Ternary Goldbach Conjecture

The Ternary Goldbach Conjecture states that every odd integer is the sum of three primes. In 1937, I.M. Vinogradov proved this statement for large odd primes by using Hardy and Littlewood’s circle method and the results related to the distribution of primes in arithmetic progression.
Theorem. There exists \(\mathfrak{S}: \mathbb{N} \rightarrow \mathbb{R}\) and positive constants \(c_{1}\) and \(c_{2}\) such that \[c_{1}<\mathfrak{S}(N)<c_{2}\] for all sufficiently large odd integers \(N\), and let

\[r(N)=\sum_{P_{1}+P_{2}+P_{3}=N} 1\]

then

\[r(N)=\mathfrak{S}(N) \frac{N^{2}}{2(\log N)^{3}}\left(1+O\left(\frac{\log \log N}{\log N}\right)\right).\]

Instead of directly estimating for \(r(N)\), we introduce

\[R(N)=\sum_{p_{1}+p_{2}+p_{3}=N} \log p_{1} \log p_{2} \log p_{3}\]

Theorem. For all sufficiently large odd integers and for every \(A>0\), \[R(N)=\mathfrak{S}(N) \frac{N^{2}}{2}+O\left(\frac{N^{2}}{(\log N)^{4}}\right)\] where the implied constant only depends on \(A\).
From the asymptotic formula for \(R(N)\), it follows the previous theorem, as \(R(N)\) is the sum of \((\log N)^{3} r(N)\) with an error term (which can be computed, it is \(O\left(\frac{N^{2} \log \log N}{\log N}\right)\) ).
By Hardy and Littlewood’s method, we rewrite \[R(N)=\int_{0}^{1} F(\alpha)^{3} e(-N \alpha) d \alpha\] where

\[F(\alpha)=\sum_{p<N}(\log p) e(p \alpha)\]

Let \(B>0\) and \(Q=(\log N)^{B}\), we define the major arcs \(\mathfrak{M}\) and the minor arcs \(\mathfrak{m}\) by the following \[\mathfrak{M}=\bigcup_{q=1}^{Q} \bigcup_{\substack{a=0 \\(a, q)=1}}^{q}\left\{\alpha \in[0,1]:\left|\alpha-\frac{a}{q}\right| \leqslant \frac{Q}{N}\right\}=\bigcup_{q=1}^{Q} \bigcup_{\substack{a=0 \\(a, q)=1}}^{q} \mathfrak{M} (q, a)\] and \[\mathfrak{m}=[0,1] \backslash \mathfrak{M} .\] Since \(\frac{1}{Q^{2}} \leq \frac{2 Q}{N}\) is impossible for \(N\) large, then it is verified that each interval in the major arcs does not overlap.
We first estimate the integral over the major arc.
Let \(u(\beta)=\sum_{m=1}^{N} e(m \beta)\), and \(\beta=\alpha-\frac{a}{q},|\beta| \leqslant \frac{Q}{N}\), then \[\begin{aligned} F(\alpha)-\frac{\mu(q)}{\varphi(q)} u(\beta) & =\sum_{p \leq N}(\log p) e(p \alpha)-\frac{\mu(q)}{\varphi(q)} \sum_{m=1}^{N} e(m \beta) \\ & =\sum_{m=1}^{N}\left(\lambda(m) e\left(\frac{m a}{q}\right)-\frac{\mu(q)}{\varphi(q)}\right) e(m \beta) \end{aligned}\] where \[\lambda(m)= \begin{cases}\log p & \text { if } m \text { is a prime, } \\ 0 & \text { otherwise. }\end{cases}\] Then, for \(1 \leqslant x \leqslant N\), let \[F_{x}\left(\frac{a}{q}\right)=\sum_{m \leq x} \lambda(m) e\left(\frac{m a}{q}\right)=\sum_{\substack{ r=1 \\(r, q)=1}}^{q} \sum_{\substack{p \leq x \\ p \ \equiv \ r (mod \ q)}}(\log p) e\left(\frac{r a}{q}\right)+O(\log q)\] by the Siegel-Walfisz theorem, for any \(C>0\), we have \[\begin{aligned} F_{x}\left(\frac{a}{q}\right) & =\sum_{r=1}^{q} e\left(\frac{r a}{q}\right)\left(\frac{x}{\varphi(q)}+O\left(\frac{x}{(\log x)^{C}}\right)\right)+O(\log Q) \\ & =\frac{\mu(q)}{\varphi(q)} x+O\left(\frac{Q N}{(\log N)^{C}}\right) \end{aligned}\] It follows that \[\begin{aligned} A(x) & =\sum_{m \leq x}\left(\lambda(m) e\left(\frac{m a}{q}\right)-\frac{\mu(q)}{\varphi(q)}\right) \\ & =O\left(\frac{\varphi N}{(\log N)^{c}}\right) \end{aligned}\] Then, by Abel’s summation, we obtain \[\begin{aligned} F(\alpha)-\frac{\mu(q)}{\varphi(q)} & =A(N) e(N \beta)-2 \pi i \beta \int_{1}^{N} A(x) e(x \beta) d x \\ & \ll \frac{Q^{2} N}{(\log N)^{C}}. \end{aligned}\] Let \(\quad v(\beta)=\frac{\mu(q)}{\varphi(q)} u(\beta)\), since \(|u(\beta)| \leqslant N, \quad F(\alpha) \ll \log N \pi(N) \ll N\), \[\begin{aligned} F(\alpha)^{3}-v(\beta)^{3} & =(F(\alpha)-v(\beta))\left(F(\alpha)^{2}+F(\alpha) v(\beta)+v(\beta)^{2}\right) \\ & \ll \frac{Q^{2} N^{3}}{(\log N)^{C}}. \end{aligned}\]
Theorem. For any positive numbers \(B, C\), and \(\varepsilon\), with \(C>2 B\), the integral over the major arcs is \[\int_{m} F(\alpha)^{3} e(-N \alpha) d \alpha=\mathfrak{S}(N) \frac{N^{2}}{2}+O\left(\frac{N^{2}}{(\log N)^{(1-\varepsilon) B}}\right)+O\left(\frac{N^{2}}{(\log N)^{C-5 B}}\right)\] where the implied constant depends on \(B, C\), and \(\varepsilon\).
Proof: The length of \(\mathfrak{M} (q, a) \ll \frac{Q}{N}\), and by previous, we have \[\begin{aligned} \int_{m}\left(F(\alpha)^{3}-v(\beta)^{3}\right) e(-N \alpha) d \alpha & \ll \sum_{q \leq Q} \sum_{\substack{a=0 \\ (a,q)=1}}^{q} \frac{Q}{N} \frac{Q^{2} N^{3}}{(\log N)^{C}} \\ & \ll \frac{Q^{5} N^{2}}{(\log N)^{C}} \ll \frac{N^{2}}{(\log N)^{C-5 B}} \end{aligned}\] Since \(\alpha=\beta+\frac{a}{q}\) with \(|\beta| \leqslant \frac{Q}{N}\), then \[\begin{aligned} \int_\mathfrak{M} v(\beta)^{3} e(-N \alpha) d \alpha & =\sum_{q \leq Q} \frac{M(q)}{\varphi(q)^{3}} \sum_{\substack{a=1 \\ (a, q)=1}}^{q} e\left(-\frac{N a}{q}\right) \int_{-Q / N}^{Q / N} u(\beta)^{3} e(-N \beta) d \beta \\ & =\mathfrak{S}(N, Q) \int_{-Q / N}^{Q / N} u(\beta)^{3} e(-N \beta) d \beta \end{aligned}\] Since for \(|\beta| \leq \frac{1}{2}\), we have \(|u(\beta)| \ll|\beta|^{-1}\), it gives \[\begin{aligned} \int_{-Q / N}^{Q / N} u(\beta)^{3} e(-N \beta) d \beta & =\int_{-\frac{1}{2}}^{\frac{1}{2}} u(\beta)^{3} e(-N \beta) d \beta+O\left(N^{2} Q^{-2}\right) \\ & =\binom{N-1}{2}+O\left(N^{2} Q^{-2}\right) \\ & =\frac{N^{2}}{2}+O\left(N^{2} Q^{-2}\right) \end{aligned}\] Let \(\mathfrak{S}(N)=\sum_{q=1}^{\infty} \frac{\mu(q) C_q(N)}{\varphi(q)^{3}}\), where \(C_{q}(N)=\sum_{\substack{a=1 \\(a, q)=1}}^{q} e\left(\frac{a N}{q}\right)\).
\(C_q(N)\) is known as Ramanujan’s sum which is multiplicative \(\left|C_{q}(N)\right| \leqslant \varphi(q)\) and \[C_{p}(N)= \begin{cases}p-1 & \text { if } p \mid N \\ -1 & \text { otherwise } .\end{cases}\] Note we can rewrite \(\mathfrak{S}(N)\) as an infinite product of primes as the summand is a multiplicative function, \[\begin{aligned} \mathfrak{S}(N) & =\prod_{P}\left(1+\sum_{j=1}^{\infty} \frac{\mu\left(p^{j}\right) C_{p^{j}}(N)}{\varphi\left(p^{j}\right)^{3}}\right) \\ & =\prod_{p}\left(1-\frac{C_{p}(N)}{\varphi(p)^{3}}\right) \\ & =\prod_{p}\left(1+\frac{1}{(p-1)^{3}}\right) \prod_{p \mid N}\left(1-\frac{1}{p^{2}-3 p+3}\right) \end{aligned}\] and so there exists \(c_{1}\) and \(c_{2}\) such that \[c_{1}<\mathfrak{S}(N)<c_{2}\] for all positive integers \(N\). By the fact that \(\varphi(q)>q^{1-\varepsilon}\), it yields \[\mathfrak{S}(N)-\mathfrak{S}(N,Q) \ll \sum_{q>Q} \frac{1}{\varphi(q)} \ll \frac{1}{Q^{1-\varepsilon}}.\] By inserting the calculation of the singular series into the previous integral, it completes the proof of the theorem.
As for the integral over the minor arcs, the proof is based on Vaughan. which involves the idea of bilinear form and the application of Weyl’s inequality.

Lemma (Vaughan’s identity) For \(u \geqslant 1\), let \[M_{u}(k)=\sum_{\substack{d\mid k\\ d\leq u}} \mu(d)\] Let \(\Phi(k, l)\) be an arithmetic function of two variables. Then \[\sum_{u<l \leqslant N} \Phi(1, l)+\sum_{u<k \leqslant N} \sum_{u<l<\frac{N}{k}} M_{u}(k) \Phi(k, l)=\sum_{d \leqslant u} \sum_{u<l \leqslant \frac{N}{d}} \sum_{m \leq \frac{N}{ld}} \mu(d) \Phi(d m, l)\] Proof: This follows directly from evaluating the sum \[S=\sum_{k=1}^{N} \sum_{u<l \leq \frac{N}{k}} M_{u}(k) \Phi(k, l)\] in two different ways.
By inserting \(u=N^{2 / 5}\) and \(\Phi(k, l)=\Lambda(l) e(\alpha k l)\), we get \[\begin{aligned} \sum_{u^{k} l \leq N} \Phi(1, l) & =\sum_{N^{2/5} \leq l \leq N} \Lambda(l) e(\alpha l) \\ & =F(\alpha)+O\left(\sum_{\substack{p^{k} \leq N \\ k \leq 2}} \log p\right)+O\left(N^{2 / 5} \log N\right) \\ & =F(\alpha)+O\left(\pi\left(N^{\frac{1}{2}}\right) \log N\right) \\ & =F(\alpha)+O\left(N^{\frac{1}{2}}\right) \end{aligned}\] Let \[S_{1}=\sum_{d \leq N^{2 / 5}} \sum_{l \leqslant \frac{N}{d}} \sum_{m \leqslant \frac{N}{l d}} \mu(d) \Lambda(l) e(\alpha d l m),\] \[S_{2}=\sum_{d \leqslant {N}^{2 / 5}} \sum_{l \leqslant {N}^{2 / 5}} \sum_{m \leqslant \frac{N}{l d}} \mu(d) \Lambda(l) e(\alpha d l m),\] and \[S_{3}=\sum_{k>N^{\frac{2}{5}}} \sum_{N^{2 / 5}<l \leqslant N / k} M_{N^{2 / 5}}(k) \Lambda(l) e(\alpha k l) .\] Then, Vaughan’s identity yields that \[F(\alpha)=S_{1}-S_{2}-S_{3}+O\left(N^{\frac{1}{2}}\right).\] For \(S_{1}\) and \(S_{2}\), since \(\sum_{l \mid r} \Lambda(l)=\log r\) and the summation of the möbius function over \(d\) can be bounded with regard to the function as 1. Then, by applying the results related to Weyl’s inequality, we establish the estimation of \(S_{1}\) and \(S_{2}\). However, \(S_{3}\) is more complicated.
Let \(lm=r\), and \[\begin{aligned} S_{1} & =\sum_{d \leqslant u} \sum_{r<N / d} \mu(d) e(\alpha d r) \sum_{l \mid r} \Lambda(l) \\ & =\sum_{d \leqslant u} \mu(d) \sum_{r<N / d} e(\alpha d r) \log r \\ & \ll \sum_{d \leqslant u}\left|\sum_{r<N / d} e(\alpha d r) \log r\right|. \end{aligned}\] The inner part is \[\begin{aligned} \sum_{r<N / d} e(\alpha d r) \log r & =\sum_{r<N / d} e(\alpha d r) \int_{1}^{r} \frac{d x}{x} \\ & =\sum_{s=2}^{[N / d]} \int_{s-1}^{s}\left(\sum_{r=s}^{[N / d]} e(\alpha d r)\right) \frac{d x}{x}. \end{aligned}\]
Lemma. Let \(\alpha\) be a real number, if \[\left|\alpha-\frac{a}{q}\right| \leqslant \frac{1}{q^{2}}\] where \(q \geq 1\) and \((a, q)=1\), then for any \(U \geq 1\) and positive integer \(n\) we have \[\sum_{1 \leqslant k \leqslant u} \min \left(\frac{n}{k},\|\alpha k\|^{-1}\right) \ll\left(\frac{n}{q}+u+q\right) \log 2 q U.\]
If \(\alpha \in \mathfrak{m}\), and \[\sum_{r=1}^{[N / d]} e(\alpha d r) \log r \ll \min \left(\frac{N}{d},\|\alpha d\|^{-1}\right) \log N\] by lemma, it yields that \[S_{1} \ll\left(\frac{N}{q}+N^{2 / 5}+q\right)(\log N)^{2}.\] Similarly, for \(S_{2}\), by making substitution \(k=d l\), we obtain \[\begin{aligned} S_{2} & =\sum_{k \leqslant N^{4 / 5}}\left(\sum_{m \leqslant N / k} e(\alpha k m)\right)\left(\sum_{\substack{k=d l \\ d, l \leq N^{2 / 5}}} \mu(d) \Lambda(l)\right) \\ & \ll\left(\frac{N}{q}+N^{4 / 5}+q\right)(\log N)^{2} \end{aligned}\] As for \(S_{3}\), we split up the range for \(k\). Let \[\mathscr{A}=\left\{2^{i} u: i=1,2, \ldots, 2^{i} u \leqslant N\right\}\] so that \[|\mathscr{A}| \ll \log N\] Then \[S_{3}=\sum_{K \in\mathscr{A}} S_{3, k}\] where

\[S_{3, k}=\sum_{K < k \leqslant 2 K} M_{u}(k) \sum_{u<l \leqslant N / k} \Lambda(l) e(\alpha k l)\] By the Cauchy-Schwarz inequality, we estimate the sum, respectively.
Note \(|\operatorname{M}_u(k)| \leqslant d(k)\), then this yields \[\sum_{k<k \leqslant 2 k}|M_u(k)|^{2} \ll \quad K(\log N)^{3} .\] As for the other sum, we have \[\begin{aligned} & \sum_{k<k \leq 2 k}\left|\sum_{u<l \leq N / k} \Lambda(l) e(\alpha k l)\right|^{2} \\ = & \sum_{u<l\leq \frac{N}{K}} \sum_{u<m<\frac{N}{K}} \Lambda(l) \Lambda(m) \sum_{k \in I(l, m)} e(\alpha k(l-m)) \end{aligned}\] where \[I(l, m)=\left(K, \min \left(2 K, \frac{N}{l}, \frac{N}{m}\right) .\right.\] Note \(|I(l, m)| \leq K\), then \[\sum_{k \in I(l, m)} e(\alpha k(l-m)) \ll \min \left(K,\|\alpha(l-m)\|^{-1}\right).\] Since \(0 \leqslant \Lambda(l), \Lambda(m) \leqslant \log N\) for \(u<l, m<\frac{N}{K}\), and let \(j=l-m\), and by the previous lemma over the sum of \(j\) we have, \[\begin{aligned} \sum_{k \leqslant k \leqslant 2 k}\left|\sum_{k<l \leqslant N / K} N(l) e(\alpha k l)\right|^{2} & \ll(\log N)^{2} \frac{N}{K} \sum_{j \leqslant \frac{N}{K}} \min \left(\frac{N}{j},\|\alpha j\|^{-1}\right) \\ & \ll \frac{N}{K}\left(\frac{N}{q}+\frac{N}{K}+q\right)(\log N)^{3}. \end{aligned}\] Thus, \[S_{3, k} \ll(\log N)^{3}\left(\frac{N}{q^{1 / 2}}+N^{4 / 5}+q^{\frac{1}{2}} N^{\frac{1}{2}}\right)\] which follows that \[S_{3} \ll(\log N)^{4}\left(\frac{N}{q^{1 / 2}}+N^{4 / 5}+q^{\frac{1}{2}} N^{\frac{1}{2}}\right) .\] Therefore, for \(\alpha \in \mathfrak{m}\), we have \[F(\alpha) \ll \frac{N}{(\log N)^{\frac{B}{2}-4}}\] and \[\begin{aligned} \int_\mathfrak{m} F(\alpha)^{3} e(-N \alpha) d \alpha &\ll \frac{N}{(\log N)^{\frac{B}{2}-4}} \int_{0}^{1}|F(\alpha)|^{2} d \alpha\\ & =\frac{N}{(\log N)^{\frac{B}{2}-4}} \sum_{p \leqslant N}(\log p)^{2} \\ & \ll \frac{N^{2}}{(\log N)^{\frac{B}{2}-5}} \end{aligned}\] Therefore, with an appropriate choice of \(A\) based on \(B, C\), and \(\varepsilon\), we complete the proof.