---
abstract: 'Suppose that all primes are colored with $k$ colors. Then there exist monochromatic primes $p_1,p_2,p_3$ such that $p_1+p_2=p_3+1$.'
address: ' Department of Mathematics, Shanghai Jiaotong University, Shanghai 200240, People’s Republic of China'
author:
- Hongze Li
- Hao Pan
title: 'A Schur-type addition theorem for primes'
---

[^1]

Introduction
============

In [@GreenTao], Green and Tao proved a celebrated result that the primes contain arbitrarily long non-trivial arithmetic progressions. In fact, they proved a Szemerédi-type [@Szemeredi75] result for primes:

Thus if all primes are colored with $k$ colors, then there exist arbitrarily long monochromatic arithmetic progressions. This is a van der Waerden-type [@vanderWaerden27] theorem for primes. (The well-known van der Waerden theorem states that for any $k$-coloring of all positive integers, there exist arbitrarily long monochromatic arithmetic progressions.)

On the other hand, Schur’s theorem [@Schur16] is another famous result in the Ramsey theory for integers. Schur’s theorem asserts that for any $k$-coloring of all positive integers, there exist monochromatic $x,y,z$ such that $x+y=z$. In this paper, we shall prove a Schur-type theorem for primes.

\[t1\] Suppose that all primes are arbitrarily colored with $k$ colors. Then there exist monochromatic primes $p_1,p_2,p_3$ such that $p_1+p_2=p_3+1$.

Furthermore, motivated by the Green-Tao theorem and Theorem \[t1\], we propose the following conjecture:

Suppose that all primes are colored with $k$ colors. Then for arbitrary $l\geq 3$, there exist monochromatic primes $p_0,p_1,p_2,\ldots,p_l$ such that $p_1,\ldots,p_l$ form an arithmetic progression with the difference $p_0-1$.

Theorem \[t1\] will be proved in the next section. And our proof uses a variant of Green’s method [@Green05] in his proof of Roth’s theorem for primes.

Proof of Theorem \[t1\]
=======================

\[schur\] Suppose that the set $\{1,2,\ldots,n\}$ is split into $A_1\cup A_2\cup\cdots\cup A_k$. Then there exists a constant $C_1(k)>0$ such that $$\sum_{1\leq i\leq k}|\{(x,y,z):\, x,y,z\in A_i, x+y=z\}|\geq
C_1(k)n^2$$ if $n$ is sufficiently large.

This result is not new. In fact, Robertson and Zeilberger [@RobertsonZeilberger98], Schoen [@Schoen99] had showed that if the integers from 1 to $n$ are colored with two colors, then there exist at least $(1/22-\epsilon)n^2$ monochromatic Schur triples $\{x,y,x+y\}$. Furthermore, Robertson and Zeilberger [@RobertsonZeilberger98] also claimed that for any $k$-coloring of $\{1,\ldots,n\}$, the number of monochromatic Schur triples is greater than $$\big(\frac{1}{2^{2k-3}11}-\epsilon\big)n^2.$$

However, for the sake of completeness, here we give a proof of Lemma \[schur\]. Suppose that $1,2,\ldots,n$ are colored with $k$ colors. Let $G$ be a complete graph with the vertex set $V=\{v_0,v_1,\ldots,v_n\}$. Then we $k$-color all edges of $G$ by giving the edge $v_sv_t$ the color of $t-s$ for every $0\leq
s<t\leq n$. Clearly for $0\leq r<s<t\leq n$, three vertices $v_r,v_s,v_t$ form a monochromatic triangle if and only if $\{s-r,t-s,t-r\}$ is a monochromatic Schur triple. And it is easy to see that one monochromatic Schur triple is corresponding to at most $n$ monochromatic triangles. Hence Lemma \[schur\] immediately follows from the next lemma:

Let $G$ be a complete graph with $n$ vertices. If all edges of $G$ are colored with $k$ colors, then there exist at least $C_1'(k)n^3$ monochromatic triangles provided that $n$ is sufficiently large, where $C_1'(k)>0$ is a constant only depending on $k$.

Since $G$ is a complete graph, $G$ contains $\binom{n}{3}$ triangles. We use induction on $k$. There is nothing to do when $k=1$. Assume that $k\geq 2$ and our assertion holds for any smaller value of $k$. Suppose that the vertex set $V$ of $G$ is $\{v_1,\ldots,v_n\}$. Then for every $1\leq s\leq n$, by the pigeonhole principle, there exist vertices $v_{t_{s,1}},\ldots,v_{t_{s,\ceil{n/k}}}$ and $1\leq c_s\leq k$ such that the edge $v_sv_{t_{s,1}},\ldots,v_sv_{t_{s,\ceil{n/k}}}$ are colored with the $c_s$-th color, where $\ceil{x}$ denotes the smallest integer not less than $x$. Let us consider the $\binom{\ceil{n/k}}{2}$ edges between $v_{t_{s,1}},\ldots,v_{t_{s,\ceil{n/k}}}$. Suppose that at most $(C_1'(k-1)/2k^3)n^2$ of these edges are colored with the $c_s$-th color. Then by the induction hypothesis on $k-1$, the remainder edges form at least $$C_1'(k-1)(n/k)^3-\frac{C_1'(k-1)}{2k^3}n^3=\frac{C_1'(k-1)}{2k^3}n^3$$ monochromatic triangles, since one edge belongs to at most $n$ triangles.

Then we may assume that for each $1\leq s\leq n$, there exist at least $(C_1'(k-1)/2k^3)n^2$ edges between $v_{t_{s,1}},\ldots,v_{t_{s,\ceil{n/k}}}$ are colored with the $c_s$-th color. Thus we get at least $(C_1'(k-1)/2k^3)n^2$ monochromatic triangles containing the vertex $v_s$. And there are totally at least $$\frac{C_1'(k-1)}{6k^3}n^3$$ monochromatic triangles, by noting that every triangles are counted three times.

Let $A$ be a subset of $\{1,2,\ldots,n\}$ with $|A|\geq
(1-C_1(k)/6)n$. Suppose that $A$ is split into $A_1\cup
A_2\cup\cdots\cup A_k$. Then $$\sum_{1\leq i\leq k}|\{(x,y,z):\, x,y,z\in A_i, x+y=z\}|\geq
\frac{C_1(k)}{2}n^2$$ provided that $n$ is sufficiently large.

Let $\bar{A}=\{1,\ldots,n\}\setminus A$. Then $$\begin{aligned}
&|\{(x,y,z):\, x,y,z\in A_1\cup\bar{A}, x+y=z\}|\\
\leq&|\{(x,y,z):\, x,y,z\in A_1, x+y=z\}|\\
&+|\{(x,y,z):\, \text{one of
}x,y,z\text{ lies in }\bar{A}, x+y=z\}|\\
\leq&|\{(x,y,z):\, x,y,z\in A_1, x+y=z\}|+3|\bar{A}|n.\end{aligned}$$ Hence by Lemma 2.1 we have $$\begin{aligned}
&\sum_{1\leq i\leq k}|\{(x,y,z):\, x,y,z\in A_i, x+y=z\}|\\
\geq&|\{(x,y,z):\, x,y,z\in A_1\cup\bar{A},
x+y=z\}|-3|\bar{A}|n\\
+&\sum_{2\leq i\leq k}|\{(x,y,z):\, x,y,z\in A_i,
x+y=z\}|\\
\geq&\frac{C_1(k)}{2}n^2.\end{aligned}$$

Let $\P$ denote the set of all primes. Assume that $\P=P_1\cup
P_2\cup\cdots\cup P_k$, where $P_i\cap P_j=\emptyset$ for $1\leq
i<j\leq k$. Let $w=w(n)$ be a function tending sufficiently slowly to infinity with $n$ (e.g., we may choose $w(n)=\floor{\frac{1}{4}\log\log n}$), and let $$W=\prod_{\substack{p\in\P\\ p\leqslant w(n)}}p.$$ Clearly we have $W\leqslant\log n$ for sufficiently large $n$. Let $$\kappa=\frac{C_1(k)}{10000k}.$$ In view of the well-known Siegel-Walfisz theorem, we may assume that $n$ is sufficiently large so that $$\sum_{\substack{x\in\P\cap[1,n]\\ x\equiv 1\pmod{W}}}\log
x\geq(1-\kappa)\frac{n}{\phi(W)},$$ where $\phi$ is the Euler totient function. Let $M=n/W$ and $N$ be a prime in the interval $[(2+\kappa)M,(2+2\kappa)M]$. (Thanks to the prime number theorem, such prime $N$ always exists whenever $M$ is sufficiently large.) Define $$\lambda_{b,W,N}(x)=\begin{cases} \phi(W)\log(Wx+b)/WN&\text{ if
}x\leq N\text{ and }Wx+b\text{ is prime},\\
0&\text{otherwise}.
\end{cases}$$ Let $$A_0=\{1\leq x\leq M:\, Wx+1\in\P\}$$ and $$A_i=\{1\leq x\leq M:\, Wx+1\in P_i\}$$ for $1\leq i\leq k$. Define $$a_i(x)=\1_{A_i}(x)\lambda_{1,W,N}(x)$$ for $0\leq i\leq k$, where we set $\1_A(x)=1$ if $x\in A$ and $0$ otherwise. Clearly we have $a_0=a_1+\cdots+a_k$ and $$\sum_{x}a_0(x)=\sum_{1\leq x\leq M}\lambda_{1,W,N}(x)\geq
(1-\kappa)\frac{M}{N}\geq \frac{1}{2}-3\kappa.$$

Below we consider $A_0,A_1,\ldots,A_k$ as the subsets of $\Z_N=\Z/N\Z$. Since $M<N/2$, if there exist $x,y,z\in A_i$ such that $x+y=z$ in $\Z_N$, then we have $p_1+p_2=p_3+1$ in $\Z$, where $p_1=Wx+1\in P_i, p_2=Wy+1\in P_i, p_3=Wz+1\in P_i$. For a complex-valued function $f$ over $\Z_N$, define $\tilde{f}$ by $$\tilde{f}(r)=\sum_{x\in\Z_N}f(x)e(-xr/N),$$ where $e(x)=e^{2\pi\sqrt{-1}x}$. And for two functions $f, g$, define $$(f*g)(x)=\sum_{y\in\Z_N}f(y)g(x-y).$$ It is easy to check that $(f*g)\,\tilde{}=\tilde{f}\tilde{g}$. Let $0<\delta, \epsilon<1/2$ be two sufficiently small real numbers which will be chosen later. Let $$R=\{r\in\Z_N:\, \max_{1\leq i\leq
k}|\tilde{a_i}(r)|\geqslant\delta\}.$$ and $$B=\{x\in\Z_N:\, x\in[-\kappa N,\kappa N],\
\|xr/N\|\leqslant2\epsilon\text{ for all }r\in R\},$$ where $ \|x\|=\min_{z\in\Z}|x-z|$. Here our definition of $B$ is slightly different from Green’s one in [@Green05 Page 1629]. As we shall see later, this modification is the key of our proof.

\[bohr\] $$|B|\geq\epsilon^{|R|}\kappa N.$$

Assume that $R=\{r_1,r_2,\ldots,r_{m}\}$. Let $d$ be the greatest integer not exceeding $1/\epsilon$. Clearly we have $1/d\leq
2\epsilon$ since $\epsilon<1/2$. Let $$G_{t_1,\ldots,t_m}=\{-\kappa N/2\leq x\leq\kappa N/2:\,
t_j/d\leq\{xr_j/N\}<(t_j+1)/d\text{ for }1\leq j\leq m\},$$ where $\{\alpha\}$ denotes the fractional part of $\alpha$. Clearly $$\sum_{0\leq t_1,\ldots,t_m\leq d-1}|G_{t_1,\ldots,t_m}|=\kappa N.$$ Hence there exists a term of $(t_1,\ldots,t_m)$ such that $$|G_{t_1,\ldots,t_m}|\geq d^{-m}\kappa N\geq \epsilon^{m}\kappa N.$$ For any given $x_0\in G_{t_1,\ldots,t_{m}}$, when $x\in
G_{t_1,\ldots,t_{m}}$, we have $x-x_0\in [-\kappa N,\kappa N]$ and $$\|(x-x_0)r_j/N\|\leq|\{xr_j/N\}-\{x_0r_j/N\}|\leq1/d\leq2\epsilon$$ for $1\leq j\leq m$. So $G_{t_1,\ldots,t_{m}}\subseteq x_0+B$. This completes the proof.

\[lambda\] $$\sup_{r\not=0}|\tilde{\lambda}_{b,W,N}(r)|\leq 2\log\log w/w$$ provided that $w$ is sufficiently large.

This is Lemma 6.2 of [@Green05].

Let $\beta=\1_{B}/|B|$ and $a_i'=a_i*\beta*\beta$ for $0\leq i\leq
k$.

\[upper\] Suppose that $\epsilon^{|R|}\geq \kappa^{-2}\log\log
w/w$. Then we have $$\sup_{x\in\Z_N}a_0'(x)\leq\frac{1+3\kappa}{N}.$$

We have $$\begin{aligned}
a_0'(x)=&a_0*\beta*\beta(x)\\
\leq&\lambda_{1,W,N}*\beta*\beta(x)\\
=&N^{-1}\sum_{r\in\Z_N}\tilde{\lambda}_{1,W,N}(r)\tilde{\beta}(r)^2e(xr/N)\\
\leq&N^{-1}\tilde{\lambda}_{1,W,N}(0)\tilde{\beta}(0)^2+N^{-1}\sup_{r\not=0}|\tilde{\lambda}_{1,W,N}(r)|\sum_{r\in\Z_N}|\tilde{\beta}(r)|^2\\
=&N^{-1}\tilde{\lambda}_{1,W,N}(0)+\sup_{r\not=0}|\tilde{\lambda}_{1,W,N}(r)|\sum_{r\in\Z_N}|\beta(r)|^2\\
\leq&\frac{1+\kappa}{N}+\frac{2\log\log w}{w|B|},\end{aligned}$$ where Lemma \[lambda\] is applied in the last step. Thus Lemma \[upper\] immediately follows from Lemma \[bohr\].

\[bourgain\] Let $\rho>2$. For any function $f:\,\Z_N\to\C$, $$\sum_{r\in\Z_N}|(f\lambda_{b,W,N})\,\tilde{}(r)|^\rho\leq
C_2(\rho)\bigg(\sum_{x=1}^N|f(x)|^2\lambda_{b,W,N}(x)\bigg)^{\frac{\rho}{2}}$$ where $C_2(\rho)$ is a constant only depending on $\rho$.

This is an immediate consequence of Theorem 2.1 and Lemma 6.5 of [@Green05].

By Lemma \[bourgain\], we have $$\sum_{r\in\Z_N}|\tilde{a}_i(r)|^\rho\leq C_2(\rho)$$ for $\rho>2$ and $1\leq i\leq k$. In particular, $$\sum_{r\in R}\delta^3\leq
\sum_{r\in\Z_N}\bigg(\sum_{i=1}^k|\tilde{a}_i(r)|^3\bigg)\leq
C_2(3)k,$$ which implies that $|R|\leq C_2(3)\delta^{-3}k$.

\[beta\] For each $r\in R$, $$|1-\tilde{\beta}(r)^4\tilde{\beta}(-r)^2|\leq 384\epsilon^2.$$

By the definition of $B$, we have $$\begin{aligned}
|1-\tilde{\beta}(r)|=\frac{1}{|B|}\bigg|\sum_{x\in
B}(1-e(-xr/N))\bigg| \leq4\pi\sup_{x\in B}\|xr/N\|^2\leq
64\epsilon^2.\end{aligned}$$ So $$\begin{aligned}
|1-\tilde{\beta}(r)^4\tilde{\beta}(-r)^2|=&\bigg|\sum_{j=0}^3\tilde{\beta}(r)^j(1-\tilde{\beta}(r))+\tilde{\beta}(r)^4\sum_{j=0}^1\tilde{\beta}(-r)^j(1-\tilde{\beta}(-r))\bigg|\\
\leq&384\epsilon^2.\end{aligned}$$ by noting that $|\tilde{\beta}(r)|\leq\tilde{\beta}(0)=1$.

\[difference\] For $1\leq i\leq k$, $$\bigg|\sum_{i=1}^k\sum_{\substack{
x,y,z\in\Z_N\\x+y=z}}a_i(x)a_i(y)a_i(z)-\sum_{i=1}^k\sum_{\substack{
x,y,z\in\Z_N\\x+y=z}}a_i'(x)a_i'(y)a_i'(z)\bigg|\leq
\frac{C_3k^2}{N}(\epsilon^2\delta^{-3}+\delta^{\frac{1}{3}}),$$ where $C_3$ is an absolute constant.

Clearly $$\sum_{\substack{
x,y,z\in\Z_N\\x+y=z}}f_1(x)f_2(y)f_3(z)=N^{-1}\sum_{r\in\Z_N}\tilde{f}_1(r)\tilde{f}_2(r)\tilde{f}_3(-r).$$ Hence $$\begin{aligned}
&\sum_{i=1}^k\sum_{\substack{
x,y,z\in\Z_N\\x+y=z}}a_i(x)a_i(y)a_i(z)-\sum_{i=1}^k\sum_{\substack{
x,y,z\in\Z_N\\x+y=z}}a_i'(x)a_i'(y)a_i'(z)\\
=&N^{-1}\sum_{i=1}^k\sum_{r\in\Z_N}\tilde{a}_i(r)^2\tilde{a}_i(-r)(1-\tilde{\beta}(r)^4\tilde{\beta}(-r)^2).\end{aligned}$$ By Lemma \[beta\], $$\begin{aligned}
&\bigg|\sum_{i=1}^k\sum_{r\in
R}\tilde{a}_i(r)^2\tilde{a}_i(-r)(1-\tilde{\beta}(r)^4\tilde{\beta}(-r)^2)\bigg|\\
\leq&384\epsilon^2k|R|\sup_{r}\max_{1\leq i\leq
k}|\tilde{a}_i(r)|^3\\\leq& 384C_2(3)\epsilon^2\delta^{-3}k^2,\end{aligned}$$ since $|\tilde{a}_i(r)|\leq\tilde{a}_i(0)\leq 1$. On the other hand, by the Hölder inequality, we have $$\begin{aligned}
&\bigg|\sum_{i=1}^k\sum_{r\not\in
R}\tilde{a}_i(r)^2\tilde{a}_i(-r)(1-\tilde{\beta}(r)^4\tilde{\beta}(-r)^2)\bigg|\\
\leq&2\sum_{i=1}^k\sum_{r\not\in
R}|\tilde{a}_i(r)|^2|\tilde{a}_i(-r)|\\
\leq&2\sup_{r\not\in R}\max_{1\leq i\leq
k}|\tilde{a}_i(r)|^{\frac{1}{3}}\bigg(\sum_{i=1}^k\sum_{r}|\tilde{a}_i(r)|^{\frac{5}{2}}\bigg)^{\frac{2}{3}}
\bigg(\sum_{i=1}^k\sum_{r}|\tilde{a}_i(-r)|^{3}\bigg)^{\frac{1}{3}}\\
\leq&2C_2(5/2)^{\frac{2}{3}}C_2(3)^{\frac{1}{3}}\delta^{\frac13}k.\end{aligned}$$ We choose $C_3=384C_2(3)+2C_2(5/2)^{\frac{2}{3}}C_2(3)^{\frac{1}{3}}$, then the Lemma follows.

Define $$X=\{x\in\Z_N:\, a_0'(x)\geq \frac{\kappa}{N} \}.$$ Then by Lemma \[upper\], we have $$\frac{1+3\kappa}{N}|X|+\frac{\kappa}{N}(N-|X|)\geq
\sum_{x\in\Z_N}a_0'(x)=\sum_{x\in\Z_N}a_0(x)\geq\frac{1}{2}-3\kappa.$$ It follows that $$|X|\geq\big(\frac{1}{2}-6\kappa\big)N.$$ Notice that $\supp(a_i)\subseteq[1,M]$ and $\supp(\beta)\subseteq[-\kappa N,\kappa N]$, where $$\supp(f)=\{x\in\Z_N:\, f(x)\not=0\}.$$ Hence $$\supp(a_i')=\supp(a_i*\beta*\beta)\subseteq[-2\kappa
N,M+2\kappa N]$$ for $0\leq i\leq k$. Thus we have $$X\subseteq\supp(a_0')\subseteq[-2\kappa N,M+2\kappa N].$$ Let $A_0'=X\cap[1,M]$. Then $$|A_0'|\geq|X|-4\kappa N\geq(1-20\kappa)M,$$ by recalling that $(2+\kappa)M\leq N\leq (2+2\kappa)M$. Since $$a_0'=a_0*\beta*\beta=(a_1+\cdots+a_k)*\beta*\beta=a_1'+\cdots+a_k',$$ we have $$\max_{1\leq i\leq k}a_i'(x)\geq\frac{\kappa}{kN}$$ for each $x\in A_0'$. Let $$X_i=\{x\in A_0':\, a_i'(x)=\max_{1\leq i\leq k}a_i'(x)\}.$$ Clearly $A_0'=X_1\cup\cdots\cup X_k$. Let $A_1'=X_1$ and $$A_i'=X_i\setminus\bigg(\bigcup_{j=1}^{i-1}X_j\bigg)$$ for $2\leq i\leq k$. Then $A_1',\ldots,A_k'$ form a partition of $A_0'$. Furthermore, for $1\leq i\leq k$ and each $x\in A_i'$, we have $$a_i'(x)\geq\frac{\kappa}{kN}.$$ Thus by Corollary \[schur\] and Lemma \[difference\] $$\begin{aligned}
\sum_{i=1}^k\sum_{\substack{ x,y,z\in\Z_N\\x+y=z}}a_i(x)a_i(y)a_i(z)
\geq&\sum_{i=1}^k\sum_{\substack{
x,y,z\in\Z_N\\x+y=z}}a_i'(x)a_i'(y)a_i'(z)-\frac{C_3k^2}{N}(\epsilon^2\delta^{-3}+\delta^{\frac{1}{3}})\\
\geq&\sum_{i=1}^k\sum_{\substack{
x,y,z\in A_i'\\x+y=z}}\bigg(\frac{\kappa}{kN}\bigg)^3-\frac{C_3k^2}{N}(\epsilon^2\delta^{-3}+\delta^{\frac{1}{3}})\\
\geq&\bigg(\frac{\kappa}{kN}\bigg)^3\frac{C_1(k)M^{2}}{2}-\frac{C_3k^2}{N}(\epsilon^2\delta^{-3}+\delta^{\frac{1}{3}})\\\end{aligned}$$ Finally, we may choose sufficiently small $\delta$ and $\epsilon$ with $$\epsilon^{-C_2(3)\delta^{-3}k}\geq \kappa^{-2}\log\log w/w$$ such that $$\epsilon^2\delta^{-3}+\delta^{\frac{1}{3}}\leq\frac{C_1(k)\kappa^3}{24C_3k^5},$$ whenever $N$ is sufficiently large. Thus $$\begin{aligned}
\sum_{i=1}^k\sum_{\substack{ x,y,z\in\Z_N\\x+y=z}}a_i(x)a_i(y)a_i(z)
\geq\frac{C_1(k)\kappa^3M^{2}}{2k^3N^3}-\frac{C_1(k)\kappa^3}{24k^3N}\geq
\frac{C_1(k)\kappa^3}{12k^3N}-\frac{C_1(k)\kappa^3}{24k^3N}>0.\end{aligned}$$ This completes the proof.

Notice that $$\sum_{i=1}^k\sum_{\substack{ x,z\in\Z_N\\
2x=z}}a_i(x)^2a_i(z)=O\bigg(\frac{k\phi(W)^3\log(WN+1)^3}{W^3N^2}\bigg)=o(N^{-1}).$$ Hence in fact there exist three distinct monochromatic primes $p_1,p_2,p_3$ satisfying $p_1+p_2=p_3+1$.

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[^1]: This work was supported by the National Natural Science Foundation of China (Grant No. 10771135).
