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Let $n\in \mathbb{N}\setminus\{1\}$, can every $n$ be written as: $ \ n=\pm p_{n}\pm p_{n-1}\cdots\pm p_{2}\pm p_{1} \,$ where $p_k$ is a prime number and :$\ \ p_{n}\neq p_{n-1}\neq\cdots \neq p_{2}\neq p_{1}$?

Examples:

$2=+7-5$

$3=+11-5-3$

$4=-17+11+7+3$

$5=+29-17-11+7-3$

$6=-29+17+13+7-5+3$

$7=+31-19-17+11-7+3+5$

$8=-31+29-21+17+13-11+7+5$

$9=+31-21-19+17+13-11+7-5-3$

$10=+37-31+21-19-17+13+11-7+5-3$

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  • $\begingroup$ @Shuri2060 Note the condition that the primes must be different. $\endgroup$
    – ajotatxe
    Jul 9, 2017 at 11:26
  • $\begingroup$ Assuming the twin prime conjecture, the even case is trivial and the odd case can be done using say ($11-5-3+(p_i+2-p_i)+...$). (but of course, it's only a conjecture) $\endgroup$
    – Shuri2060
    Jul 9, 2017 at 11:36
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    $\begingroup$ @Peter $2n=\sum^{n}_{i=1}((p_i+2)-p_i)$ where $(\forall i) (p_i, p_i+2$ are primes and $p_{i+1}>p_i+2)$. Similarly, $2n+1=11-5-3+\sum^{n-1}_{i=1}((p_i+2)-p_i)$ and $p_1>11$. $\endgroup$
    – Shuri2060
    Jul 9, 2017 at 12:16
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    $\begingroup$ Is it necessary to exclude 1? Can't we do $3 - 2 = 1$? Did you mean to exclude 0 instead? $\endgroup$ Jul 9, 2017 at 16:55
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    $\begingroup$ The sum of 8 and 9 contains composite number 21(21 is composite) $\endgroup$ Oct 8, 2017 at 14:31

2 Answers 2

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Yitang Zhang has been proved that there exist infinitely many primes such that their difference is at most $70,000,000$.

By the pigeonhole principle, this means that there exists at least one even fixed number, namely $2k$, less than $70,000,000$ such that there exists infinitely many primes whose difference is equal to $2k$.

Using computers we can list $k$ pairs of twin primes which are greater than 1000. Assume that the largest prime appears in this list in equal to $M$.

Now given $2n$ arrbitrary, by the Euclid's algorithm, there exist $t$ and $r$, such that we have such that $n=tk+r$ and $0 \leq r\leq k-1$.

Write $2n=(2k+...+2k)+(2+...+2)$, in which $2k$ apears $t$-times and $2$ appears $r$-times. Now replace every $2$ by the difference of a distinct twin primes, and replace every $2k$ by the difference of two new distinct primes which are both greater than $M$.

But notice that we have write $2n$ as the sum of $2t+2r$ primes, let $l:=n-t-r$,

  • if $l$ is divisible by 2, i.e. $l=2 l'$, then we can write $2n=(2k+...+2k)+(2+...+2)+\Big((2k-2k)+...+(2k-2k)\Big)$ in which $2k-2k$ apears $l'$-times, now replace every new $2k$ by the difference of two new primes.
  • if $l$ has the form $l=2l'+1$, i.e. $l=2l''+3$, then we can write $2n=(2k+...+2k)+(2+...+2)+\Big((2k-2k)+...+(2k-2k)\Big)+(4-2-2)$ in which $2k-2k$ apears $l''$-times, now replace every new $2k$ by the difference of two new primes, and replace $4$ by $101-97$ and replace each of the two new $2$s by $73-71$ and $61-59$.

Also notice that $2n+1=(2n-2)+(11-5-3)$.

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  • $\begingroup$ 'By the pigeonhole principle, this means that there exists an even number, namely $2k$, less than $70,000,000$ such that there exists infinitely many primes whose difference is equal to k' - Could you elaborate? If such an argument existed surely you could use inductive logic to prove the twin prime conjecture? $\endgroup$
    – Shuri2060
    Jul 9, 2017 at 11:48
  • $\begingroup$ All difference of odd primes is equal to $2k$..... $\endgroup$
    – Piquito
    Jul 9, 2017 at 12:07
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    $\begingroup$ The bound has been substantially lowered $\endgroup$
    – Asinomás
    Jul 9, 2017 at 12:13
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    $\begingroup$ The last news I have heard about the bound is that it has been lowered to 246. $\endgroup$
    – Davood
    Jul 9, 2017 at 12:19
  • $\begingroup$ Dear @Shuri2060, you are right and I have modified the answer. $\endgroup$
    – Davood
    Jul 9, 2017 at 12:20
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Well, this is one that we could look at through the lens of a restatement of Goldbach's conjecture $$\{p+q: p, q \in \mathbb{P} \cap [1]_2 \} = \{0 \pmod{2}\} \cap [6,\infty)$$ and a restatement of De Polignac's Conjecture: $$\{p-q: p, q \in \mathbb{P} \cap [1]_2 \} = \{0 \pmod{2}\} \cap [2,\infty)$$

I wouldn't be answering if I didn't believe there was a compelling argument for both, in terms of compound arithmetic progressions and composite topologies.

Goldbach's Conjecture loses its teeth, when the primes are unique.

Assume both are true. Then with 2 primes, all the positive even numbers are covered under De Polignac's Conjecture, so it works for $n = 2$. But the covering in this case contains all the even numbers except 0, because the primes in the sum were specified to be unique.

So therefore with three, we can start at any non-zero even number, and find primes that formed it through either addition or subtraction. Restricting ourselves to positive even numbers formed by the sum of two primes, allows that both are less than $n$, and that we can negate the sum and arrive at: $$ n = p_3 - (p_1 + p_2) $$

Uniqueness implies that the Goldbach-type form $p_1 + p_2$ can only reach $[8, \infty) \cap [0]_2$, assuming it still holds when $p \ne q$ in the expression above (I wouldn't ask you to do this if I didn't have a way to either verify it, or prove that it was wrong). If we restrict $n$ to the positive numbers, $p_3 > p_1 + p_2$, such that $p_3$ is unique and we can subtract any even number in $[8, \infty)$ less than $p_3$, arriving at all the odd numbers greater than 1, so of course it works for $n = 3$.

To maintain uniqueness, order the primes by subscript. Because we've assumed De Polignac's Conjecture, then we've enabled infinitely many gaps of each even size. Therefore, subtract the previous expression from $p_4$ and observe that the following expression allows $n$ to be any positive even number or zero: $$ n = p_4-p_3 + p_2 - p_1 $$

Coset Property.

If $S_n := \{\sum_{k=1}^n \pm p_k : p_i \in \mathbb{P} \subset \mathbb{Z} \}$ is a set, $\{n \pmod {2} \} \cap \mathbb{Z}^+ \subset S_n$.

Proof.

If the coset property holds for any $n-1$, then $S_{n-1}$ contains the arithmetic progression of evens or odds according to whether $n-1$ is even or odd. If you subtract this set of numbers from some prime $p_n > p_{n-1}$, because the prime $p_n$ could be arbitrarily large, then $\{n \pmod {2} \} \cap \mathbb{Z}^+ \subset S_n$

Then because the coset property can be observed for $n = 3$, it can be observed for $n \ge 3$. Since we already have examples for $2 \le n \le 10$ and $n \in \{n\pmod{2}\}$, the proof is complete assuming De Polignac's Conjecture and the modified form of Goldbach's Conjecture.

I'm not entirely sure how the uniqueness requirement affects Goldbach, though.

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