arc \(\mathfrak{M}(q, a)\) is the interval consisting of all real numbers \(\alpha \in[0,1]\) so that \[\left|\alpha-\frac{a}{q}\right| \leq \frac{Q}{N}.\]
If \(\alpha \in \mathfrak{M}(q, a) \cap \mathfrak{M}\left(q^{\prime}, a^{\prime}\right)\) and \(a / q \neq a^{\prime} / q^{\prime}\), then \(\left|a q^{\prime}-a^{\prime} q\right| \geq 1\) and \[\begin{aligned} \frac{1}{Q^{2}} & \leq \frac{1}{q q^{\prime}} \leq \frac{\left|a q^{\prime}-a^{\prime} q\right|}{q q^{\prime}}=\left|\frac{a}{q}-\frac{a^{\prime}}{q^{\prime}}\right| \\ & \leq\left|\frac{a}{q}-\alpha\right|+\left|\alpha-\frac{a^{\prime}}{q^{\prime}}\right| \leq \frac{2 Q}{N} \end{aligned}\] or, equivalently, \[N \leq 2 Q^{3}=2(\log N)^{3 B},\]
This is impossible for \(N\) sufficiently large.
Therefore, the major arcs \(\mathfrak{M}(q, a)\) are pairwise disjoint for large \(N\). The set of major arcs is \[\mathfrak{M}=\bigcup_{q=1}^{Q} \bigcup_{\substack{a=0 \\(a, q)=1}}^{q} \mathfrak{M}(q, a) \subseteq[0,1],\] and the set of minor arcs is \[\mathfrak{m}=[0,1] \backslash \mathfrak{M},\]
We consider a weighted sum over the representations of \(N\) as a sum of three primes: \[R(N)=\sum_{p_{1}+p_{2}+p_{1}=N} \log p_{1} \log p_{2} \log p_{3}\]
Vinogradov obtained an asymptotic formula for \(R(N)\), from which the ternary Goldbach problem will follow by an elementary argument.
We can use the circle method to express the representation function \(R(N)\) as the integral of a trigonometric polynomial over the major and minor arcs. Let \[F(\alpha)=\sum_{p \leq N}(\log p) e(p \alpha).\]
This exponential sum over primes is the generating function for \(R(N)\): \[\begin{aligned} R(N) & =\sum_{p_{1}+p_{2}+p_{3}=N} \log p_{1} \log p_{2} \log p_{3}\\ &=\int_{0}^{1} F(\alpha)^{3} e(-N \alpha) d \alpha \\ & =\int_{\mathfrak{M}} F(\alpha)^{3} e(-N \alpha) d \alpha+\int_{\mathfrak{m}} F(\alpha)^{3} e(-N \alpha) d \alpha. \end{aligned}\]
The main term in Vinogradov’s theorem will come from the integral over the major arcs, and the integral over the minor arcs will be negligible.
Just as in the Hardy–Littlewood asymptotic formula, the integral over the major arcs in Vinogradov’s theorem is (except for a small error term) the product of the singular series \(\mathfrak{S}(N)\) and an integral \(J(N)\). In this case, the integral \(J(N)\) is very easy to evaluate.
Let \[u(\beta)=\sum_{m=1}^{N} e(m \beta).\]
Then \[J(N)=\int_{-1 / 2}^{1 / 2} u(\beta)^{3} e(-N \beta) d \beta=\frac{N^{2}}{2}+O(N).\]
Proof. The number of representations of \(N\) as the sum of three positive integers is \[\begin{aligned} J(N) & =\int_{-1 / 2}^{1 / 2} u(\beta)^{3} e(-N \beta) d \beta \\ & =\int_{-1 / 2}^{1 / 2} \sum_{m_{1}=1}^{N} \sum_{m_{2}=1}^{N} \sum_{m_{3}=1}^{N} e\left(\left(m_{1}+m_{2}+m_{3}-N\right) \beta\right) d \beta \\ & =\binom{N-1}{2} =\frac{N^{2}}{2}+O(N). \end{aligned}\]$$\tag*{$\blacksquare$}$$
If \(q \in \mathbb Z_+\) and \((q, a)=1\), then, for any \(C>0\), one has \[\vartheta(x ; q, a)=\sum_{\substack{p \leq x \\ p=a(\bmod q)}} \log p=\frac{x}{\varphi(q)}+O\left(\frac{x}{(\log x)^{C}}\right)\] for all \(x \geq 2\), where the implied constant depends only on \(C\).
Let \[F_{x}(\alpha)=\sum_{p=1}(\log p) e(p \alpha).\]
Let \(B, C\in\mathbb R_+\). If \(1 \leq q \leq Q=(\log N)^{B}\) and \((q, a)=1\), then \[F_{x}(a / q)=\frac{\mu(q)}{\varphi(q)} x+O\left(\frac{Q N}{(\log N)^{C}}\right)\] for \(1 \leq x \leq N\), where the implied constant depends only on \(B\) and \(C\).
Remember that \[u(\beta)=\sum_{m=1}^{N} e(m \beta).\]
Let \(B, C\in\mathbb R_+\) be such that \(C>2 B\). If \(\alpha \in \mathfrak{M}(q, a)\) and \(\beta=\alpha-a / q\), then \[F(\alpha)=\frac{\mu(q)}{\varphi(q)} u(\beta)+O\left(\frac{Q^{2} N}{(\log N)^{C}}\right),\] and \[F(\alpha)^{3}=\frac{\mu(q)}{\varphi(q)^{3}} u(\beta)^{3}+O\left(\frac{Q^{2} N^{3}}{(\log N)^{C}}\right),\] where the implied constants depend only on \(B\) and \(C\).
For any \(B, C, \varepsilon\in\mathbb R_+\) with \(C>2 B\), the integral over the major arcs is \[\begin{aligned} \int_{\mathfrak{M}} F(\alpha)^{3} e(-N \alpha) d \alpha &=\mathfrak{S}(N) \frac{N^{2}}{2}\\ &+O\left(\frac{N^{2}}{(\log N)^{(1-\varepsilon) B}}\right)+O\left(\frac{N^{2}}{(\log N)^{C-5 B}}\right), \end{aligned}\] where the implied constants depend only on \(B, C\), and \(\varepsilon\).
Recall that \[F(\alpha)=\sum_{p \leq N}(\log p) e(p \alpha).\]
If \[\left|\alpha-\frac{a}{q}\right| \leq \frac{1}{q^{2}},\] where \(a, q\in\mathbb Z\) are such that \(1 \leq q \leq N\) and \((a, q)=1\), then \[|F(\alpha)| =O\bigg(\left(\frac{N}{q^{1 / 2}}+N^{4 / 5}+N^{1 / 2} q^{1 / 2}\right)(\log N)^{4}\bigg).\]
For \(u \geq 1\), let \[M_{u}(k)=\sum_{\substack{d, k \\ d \leq u}} \mu(d) .\]
Let \(\Phi(k, l)\) be an arithmetic function of two variables. Then \[\sum_{u<l \leq N} \Phi(1, l)+\sum_{u<k \leq N} \sum_{u<l \leq \frac{N}{k}} M_{u}(k) \Phi(k, l)=\sum_{d \leq u} \sum_{u<l \leq \frac{N}{d}} \sum_{m \leq \frac{N}{ld}} \mu(d) \Phi(d m, l).\]
Proof. Hint: evaluate the sum \[S=\sum_{k=1}^{N} \sum_{u<l \leq \frac{N}{k}} M_{u}(k) \Phi(k, l)\] in two different ways.$$\tag*{$\blacksquare$}$$
Let \(\Lambda(l)\) be the von Mangoldt function. For every \(\alpha\in\mathbb R\), one has \[F(\alpha)=\sum_{p \leq N}(\log p) e(p \alpha)=S_{1}-S_{2}-S_{3}+O\big(N^{1 / 2}\big),\] where \[\begin{aligned} & S_{1}=\sum_{d \leq N^{2 / 5}} \sum_{l \leq \frac{N}{d}} \sum_{m \leq \frac{N}{l d}} \mu(d) \Lambda(l) e(\alpha d l m), \\ & S_{2}=\sum_{d \leq N^{2 / 5}} \sum_{l \leq N^{2 / 5}} \sum_{m \leq \frac{N}{ld}} \mu(d) \Lambda(l) e(\alpha d l m), \end{aligned}\] and \[S_{3}=\sum_{k>N^{2 / S}} \sum_{N^{2 / 5}<l \leq N / k} M_{N^{2 / 5}}(k) \Lambda(l) e(\alpha k l).\]
Proof. Apply Vaughan’s identity with \(u=N^{2/5}\) and \(\Phi(k, l)=\Lambda(l)e(\alpha kl)\), and note \[\sum_{u<l \leq N} \Phi(1, l)=F(\alpha)+O\big(N^{1 / 2}\big). \]$$\tag*{$\blacksquare$}$$
If \[\left|\alpha-\frac{a}{q}\right| \leq \frac{1}{q^{2}},\] where \(1 \leq q \leq N\) and \((a, q)=1\), then \[\left|S_{1}\right| =O\bigg(\left(\frac{N}{q}+N^{2 / 5}+q\right)(\log N)^{2} \bigg).\]
If \[\left|\alpha-\frac{a}{q}\right| \leq \frac{1}{q^{2}},\] where \(1 \leq q \leq N\) and \((a, q)=1\), then \[\left|S_{2}\right| =O\bigg(\left(\frac{N}{q}+N^{4 / 5}+q\right)(\log N)^{2}\bigg).\]
If \[\left|\alpha-\frac{a}{q}\right| \leq \frac{1}{q^{2}},\] where \(1 \leq q \leq N\) and \((a, q)=1\), then \[\left|S_{3}\right| =O\bigg(\left(\frac{N}{q^{1 / 2}}+N^{4 / 5}+N^{1 / 2} q^{1 / 2}\right)(\log N)^{4}\bigg).\]
For any \(B>0\), we have \[\Big|\int_{\mathfrak{m}} F(\alpha)^{3} e(-\alpha N) d \alpha\Big| =O \bigg(\frac{N^{2}}{(\log N)^{(B / 2)-5}}\bigg).\] where the implied constant depends only on \(B\).
Recall that \(Q=(\log N)^B\). Let \(\alpha \in \mathfrak{m}=[0,1] \backslash \mathfrak{M}\). By Dirichlet’s theoremfor any real number \(\alpha\) there exists a fraction \(a / q \in[0,1]\) with \(1 \leq q \leq N / Q\) and \((a, q)=1\) such that \[\left|\alpha-\frac{a}{q}\right| \leq \frac{Q}{q N} \leq \min \left(\frac{Q}{N}, \frac{1}{q^{2}}\right).\]
If \(q \leq Q\), then \(\alpha \in \mathfrak{M}(q, a) \subseteq \mathfrak{M}\), which is false. Therefore, \[Q<q \leq \frac{N}{Q}.\]
Then by the previous theorem we obtain \[\begin{aligned} |F(\alpha)| & =O\bigg(\left(\frac{N}{q^{1 / 2}}+N^{4 / 5}+N^{1 / 2} q^{1 / 2}\right)(\log N)^{4}\bigg) \\ & =O\bigg(\left(\frac{N}{(\log N)^{B / 2}}+N^{4 / 5}+N^{1 / 2}\left(\frac{N}{(\log N)^{B}}\right)^{1 / 2}\right)(\log N)^{4}\bigg) \\ & =O\bigg(\frac{N}{(\log N)^{(B / 2)-4}}\bigg). \end{aligned}\]
Since \[\vartheta(N)=\sum_{p \leq N} \log p = O(N),\] we have \[\int_{0}^{1}|F(\alpha)|^{2} d \alpha=\sum_{p \leq N}(\log p)^{2} \leq \log N \sum_{p \leq N} \log p = O(N \log N).\]
Thus \[\begin{aligned} \int_{\mathfrak{m}}|F(\alpha)|^{3} d \alpha & =O \bigg(\sup \{|F(\alpha)|: \alpha \in \mathfrak{m}\} \int_{\mathfrak{m}}|F(\alpha)|^{2} d \alpha \bigg)\\ & =O\bigg( \frac{N}{(\log N)^{(B / 2)-4}} \int_{0}^{1}|F(\alpha)|^{2} d \alpha \bigg)\\ & =O \bigg(\frac{N^{2}}{(\log N)^{(B / 2)-5}}\bigg). \end{aligned}\] This completes the proof. $$\tag*{$\blacksquare$}$$
Let \(\mathfrak{S}(N)\) be the singular series for the ternary Goldbach problem, i.e. \[\mathfrak{S}(N)=\prod_{p}\left(1+\frac{1}{(p-1)^{3}}\right) \prod_{p \mid N}\left(1-\frac{1}{p^{2}-3 p+3}\right).\]
For all sufficiently large odd integers \(N\) and for every \(A>0\), we have \[R(N)=\sum_{p_{1}+p_{2}+p_{1}=N} \log p_{1} \log p_{2} \log p_{3}=\mathfrak{S}(N) \frac{N^{2}}{2}+O\left(\frac{N^{2}}{(\log N)^{A}}\right),\] where the implied constant depends only on \(A\).
In particular, we have \[r(N)=\sum_{p_{1}+p_{2}+p_{3}=N} 1=\mathfrak{S}(N) \frac{N^{2}}{2(\log N)^{3}}\left(1+O\left(\frac{\log \log N}{\log N}\right)\right).\]