On the Weakly Prime-Additive Numbers with Length 4

TL;DR

This paper proves the existence of infinitely many length 4 weakly prime-additive numbers divisible by any positive integer m under GRH, extending previous results from length 3 to length 4.

Contribution

It establishes the existence of infinitely many length 4 weakly prime-additive numbers divisible by any positive integer m under GRH, generalizing earlier length 3 results.

Findings

Existence of infinitely many length 4 weakly prime-additive numbers divisible by any m under GRH.

Extension of length 3 results to length 4 in weakly prime-additive numbers.

Related results analogous to the length 3 case are also presented.

Abstract

In 1992, Erd$\H{o}$s and Hegyv$\'{a}$ri showed that for any prime p, there exist infinitely many length 3 weakly prime-additive numbers divisible by p. In 2018, Fang and Chen showed that for any positive integer m, there exists infinitely many length 3 weakly prime-additive numbers divisible by m if and only if 8 does not divide m. Under the assumption (*) of existence of a prime in certain arithmetic progression with prescribed primitive root, which is true under the Generalized Riemann Hypothesis (GRH), we show for any positive integer m, there exists infinitely many length 4 weakly prime-additive numbers divisible by m. We also present another related result analogous to the length 3 case shown by Fang and Chen.

Equations

$$pqrs|p^a+q^b+r^c+s^d.$$

$$a^{\varphi (n)}\equiv 1(\mathrm{mod}n)$$

$$1.$$

$$2.$$

$$3.$$

$$4.$$

$$5.$$

$$Ifp\equiv q\equiv 3(\mathrm{mod}4),\left(\frac{p}{q}\right)=-\left(\frac{q}{p}\right)$$

$$\beta =\frac{g_1}{(g_1,f)}\text{ and }{\gamma }_1=\{\begin{array}{ll}(-1{)}^{\frac{\beta -1}{2}}(f,g_1)& \text{if }\beta \text{ is odd;}\\ 1& \text{otherwise.}\end{array}$$

$${\pi }_g(x;f,a)=\delta (a,f,g)\frac{x}{\log x}+O_{f,g}\left(\frac{x\log \log x}{{\log }^{2}x}\right)$$

$$\delta (a,f,g)=\frac{A(a,f,h)}{\varphi (f)}\left(1-\left(\frac{{\gamma }_1}{a}\right)\frac{\mu (|\beta |)}{{\prod }_{\begin{array}{c}p|\beta \\ p|h\end{array}}(p-1){\prod }_{\begin{array}{c}p|\beta \\ p\nmid h\end{array}}(p^2-p-1)}\right)$$

$$\delta (a,f,g)=\frac{A(a,f,h)}{\varphi (f)}$$

$$A(a,f,h)=\prod _{p|(a-1,f)}\left(1-\frac{1}{p}\right)\prod _{\begin{array}{c}p\nmid f\\ p|h\end{array}}\left(1-\frac{1}{p-1}\right)\prod _{\begin{array}{c}p\nmid f\\ p\nmid h\end{array}}\left(1-\frac{1}{p(p-1)}\right)$$

$$\delta (a,f,g)=\frac{2}{\varphi (f)}\prod _{p|(a-1,f)}\left(1-\frac{1}{p}\right)\prod _{p\nmid f}\left(1-\frac{1}{p(p-1)}\right)>0$$

$$q\equiv 1(\mathrm{mod}{2}^{k}p)\text{ and }q\equiv -1(\mathrm{mod}{m}_1)$$

$$r\equiv 3(\mathrm{mod}{2}^k)\text{ and }r\equiv 1(\mathrm{mod}pq{m}_1)$$

$$s_0\equiv -5(\mathrm{mod}{2}^k)$$

$$s_0\equiv -1(\mathrm{mod}{m}_1)$$

$$s_0\equiv -2(\mathrm{mod}pqr)$$

$$\left(\frac{r}{s_0}\right)=\left(\frac{s_0}{r}\right)(-1{)}^{\frac{s_0-1}{2}\frac{r-1}{2}}=-\left(\frac{s_0}{r}\right)=-\left(\frac{-2}{r}\right)=-1$$

$$r^{c_0}\equiv -2(\mathrm{mod}s)$$

$$c=(p-1)(q-1)(r-1)\varphi (m)c^{\prime }+c_0$$

$$b=\frac{1}{4}(r-1)(s-1)b^{\prime }$$

$$a=(q-1)(r-1)(s-1)\varphi (m)a^{\prime }$$

$$n=p^a+q^b+r^c+s$$

$$n\equiv p^a+q^b+r^c+s\equiv 0+1+1-2\equiv 0(\mathrm{mod}p)$$

$$n\equiv p^a+q^b+r^c+s\equiv 1+0+1-2\equiv 0(\mathrm{mod}q)$$

$$q^b\equiv (q^{\frac{1}{2}(r-1)}{)}^{\frac{1}{2}(s-1)b^{\prime }}\equiv {\left(\frac{q}{r}\right)}^{\frac{1}{2}(s-1)b^{\prime }}\equiv \left(\frac{r}{q}\right)\equiv \left(\frac{1}{q}\right)\equiv 1(\mathrm{mod}r)$$

$$n\equiv p^a+q^b+r^c+s\equiv 1+1+0-2\equiv 0(\mathrm{mod}r)$$

$$n\equiv p^a+q^b+r^c+s\equiv 1+1-2+0\equiv 0(\mathrm{mod}s)$$

$$n\equiv p^a+q^b+r^c+s\equiv 1-1+1-1\equiv 0(\mathrm{mod}{m}_1)$$

$$n\equiv p^a+q^b+r^c+s\equiv 1+1+3-5\equiv 0(\mathrm{mod}{2}^k)$$

$$s_0\equiv 3(\mathrm{mod}8)$$

$$s_0\equiv -2(\mathrm{mod}Apqr)$$

Full text

On the weakly prime-additive numbers with length 4

Wing Hong Leung

Abstract: In 1992, Erdős and Hegyvári showed that for any prime $p$, there exist infinitely many length 3 weakly prime-additive numbers divisible by $p$. In 2018, Fang and Chen showed that for any positive integer $m$, there exists infinitely many length 3 weakly prime-additive numbers divisible by $m$ if and only if $8$ does not divide $m$. Under the assumption $(\ast)$ of existence of a prime in certain arithmetic progression with prescribed primitive root, which is true under the Generalized Riemann Hypothesis (GRH), we show for any positive integer $m$, there exists infinitely many length 4 weakly prime-additive numbers divisible by $m$. We also present another related result analogous to the length 3 case shown by Fang and Chen.

1. Introduction

A number $n$ with at least 2 distinct prime divisors is called prime-additive if $n=\sum_{p|n}p^{a_{p}}$ for some $a_{p}>0$. If additionally $p^{a_{p}}<n\leq p^{a_{p}+1}$ for all $p|n$, then $n$ is called strongly prime-additive. In 1992, Erdős and Hegyvári [2] stated a few examples and conjectured that there are infinitely many strongly prime-additive numbers. However, this problem was and is still far from being solved. For example, not even the infinitude of prime-additive numbers is known. Therefore they introduced the following weaker version of prime-additive numbers.

Definition**.**

A positive integer $n$ is said to be weakly prime-additive if $n$ has 2 distinct prime divisors, and there exists distinct prime divisors $p_{1},...,p_{t}$ of $n$ and positive integers $a_{1},...,a_{t}$ such that $n=p_{1}^{a_{1}}+\cdots+p_{t}^{a_{t}}$. The minimal value of such $t$ is defined to be the length of $n$, denoted as $\kappa_{n}$.

Note that if $n$ is a weakly prime-additive number, then $\kappa_{n}\geq 3$. So we call a weakly prime-additive number with length 3 a shortest weakly prime-additive number.

Erdős and Hegyvári [2] showed that for any prime $p$, there exist infinitely many weakly prime-additive numbers divisible by $p$. In fact, they showed that these weakly prime-additive numbers can be taken to be shortest weakly prime-additive in their proof. They also showed that the number of shortest weakly prime-additive numbers up to some integer $N$ is at least $c(\log N)^{3}$ for sufficiently small constant $c>0$.

In 2018, Fang and Chen [1] showed that for any positive integer $m$, there exists infinitely many shortest weakly prime-additive numbers divisible by $m$ if any only if $8$ does not divide $m$. This is Theorem 7 stated in this paper. Also, they showed that for any positive integer $m$, there exists infinitely many weakly prime-additive numbers with length $\kappa_{n}\leq 5$ and divisible by $m$.

In the same paper, Fang and Chen posted 4 open problems, the first one asking for any positive integer $m$, if there are infinitely many weakly prime-additive numbers $n$ with $m|n$ and $\kappa_{n}=4$. In Theorem 1 of this paper, we confirm this is true under the assumption $(\ast)$ of existence of a prime in certain arithmetic progression with prescribed primitive root, and such assumption is true under the Generalized Riemann Hypothesis (GRH).

Finally, it was also shown in [1] that for any distinct primes $p,q$, there exists a prime $r$ and infinitely many $a,b,c$ such that $pqr|p^{a}+q^{b}+r^{c}$. In Theorem 2, we showed an analogous result for 4 primes with a mild congruence conditions assuming $(\ast)$, the same assumption as above.

2. Main Results

Assumption ($\ast$)****.

Let $1\leq a\leq f$ be positive integers with $(a,f)=1$ and $4|f$. Let $g$ be an odd prime dividing $f$ such that $\left(\frac{g}{a}\right)=-1$ with $\left(\frac{\cdot}{\cdot}\right)$ being the Kronecker symbol. Then there exists a prime $p$ such that $p\equiv a\ (\mathrm{mod}\ f)$ and $g$ is a primitive root of $p$.

It is known that ($\ast$) is a consequence of the Grand Riemann Hypothesis (GRH), see Corollary 6.1 in the next section for details. Under the assumption ($\ast$), we have the following.

Theorem 1**.**

Assume ($\ast$). For any positive integer $m$, there exist infinitely many weakly prime-additive numbers $n$ with $m|n$ and $\kappa_{n}=4$.

Note that if a positive integer $n=p^{a}+q^{b}+r^{c}+s^{d}$ for some distinct primes $p,q,r,s$, and positive integers $a,b,c,d$ such that $p,q,r,s|n$, then $p,q,r,s$ are all odd primes. We have the following partial converse and analog to Theorem 1.4 in [1].

Theorem 2**.**

Assume ($\ast$). For any distinct odd primes $p,q,r$ with one of them $\equiv 3$ or $5\ (\mathrm{mod}\ 8)$, there exists infinitely many prime $s$, infinitely many positive integers $a,b,c,d$ such that

[TABLE]

3. Preliminaries

Lemma 3**.**

(The Fermat-Euler Theorem, Theorem 72, [5]) Let $a,n$ be coprime positive integers, then

[TABLE]

where $\phi$ is the Euler totient function.

We will need the following properties of the Kronecker Symbol $(\frac{\cdot}{\cdot})$, a generalization of the Legendre symbol. Whenever we write $(\frac{a}{b})$ for some integers $a,b$, it means the Kronecker symbol.

Lemma 4**.**

For any nonzero integers $a,b,c$, and any odd primes $p,q$. Let $a^{\prime}$, $b^{\prime}$ be the odd part of $a$ and $b$ respectively, then we have:

[TABLE]

Proof.

See [4], p. 289-290. ∎

Theorem 5**.**

(Dirichlet’s Theorem) Let $a,d$ be coprime positive integers, then there are infinitely many primes $p$ such that $p\equiv a\ (\mathrm{mod}\ d)$.

Proof.

See [4], chapter 1. ∎

Under GRH, we have the following generalization.

Theorem 6**.**

(Theorem 1.3, [3]) Let $1\leq a\leq f$ be positive integers with $(a,f)=1$. Let $g$ be an integer that is not equal to $-1$ or a square, and let $h\geq 1$ be the largest integer such that $g$ is a $h$th power. Write $g=g_{1}g_{2}^{2}$ with $g_{1}$ square free, $g_{1},g_{2}$ are integers. Let

[TABLE]

Let $\pi_{g}(x;f,a)$ be the number of primes $p\leq x$ such that $p\equiv a\ (\mathrm{mod}\ f)$ and $g$ is a primitive root $\ (\mathrm{mod}\ p)$. Then, assuming GRH, we have

[TABLE]

where

[TABLE]

if $g_{1}\equiv 1\ (\mathrm{mod}\ 4)$ or $(g_{1}\equiv 2\ (\mathrm{mod}\ 4)$ and $8|f)$ or $(g_{1}\equiv 3\ (\mathrm{mod}\ 4)$ and $4|f$), and

[TABLE]

otherwise. Here $\mu$ is the mobius function, $\left(\frac{\cdot}{\cdot}\right)$ is the Kronecker symbol, and

[TABLE]

if $(a-1,f,h)=1$ and $A(a,f,h)=0$ otherwise.

Corollary 6.1**.**

Assume GRH. Let $a,f,g$ as above and $\left(\frac{g}{a}\right)=-1$. Then there exists a prime $p$ such that $p\equiv a\ (\mathrm{mod}\ f)$ and $g$ is a primitive root of $p$, i.e. ($\ast$) is true under GRH.

Proof.

This is a special case of Theorem 6, where with our conditions on $a,f,g$, $\beta=h=1$, $\gamma_{1}=g$.

[TABLE]

∎

Remark**.**

This shows that our result also follows from GRH, which is a much stronger assumption than ($\ast$).

Theorem 7**.**

(Corollary 1.1, [1]) Let $m$ be a positive integer, then there exists infinitely many shortest weakly prime-additive numbers $n$ with $m|n$ if and only if 8 does not divide $m$.

4. Proof of Theorem 1

We first prove the following weaker version of Theorem 1.

Theorem 8**.**

Assume ($\ast$). For any positive integer $m$, there exist infinitely many weakly prime-additive numbers $n$ with $m|n$ and $\kappa_{n}\leq 4$.

Proof.

Let $m$ be a positive integer. Write $m=2^{k}m_{1}$ with $(m_{1},2)=1$ and $k\geq 0\in\mathbb{Z}$. Without loss of generality, we assume $k\geq 3$. We will construct a family of distinct primes $p,q,r,s$ and positive integers $a,b,c$ such that $m,p,q,r,s|n:=p^{a}+q^{b}+r^{c}+s$.

Let $p$ be an odd prime such that $(p,m)=1$. By the Chinese Remainder Theorem and Theorem 5, there exists an odd prime $q$ such that

[TABLE]

Again by the same two theorems, there exists an odd prime $r$ such that

[TABLE]

By the Chinese Remainder Theorem, let $s_{0}$ be the unique integer such that $1\leq s_{0}\leq pqrm$ and

[TABLE]

Then we can see that $(s_{0},pqrm)=1$.

Now note that as $k\geq 3$, $r\equiv 3\ (\mathrm{mod}\ 2^{k})$, $s_{0}\equiv-5\mod{2^{k}}$ gives $r\equiv 3\ (\mathrm{mod}\ 8)$ and $s_{0}\equiv 3\ (\mathrm{mod}\ 8)$ respectively, so we have by Lemma 4,

[TABLE]

where we used $s_{0}\equiv-2\ (\mathrm{mod}\ r)$ as $s_{0}\equiv-2\ (\mathrm{mod}\ pqr)$.

Hence applying Corollary 6.1 with $a=s_{0}$, $f=pqrm$ and $g=r$, there exists an odd prime $s$ such that $s\equiv s_{0}\ (\mathrm{mod}\ pqrm)$ and $r$ is a primitive root of $s$. So $s$ satisfies all of the above congruence relations satisfied by $s_{0}$, and there exists a positive integer $c_{0}$ s.t.

[TABLE]

Note by construction, $p,q,r,s$ are all distinct odd primes.

Now for any positive integer $c^{\prime}$, take

[TABLE]

and for any positive odd integer $b^{\prime}$, take

[TABLE]

Note that since $r\equiv 3\ (\mathrm{mod}\ 2^{k})$ and $s\equiv-5\ (\mathrm{mod}\ 2^{k})$, we have $r,s\equiv 3\ (\mathrm{mod}\ 4)$. So $b$ is odd.

Finally, for any positive integer $a^{\prime}$, take

[TABLE]

where $\phi$ is the euler $\phi$ function, and let

[TABLE]

Then we have the following congruence conditions:

1. As $q\equiv r\equiv 1\ (\mathrm{mod}\ p)$, $s\equiv-2\ (\mathrm{mod}\ p)$, we have

[TABLE]

2. As $q-1|a$, by Lemma 3, $p^{a}\equiv 1\mod q$. So we have

[TABLE]

3. Similarly, $p^{a}\equiv 1\ (\mathrm{mod}\ r)$ as $r-1|a$. Since $q\equiv 1\ (\mathrm{mod}\ 2^{k}p)$, $q\equiv 1\ (\mathrm{mod}\ 8)$. By Lemma 4, with $r\equiv 3\ (\mathrm{mod}\ 8)$ and $r\equiv 1\ (\mathrm{mod}\ q)$,

[TABLE]

So we have

[TABLE]

4. Similarly, $p^{a}\equiv q^{b}\equiv 1\ (\mathrm{mod}\ s)$. As $r^{c}\equiv-2\ (\mathrm{mod}\ s)$, we have

[TABLE]

5. As $\phi(m)|a$, by Lemma 3, $p^{a}\equiv 1\ (\mathrm{mod}\ m)$. Since $b$ is odd and $q\equiv-1\ (\mathrm{mod}\ m_{1})$, we get $q^{b}\equiv-1\ (\mathrm{mod}\ m_{1})$. Together with $r\equiv 1\ (\mathrm{mod}\ m_{1})$ and $s\equiv-1\ (\mathrm{mod}\ m_{1})$, we have

[TABLE]

6. As $p^{a}\equiv 1\ (\mathrm{mod}\ m)$, $q\equiv 1\ (\mathrm{mod}\ 2^{k})$, $r\equiv 3\ (\mathrm{mod}\ 2^{k})$ and $s\equiv-5\ (\mathrm{mod}\ 2^{k})$, we have

[TABLE]

Hence $n=p^{a}+q^{b}+r^{c}+s$ is weakly prime additive and is divisible by $m$. Since $a^{\prime},c^{\prime}$ can be any positive integer, $b^{\prime}$ can be any positive odd integer and $p$ can be any arbitrary odd prime coprime to $m$, we have constructed infinitely many weakly prime-additive $n$ with length $\leq 4$. ∎

Remark**.**

In the above construction, $s$ can be raised to any $d$-th power for any positive integer $d\equiv 1\ (\mathrm{mod}\ \phi(pqrm))$.

Together with Theorem 7, we can prove Theorem 1.

Proof of Theorem 1.

Let $m$ be a positive integer. Then by Theorem 8, there exists infinitely many weakly prime-additive numbers with length $\leq 4$ such that they are divisible by $8m$. By Theorem 7, as $8|8m$, these numbers cannot be shortest weakly prime-additive, hence they are all weakly prime-additive numbers with length 4. ∎

5. Proof of Theorem 2

Let $p,q,r$ be distinct odd primes with one of them, WLOG say, $r\equiv 3$ or $5\ (\mathrm{mod}\ 8)$. Let $k\in\mathbb{N}$ such that $\frac{(p-1)(q-1)}{2^{k}}$ is odd, and let $A=\frac{(p-1)(q-1)}{2^{k}}$. Let $f=8Apqr$ and by the Chinese Remainder Theorem, let $s_{0}$ be the unique integer such that $1\leq s_{0}\leq f$ and

[TABLE]

Note that by Lemma 4, using $r\equiv 3$ or $5\ (\mathrm{mod}\ 8)$, we have

[TABLE]

Apply Theorem 5 with the above $a,f$, and $g=r$, there exists an odd prime $s$ such that

$s\equiv s_{0}\ (\mathrm{mod}\ 8Apqr)$ and $r$ is a primitive root of $s$. So there exists $0<c_{0}<s-1$ such that

[TABLE]

Now note that as $s\equiv 3\ (\mathrm{mod}\ 8)$, we have $\left(\frac{-2}{s}\right)=1$. So $c_{0}$ is even.

Now by the Chinese Remainder Theorem, take any positive integer $c$ such that

[TABLE]

This is possible as $s\equiv 3\ (\mathrm{mod}\ 8)$ and $s\equiv-2\ (\mathrm{mod}\ A)$, so $(\frac{s-1}{2},(p-1)(q-1))=1$. Since $c_{0}$ is even and $\frac{s-1}{2}$ is odd, this makes $c\equiv c_{0}\ (\mathrm{mod}\ s-1)$. So we have $r^{c}\equiv-2\ (\mathrm{mod}\ s)$ and $r^{c}\equiv 1\ (\mathrm{mod}\ pq)$.

Finally, for any positive integers $a,b,d$ such that $(q-1)(r-1)(s-1)|a$, $(p-1)(r-1)(s-1)|b$, $d\equiv 1\ (\mathrm{mod}\ (p-1)(q-1)(r-1))$, we have the following:

[TABLE]

So we get for any positive integers $a,b,c,d$ as above,

[TABLE]

∎