Goldbach's like conjectures arising from arithmetic progressions whose first two terms are primes

TL;DR

This paper introduces conjectures extending Goldbach's conjecture by exploring sums of terms from specific arithmetic progressions starting with primes, suggesting these sums can represent all integers greater than one.

Contribution

It proposes new Goldbach-like conjectures involving sums of arithmetic progression terms defined by prime pairs, expanding the scope of prime-based additive conjectures.

Findings

Conjecture that every integer greater than one can be expressed as a sum of an even number of terms from certain prime-based arithmetic progressions.

Connection established between these conjectures and the classical Goldbach's conjecture for even integers.

Introduction of analogous conjectures for odd integers and related prime sum problems.

Abstract

For two odd primes $p$ and $q$ such that $p<q$, let $A(p,q):=(a_k)_{k=1}^{\infty}$ be the arithmetic progression whose $k$th term is given by $a_k=(k-1)(q-p)+p$ (i.e., with $a_1=p$ and $a_2=q$). Here we conjecture that for every positive integer $a>1$ there exist a positive integer $n$ and two odd primes $p$ and $q$ such that $a$ can be expressed as a sum of the first $2n$ terms of the arithmetic progression $A(p,q)$. Notice that in the case of even $a$, this conjecture immediately follows from Goldbach's conjecture. We also propose the analogous conjecture for odd positive integers $a>1$ as well as some related Goldbach's like conjectures arising from the previously mentioned arithmetic progressions.

Equations

$$a_k=(k-1)(q-p)+p,k=1,2,\dots$$

$$n(n+2m+3)=(2^{u+1}-1)2^{u+1}.$$

$$n(n+2m+3)=(2^{u+1}+1)2^{u+1}.$$

$$S_2(p,q)=p+q,$$

$$\{S_2(p,q):p\mathrm{and}q\mathrm{are}\mathrm{odd}\mathrm{primes}\}=\{2n:n\in N\setminus \{1,2\}\}.$$

$$q=p-\frac{2(p-1)}{2n-1},$$

$$q=p-\frac{p-1}{n},$$

Full text

1. Conjectures

on arithmetic progressions whose first two terms are primes

Let $p$ and $q$ be two primes such that $p<q$ and let $A(p,q):=(a_{k})_{k=1}^{\infty}$ be the arithmetic progression whose $k$th term is given by

[TABLE]

In other words, $A(p,q)$ is an arithmetic progression whose first two terms are $p$ and $q$ (i.e., $a_{1}=p$ and $a_{2}=q$). The sum $S_{n}(p,q)=S_{n}$ of the first $n$ terms of the progression $A(p,q)$ is equal to

[TABLE]

From (1) we have that for all $n=1,2,\ldots$ and $m=0,1,2,\ldots$ the sum $S_{n,m}(p,q):=\sum_{i=m+1}^{n+m}a_{i}$ of some $n$ consecutive terms of progression $A(p,q)$ is equal to

[TABLE]

We start with following example.

Example 1.1* (An extension of a Sylvester’s result).*

Here we examine positive integers $a$ which can be written as a sum $S_{n,m}(2,3)$ (given by (2) with $p=2$ and $q=3$) for some $n\geq 2$ and $m\geq 1$. The sum of $k$th term and $(k+1)$th term of the progression $A(2,3)=(k+1)_{k=1}^{\infty}$ is equal to $2k+3$. Therefore, every odd integer greater than 3 is a sum of some two consecutive terms of $A(2,3)$. Furthermore, by (2) we have

[TABLE]

If $a$ is an even positive integer which is not a power of 2, then $a=(2d+1)2^{u}$ for some positive integers $d\geq 1$ and $u\geq 1$. If $1\leq d\leq 2^{u}-2$ for such a $a$, we have $S_{2d+1,2^{u}-d-2}=(2d+1)2^{u}=a$. (If $d=0$ then $n=1$ and $S_{1,m}(2,3)=m+2$ is in fact the $(m+1)$th term of $A(2,3)$). Similarly, if $d\geq 2^{u}+1$, then $S_{2^{u+1},d-2^{u}-1}=(2d+1)2^{u}=a$. This shows that each even positive integer $a=(2d+1)2^{u}$ with $1\leq d\leq 2^{u}-2$ or $d\geq 2^{u}+1$ can be expressed as a sum of at least two consecutive terms of the arithmetic progression $A(2,3)$.

It remains to consider the cases when $a$ is of the form $2^{u}$, $(2^{u+1}-1)2^{u}$ or $(2^{u+1}+1)2^{u}$ with some positive integer $u$. If $a=2^{u}$, then by (3) the equality $S_{n,m}(2,3)=a$ is equivalent to $n(n+2m+3)=2^{u+1}$, which is impossible in view of the fact that one among numbers $n$ and $n+2m+3$ is an odd integer.

If $a=(2^{u+1}-1)2^{u}$ for a positive integer $u$, then the equality $S_{n,m}(2,3)=a$ is equivalent to

[TABLE]

If $2^{u+1}-1$ is a composite number, then it can be written as a product $2^{u+1}-1=tv$ with odd integers $t\geq 3$ and $v\geq 3$. Then the above equality holds for $n=v\geq 3$ and $m=(t(tv+1))-v-3)/2=((t^{2}-1)v+t-3)/2\geq 12$. If $2^{u+1}-1$ is a prime number, then easily follows that the above equality holds only for $n=1$ and $m=(2^{u+1}-1)2^{u}-2$.

Now consider the last case, i.e., when $a=(2^{u+1}+1)2^{u}$ for a positive integer $u$. Then the equality $S_{n,m}(2,3)=a$ is equivalent to

[TABLE]

If $2^{u+1}+1$ is a composite number, then it can be written as a product $2^{u+1}+1=tv$ with odd integers $t\geq 3$ and $v\geq 3$. Then the above equality holds for $n=v\geq 3$ and $m=(t(tv-1))-v-3)/2=((t^{2}-1)v-t-3)/2\geq 9$. If $2^{u+1}+1$ is a prime number, then easily follows that the above equality holds only for $n=1$ and $m=(2^{u+1}+1)2^{u}-2$.

In view ot the above considerations, we have shown that every integer $a\geq 4$ is equal to $S_{n,m}(2,3)$ for some integers $n\geq 2$ and $m\geq 1$ in all the cases excluding the following ones:

1. $a$ is not a power of 2;

2. $a$ is not of the form $(2^{u+1}-1)2^{u}$, where $2^{u+1}-1$ is a prime number and

3. $a$ is not of the form $(2^{u+1}+1)2^{u}$, where $2^{u+1}+1$ is a prime number.

Remark 1.2*.*

Notice that if $a=2^{u}(2^{u+1}+1)$ for an integer $u\geq 1$, then $a=\sum_{i=1}^{2^{u+1}}i$, while if $a=2^{u}(2^{u+1}-1)$ for an integer $u\geq 1$, then $a=\sum_{i=1}^{2^{u+1}-1}i$. These two identities together with Example 1.1 imply the well known fact that every integer $a>1$ which is not a power of $2$, is a sum of two or more consecutive integers (see, e.g., Dickson’s History [1, 1, Ch. III, p. 139], where this result was attributed to Sylvester).

Remark 1.3*.*

Note that it is well known (see, e.g., [4, Subsections 2.2 and 2.3]) that in order to the so-called a Mersenne number $M_{u+1}:=2^{u+1}-1$ to be prime, $u+1$ must itself be prime. A Mersenne number which is prime is called Mersenne prime (this is Sloane’s sequence A000668 in [6] corresponding to indices given by Sloane’s sequence A000043). Moreover, it is easy to show that in order to $2^{u+1}+1$ to be prime, $u+1$ must be a power of 2. Such numbers are in fact Fermat numbers $F_{s}:=2^{2^{s}}+1$ ($s=0,1,2,\ldots$; this is Sloane’s sequence A000215 in [6]). Fermat conjectured in 1650 that every Fermat number is prime and Eisenstein proposed as a problem in 1844 the proof that there are an infinite number of Fermat primes (i.e., Fermat numbers which are primes) (see [5, p. 88]). However, the only known Fermat primes are $F_{0}=3$, $F_{1}=5$, $F_{2}=17$, $F_{3}=257$ and $F_{4}=65537$ (Sloane’s sequence A019434 in [6]). For more information on classical and alternative approaches to the Mersenne and Fermat numbers, see [3].

Note that the conclusion at the end of Example 1.1 immediately yields the following interesting assertion.

Proposition 1.4**.**

The following two statements are equivalent:

$(i)$*  There are infinitely many Fermat primes or there are infinitely many Mersenne primes;*

$(ii)$* The set $\{S_{n,m}(2,3):\,n=2,3,\ldots;m=1,2,\ldots\}$ omits infinitely many positive integer values which are not powers of 2.*

Example 1.5*.*

For the progression $A(3,5)=(2k+1)_{k=1}^{\infty}$ we have $S_{n,m}(3,5)=2(2m+4)$. From this it can be easily seen that a positive integer $a\geq 8$ is equal to some sum $S_{n,m}(3,5)$ with $n\geq 2$ if and only if $a$ is divisible by $4$ or $a$ is an odd composite integer greater than $14$ which is not a square of a prime.

More generally, if $q=p+2$, then $S_{n,m}(p,q)=n(n+2m+p-1)$. From this it follows that a positive integer $a$ is equal to some sum $S_{n,m}(3,5)$ with $n\geq 2$ if and only if $a=4s$ with $s\geq(p+1)/2$ or $a$ is an odd composite integer which can be expressed as a product $n=ab$ with odd integers $a$ and $b$ such that $a\geq 3$ and $b\geq a+p-1$.

From Examples 1.1 and 1.5 it follows that every integer greater than 10 can be expressed as a sum of two or more consecutive terms of the progression $A(2,3)$ or $A(3,5)$. Accordingly, it can be of interest to consider a problem of representation of a positive integer as a sum of two or more first consecutive integers in some progression $A(p,q)$. Notice that

[TABLE]

and even Goldbach’s conjecture states that every even positive integer greater than $2$ can be expressed as a sum of two primes. This famous conjecture was proposed on 7 June 1742 by the German mathematician Christian Goldbach in a letter to Leonhard Euler [2] (cf. [1]). This conjecture has been shown to hold for all integers less than $4\times 10^{18}$, but remains unproven despite considerable effort.

In view of the above equality, this conjecture is equivalent with the following set equality:

[TABLE]

This fact suggests the investigations of the values of $S_{n}(p,q)$ given by (1). Namely, for each positive integer $n$, we will consider the values

[TABLE]

where $p$ and $q$ are odd primes.

Using some heuristic arguments and computational results, we propose the following “weak even Goldbach conjecture”.

Conjecture 1.6** (“weak even Goldbach conjecture”).**

For each even positive integer $a$ greater than $2$ there exist a positive integer $n$ and odd primes $p$ and $q$ such that $a=S_{2n}(p,q)$; or equivalently, that

[TABLE]

Clearly, the following conjecture is stronger than Conjecture 1.6.

Conjecture 1.7**.**

For any positive integer $n>1$ there exist odd primes $p$ and $q$ such that

[TABLE]

Note that the equality (6) can be written as

[TABLE]

whence it follows that $p=2k(2n-1)+1$ and $q=2k(2n-3)+1$ for a positive integer $k$. Hence, Conjecture 1.7 is equivalent to the following one.

Conjecture 1.7’. For any integer $n>1$ there exists a positive integer $k$ such that both numbers $p=2k(2n-1)+1$ and $q=2k(2n-3)+1$ are primes.

If $p$ and $q$ are odd primes, then from the expression (1) we see that $S_{n}(p,q)$ is odd if and only if $n$ is even. The following conjecture is the odd analogue of Conjecture 1.6.

Conjecture 1.8** (“weak odd Goldbach conjecture”).**

For each odd positive integer $a$ greater than $2$ there exist a positive integer $n$ and odd primes $p$ and $q$ such that $a=S_{2n+1}(p,q)$; or equivalently, that

[TABLE]

Clearly, the following conjecture is stronger than Conjecture 1.8.

Conjecture 1.9**.**

For any positive integer $n>1$ there exist odd primes $p$ and $q$ such that

[TABLE]

From the equality (8) we have

[TABLE]

whence we conclude that $p=nk+1$ and $q=(n-1)k+1$ for a positive integer $k$. This together with the fact that $k=p-q$ is even shows that Conjecture 1.9 is equivalent to the following one.

Conjecture 1.9’. For any integer $n>1$ there exists a positive integer $k$ such that both numbers $p=2kn+1$ and $q=2k(n-1)+1$ are primes.

Finally, notice that Conjectures 1.6 and 1.8 can be joined into the following conjecture.

Conjecture 1.10** (“weak Goldbach conjecture”).**

Conjectures $1.6$ and $1.8$ are true if and only if the following statement holds true$:$

For each positive integer $a$ greater than $2$ there exist a positive integer $n$ and odd primes $p$ and $q$ such that

[TABLE]