Are there always prime numbers that verify these parameters?

Question

Let $p_1$, $p_2$, $p_3$, $p_4$ be prime numbers such that :

$p_k\ne 2$ or $3$ for $k=1$ or $4$
$p_1\gt p_3$, $ p_4\gt p_2$
$p_1 - p_3 +2=- p_2+p_4$

Prove that for any $p_1$ and $p_2$, they exist a $p_3$ and a $p_4$.

I hope you'll understand ! It is, i think, really difficult. I'm not sure if this is always true honestly.

This problem was made by my mind. There is a link with Goldbach's conjecture. I just found it through some reflexion. — Jotadiolyne Dicci, Jul 10 '20 at 21:21
In the light of what is givem, rewrite condition 4 as $p_1+p_2+2=p_3+p_4$. Then the claim is immediately equivalent to Goldbach's conjecture. Good luck. — Hagen von Eitzen, Jul 10 '20 at 21:23

score 2 · Accepted Answer · answered Jul 10 '20 at 21:32

2

If I understand it correctly, the statement isn't true. Let $p_1=5,\ p_2=17$. Then $p_3\in\{2,3\}$, and $$p_1+p_2+2-p_3=24-p_3\in\{22,21\},$$ so there is no prime $p_4$ satisfying the conditions.

answered Jul 10 '20 at 21:32

saulspatz

53,131

Oh interesting ! I didn't see that coming ! – Jotadiolyne Dicci Jul 10 '20 at 21:38
$p_1=p_2=5$ would also work. In fact there are probably infinitely many counterexamples with $p_1=5$. – Barry Cipra Jun 22 '21 at 01:09

score 1 · Answer 2 · answered Jul 10 '20 at 21:23

As you commented, the third point is connected to the Goldbach conjecture when written as $p_1+p_2+2=p_3+p_4$; in other words, two consecutive even numbers are Goldbach numbers. It is known that almost all even numbers are Goldbach numbers, so this equation alone has lots of solutions; I don't think it would be hard to show that most such solutions satisfy the first two bullet points as well.

Are there always prime numbers that verify these parameters?

2 Answers2