Symmetric primes revisited

TL;DR

This paper improves a 1996 theorem on symmetric prime pairs, establishing a likely optimal upper bound and proving the infinitude of such pairs, including strings of consecutive primes all forming symmetric pairs.

Contribution

It refines the upper bound on symmetric prime pairs and proves the existence of infinitely many such pairs, including consecutive prime strings.

Findings

Improved the upper bound on symmetric prime pairs.

Proved infinitely many symmetric prime pairs exist.

Established existence of consecutive prime strings with symmetric pairs.

Abstract

A pair of odd primes is said to be symmetric if each prime is congruent to one modulo their difference. A theorem from 1996 by Fletcher, Lindgren, and the third author provides an upper bound on the number of primes up to x that belong to a symmetric pair. In the present paper, that theorem is improved to what is likely to be the best possible result. We also establish that there exist infinitely many symmetric pairs of primes. In fact, we show that for every integer m at least 2 there is a string of m consecutive primes, any two of which form a symmetric pair.

Tables

Table 1. Table 1. Tabulation of S ​ ( p n ) 𝑆 subscript 𝑝 𝑛 S(p_{n}) , the number of symmetric primes to the n 𝑛 n th prime.

$n$	$S\left(p_n\right)$	$S\left(p_n\right)/n$	$1/{\left(\log p_n\right)}^{\eta }$
$10$	9	0.9000	0.9008
${10}^2$	86	0.8600	0.8536
${10}^3$	864	0.8640	0.8279
${10}^4$	8473	0.8473	0.8101
${10}^5$	83263	0.8326	0.7964
${10}^6$	819848	0.8198	0.7854
${10}^7$	8098086	0.8098	0.7761
${10}^8$	80112625	0.8011	0.7681

Equations

$$\gcd (p-1,q-1)=|p-q|.$$

$$\left(\frac{q}{p}\right)=(-1{)}^{S(q,p)},\text{and}(\frac{p}{q})=(-1{)}^{S(p,q)}.$$

$$\eta :=1-\frac{1+\log \log 2}{\log 2}=0.08607\cdots .$$

$$S(x)⩽\frac{\pi (x)}{(\log x{)}^{\eta }}(\log \log x{)}^{O(1)}.$$

$$S_2(x)⩽\frac{\pi (x)}{(\log x{)}^{\eta }}(\log \log x{)}^{O(1)}.$$

$$\frac{x}{a(\log x{)}^2}(\log \log x{)}^{O(1)}.$$

$$R(x,d,m)⩽\frac{x}{dm(\log x{)}^3}(\log \log x{)}^{O(1)}.$$

$$\sum _{\begin{array}{c}dm<x^{1-1/\log \log x}\\ \Omega (dm)<L\end{array}}\frac{1}{dm}$$

$$⩽\sum _{i+j<L}\frac{1}{i!}{E}^i\frac{1}{j!}{E}^j=\sum _{k<L}\frac{1}{k!}{E}^k\sum _{i+j=k}\frac{k!}{i!j!}$$

$$=\sum _{k<L}\frac{1}{k!}(2E{)}^k\ll \frac{1}{L!}(2E{)}^L,$$

$$g_1,\dots ,g_k>0,\text{and}{\prod }_{1⩽i<j⩽k}(g_i{h}_j-g_j{h}_i)\ne 0,$$

$$b_j+1⩽b_i-b_j=\gcd (b_i-1,b_j-1)⩽b_j-1,$$

$$|P_i-P_j|=|h_i-h_j|=\gcd (h_i-1,h_j-1)(1⩽i<j⩽m),$$

Full text

In memory of Peter Fletcher (1939–2019)

1. Introduction

A pair of distinct odd primes $\{p,q\}$ is said to be a symmetric pair if

[TABLE]

For example, every twin prime pair $\{p,p+2\}$ is a symmetric pair. We say that a prime is symmetric if it is a member of some symmetric pair; otherwise, we say that it is asymmetric.

Symmetric primes arise naturally when ruminating on a common textbook proof of Gauss’s quadratic reciprocity law (QRL). Consider the rectangle $S$ with $(0,0)$ and $(p/2,q/2)$ as opposite corners, and let $l$ be the diagonal joining those corners. Let $S(q,p)$ (resp., $S(p,q)$) be the number of interior lattice points below (resp., above) $l$. Eisenstein, in his version of Gauss’s third QRL proof, showed that

[TABLE]

Since $\ell$ has no interior lattice points, $S(q,p)+S(p,q)$ is the total number of lattice points interior to $S$, which is $\frac{p-1}{2}\frac{q-1}{2}$; the law of quadratic reciprocity follows immediately. It is shown in §2 of Fletcher et al. [4] that $\{p,q\}$ is symmetric precisely when $S(q,p)=S(p,q)$.

Most primes are asymmetric; this is a consequence of [4, Theorem 3.1], which asserts that the number $S(x)$ of symmetric primes $p\leqslant x$ is $O(\pi(x)/(\log x)^{0.027})$. It is conjectured in [4] that the exponent $0.027$ can be improved to $\eta+o(1)$, where

[TABLE]

In this note we prove the conjecture.

Theorem 1.1**.**

For all large $x$, we have

[TABLE]

The constant $\eta$ appears at several places in the literature. An early appearance is in connection with the Erdős multiplication table problem where, thanks to the work of Erdős, Tenenbaum, and Ford, we now know that the number $M(N)$ of distinct entries in the $N\times N$ multiplication table is $N^{2}(\log N)^{-\eta}(\log\log N)^{O(1)}$. (In fact, Ford [5] has further shown that if the implied constant $O(1)$ is replaced with $-3/2$, the resulting expression has the same magnitude as $M(N)$.) A more recent appearance of $\eta$ occurs in Chow and Pomerance [3], where the odd legs in integer-sided right triangles with prime hypotenuse are considered. (The present note uses some techniques from [3].)

It was left as an open problem in [4] to prove that there are infinitely many symmetric primes. The next theorem uses an old result of Heath-Brown [8] together with the framework of the recent results of Zhang, Maynard, Tao, et al. on small gaps between primes.

Theorem 1.2**.**

For every integer $m\geqslant 2$, there exists a string of $m$ consecutive primes, any two of which form a symmetric pair.

Of course, Theorem 1.2 implies the infinitude of symmetric primes. Our proof of Lemma 3.2 below, in conjunction with the methods of [11, 12, 13], could be developed to prove that $S(x)\gg\pi(x)/(\log x)^{49}$. Comparing this lower bound with the upper bound of Theorem 1.1, it is tempting to conjecture that $S(x)=\pi(x)/(\log{x})^{c+o(1)}$, as $x\to\infty$, for some positive constant $c$. In [4] a heuristic argument is presented suggesting that this conjecture holds with $c=\eta$; that is, the inequality of Theorem 1.1 is actually an equality.

In Sections 2 and 3 we prove the theorems. In Section 4 we present some new computations of symmetric primes. In Section 5 we close with a few problems of a somewhat different nature.

2. The proof of Theorem 1.1

Let $\omega(n)$ denote the number of distinct primes that divide $n$, and let $\Omega(n)$ denote the number of prime factors of $n$ counted with multiplicity. Let $P^{+}(n)$ denote the largest prime factor of $n>1$, and put $P^{+}(1)=0$.

Let $S_{1}(x)$ denote the number of primes $p\leqslant x$ with $P^{+}(p-1)\leqslant x^{1/\log\log x}$. Since the number of integers $n\leqslant x$ with $P^{+}(n)\leqslant x^{1/\log\log x}$ is $O(x/(\log x)^{2})$ (see de Bruijn [2, Eq. (1.6)]) it follows that $S_{1}(x)=O(\pi(x)/\log x)$.

Next, let $S_{2}(x)$ denote the number of primes $p\leqslant x$ with $P^{+}(p-1)>x^{1/\log\log x}$ and $\Omega(p-1)>L$, where $L=\lfloor(1/\log 2)\log\log x\rfloor$. We claim that

[TABLE]

For any prime counted by $S_{2}(x)$, write $p=ar+1$, where $r=P^{+}(p-1)>x^{1/\log\log x}$. For any fixed choice of $a<x^{1-1/\log\log x}$, the number of primes $r\leqslant x/a$ with $ar+1$ prime is (by Brun’s method; see [6, Eq. (6.1)]) at most

[TABLE]

We sum this expression over $a$ assuming $\Omega(a)\geqslant L$. For $L\leqslant\Omega(a)\leqslant 1.9\log\log x$ we use [7, Theorem 08], finding that $\sum 1/a\leqslant(\log x)^{1-\eta}(\log\log x)^{O(1)}$; this is consistent with our goal (2.1). For larger values of $\Omega(a)$ we use [7, Exercise 05], getting $\sum 1/a\ll(\log x)^{0.69}$. This establishes the claim (2.1).

To finish the proof, we bound the number of symmetric primes $p\leqslant x$ with $P^{+}(p-1)>x^{1/\log\log x}$ and $\Omega(p-1)\leqslant L$. For any such prime, write $p=ar+1$ with $r=P^{+}(p-1)>x^{1/\log\log x}$ and $\Omega(a)<L$. Since $p$ is symmetric, there is some $d\mid a$ with at least one of $p+d$, $p-d$, $p+dr$, $p-dr$ prime. Write $a=dm$. For a given pair $d,m$ with $dm<x^{1-1/\log\log x}$, let $R(x,d,m)$ denote the number of primes $r\leqslant x/dm$ with $dmr+1$ prime and at least one of $dmr+d+1$, $dmr-d+1$, $dmr+dr+1$, $dmr-dr+1$ prime. Again by Brun’s method we have uniformly for $x$ large that

[TABLE]

It remains to sum this expression over pairs $d,m$ with $dm<x^{1-1/\log\log x}$ and $\Omega(dm)<L$. Let $E$ denote the reciprocal sum of all primes and prime powers less than $x$. We have

[TABLE]

since $E=\log\log x+O(1)$. A short calculation then shows that this expression is $(\log x)^{2-\eta}(\log\log x)^{O(1)}$. Thus, the sum of $R(x,d,m)$ over pairs $d,m$ is at most $\pi(x)(\log x)^{-\eta}(\log\log x)^{O(1)}$, so completing the proof.

3. The proof of Theorem 1.2

In her dissertation, Spiro [14] showed that the equation $d(n)=d(n+5040)$ has infinitely many solutions, where $d(n)$ is the divisor function. Heath-Brown [8] has shown that one can replace 5040 with 1 in this theorem, a key ingredient (see also [9, 10]) being the existence of sets with the property described in the next lemma (and another property that is not needed here).

Lemma 3.1**.**

For every $k\geqslant 2$ there is a set $\mathcal{A}_{k}\subset\mathbb{N}$ with $k$ elements such that $\gcd(a,b)=|a-b|$ for all $a,b\in\mathcal{A}_{k}$, $a\neq b$.

An example of such a set when $k=4$ is $\{6,8,9,12\}$.

If we have two numbers $a<b$ with $\gcd(a,b)=b-a$ and an integer $n$ for which $p=an+1$ and $q=bn+1$ are both prime, then $\{p,q\}$ is a symmetric pair. Thus, under the prime $k$-tuples conjecture we obtain infinitely many symmetric pairs. Alternatively, the prime $k$-tuples conjecture implies the existence of infinitely many twin prime pairs $\{p,p+2\}$, which are symmetric.

The preceding statements are still conjectural, but the Maynard-Tao theorem gives us a path for producing infinitely many symmetric primes.

Lemma 3.2**.**

For every $m\geqslant 2$ there is a set $\mathcal{P}_{m}$ of $m$ primes such that for all $p,q\in\mathcal{P}_{m}$, $p\neq q$, we have $\gcd(p-1,q-1)=|p-q|$. Moreover, one can find such a set whose least element exceeds $m$.

Proof.

Recall that a $k$-tuple of linear forms $\{g_{i}t+h_{i}\}_{i=1}^{k}$ is said to be admissible if the associated polynomial $\prod_{i}(g_{i}t+h_{i})$ has no fixed prime divisor, i.e., for each prime $p$ there is an integer $t$ with none of $g_{i}t+h_{i}$ divisible by $p$. To prove the lemma, we apply a remarkable theorem of Maynard (see, e.g., [11, Theorem 3.4]) and Tao (unpublished) in the direction of the prime $k$-tuples conjecture.

Theorem 3.3** (Maynard–Tao).**

For every $m\geqslant 2$ there is an integer $k=k_{m}$, depending only on $m$, such that if $\{g_{i}t+h_{i}\}_{i=1}^{k}$ is admissible,

[TABLE]

then $\{g_{i}n+h_{i}\}_{i=1}^{k}$ contains $m$ primes for infinitely many $n\in\mathbb{N}$. In fact, the number of such $n\leqslant x$ is $\gg x/(\log x)^{k}$.

We apply Theorem 3.3 to the linear forms $\{a_{i}t+1\}_{i=1}^{k}$, where $k=k_{m}$ and the integers $a_{i}$ are the elements of a set $\mathcal{A}_{k}$ of the type described in Lemma 3.1. Then $\{a_{i}n+1\}_{i=1}^{k}$ contains $m$ primes for infinitely many $n\in\mathbb{N}$, and Lemma 3.2 follows at once. ∎

One can adapt the work of [12, 13] to show that $50$ is an acceptable value of $k_{2}$ in Theorem 3.3. This explains the “49” in the lower bound $S(x)\gg\pi(x)/(\log{x})^{49}$ claimed in the introduction.

To prove Theorem 1.2, which asserts that some of the sets $\mathcal{P}_{m}$ in Lemma 3.2 consist of consecutive primes, we use the following result from Banks et al. [1].

Theorem 3.4** (Banks–Freiberg–Turnage-Butterbaugh).**

Let $m\geqslant 2$ and $k=k_{m}$, where $k_{m}$ is as in the Maynard–Tao theorem. Let $b_{1},\ldots,b_{k}$ be distinct integers such that $\{t+b_{j}\}_{j=1}^{k}$ is admissible, and let $g$ be an arbitrary positive integer that is coprime to $b_{1}\cdots b_{k}$. Then, for some subset $\{h_{1},\ldots,h_{m}\}\subseteq\{b_{1},\ldots,b_{k}\},$ there are infinitely many $n\in\mathbb{N}$ such that $gn+h_{1},\ldots,gn+h_{m}$ are consecutive primes.

Now, let $m\geqslant 2$ and $k\geqslant k_{m}$. By Lemma 3.2 there exists a set of primes $\mathcal{P}_{k}=\{b_{1},\ldots,b_{k}\}$ such that $\gcd(b_{i}-1,b_{j}-1)=|b_{i}-b_{j}|$ for all $1\leqslant i<j\leqslant k$, and each $b_{i}$ exceeds $k$ (thus, the $k$-tuple $\{t+b_{j}\}_{j=1}^{k}$ is admissible).

Notice that $g=\prod_{i}(b_{i}-1)$ is coprime to $b_{1}\cdots b_{k}$. Otherwise, there are indices $i,j$, with $i\neq j$, for which $b_{j}\mid b_{i}-1$. But then $b_{i}\geqslant 2b_{j}+1$, and

[TABLE]

which is absurd.

By Theorem 3.4 there is a subset $\{h_{1},\ldots,h_{m}\}\subseteq\{b_{1},\ldots,b_{k}\}$ with the property that $P_{1}=gn+h_{1},\ldots,P_{m}=gn+h_{m}$ are consecutive primes for infinitely many $n\in\mathbb{N}$. Since $h_{i}-1\mid P_{i}-1$ for each $i$, and

[TABLE]

it follows that $|P_{i}-P_{j}|=\gcd(P_{i}-1,P_{j}-1)$ when $i\neq j$, i.e., $\{P_{i},P_{j}\}$ is a symmetric pair. This completes the proof of Theorem 1.2

4. Computations

In [4] some values of $S(x)$ for $x$ up to the $10^{5}$th prime were given. The data did not strongly suggest that $S(x)=o(\pi(x))$; in fact, it seemed more plausible that $S(x)/\pi(x)\approx 0.83$. Using Mathematica we have extended the calculation to the $10^{8}$th prime and we see that $S(x)/\pi(x)$ continues to be in no hurry to get to zero, but progress towards this limit is somewhat discernible. The descent to zero does indeed appear to be not so different than the main term in our upper bound.

5. Graph problems

Consider a graph on the odd primes where two primes are connected by an edge if they form a symmetric pair. The asymmetric primes are isolated nodes. Must every connected component be finite? At the other extreme, removing the asymmetric primes, is the graph connected? If not, what is the least symmetric prime that is not in the component containing the prime 3? Does the graph have infinitely many components? Does it contain a complete graph $K_{m}$ on $m$ vertices for every $m$? The answer to this last question is “yes”, from Theorem 1.2. Clearly there cannot exist an infinite complete subgraph since if $p<q$ are a symmetric pair, then $q<2p$. Say a prime $p$ is $m$-symmetric if it is in a $K_{m}$. It would be interesting to investigate the distribution of $m$-symmetric primes; the number of them to $x$ is $\pi(x)/(\log x)^{O_{m}(1)}$, but what can be said about the exponent here?

Acknowledgment. We thank James Maynard for some helpful comments.