The product of the first $k$ prime numbers except the prime number $2$ $\pm 2$ is a prime number?

Question

If $p_i$ is the $i^{th}$ prime number, let $\forall k \in \mathbb{N}, k>2, P_k := \prod\limits_{i = 2}^{k}p_i.$ Then, at least one of $(P_k + 2, P_k -2)$ is a prime number.

I tested it up to $k = 10$. Does it hold for all $k$, really? Is what I'm proposing nonsense? It's just an idea that came to my mind. I am new here.

As a general remark....there are lots and lots of small prime numbers. If you only search amongst small numbers, you will get a lot of false patterns. Any general claim, such as you are making, merits a deeper search. Wolfram Alpha is good enough in this case. — lulu, Feb 16 '23 at 17:54
Unfortunately, your proposition is false. The first counterexample occurs when $k = 14,$ since both $6541380665835015 - 2$ and $6541380665835015 + 2$ are composite numbers. — Luca Armstrong, Feb 16 '23 at 17:59
Note that $P_k\pm2$ won't have $p_1, p_2, \dots p_k$ as factors — J. W. Tanner, Feb 16 '23 at 18:09

lhf · Answer 1 · 2023-02-16T18:01:47.777

2

The conjecture is false. For $k=14$, we get:

$ 3 \cdot 5 \cdot 7 \cdot 11 \cdot 13 \cdot 17 \cdot 19 \cdot 23 \cdot 29 \cdot 31 \cdot 37 \cdot 41 \cdot 43 = 6541380665835015 $

$ 6541380665835015 - 2 = 6541380665835013 = 431 \cdot 125887 \cdot 120562229 $

$ 6541380665835015 + 2 = 6541380665835017 = 613 \cdot 7841 \cdot 1360935349 $

edited Feb 16 '23 at 18:01

answered Feb 16 '23 at 17:54

lhf

216,483

you forgot to type $41$ -- I believe that makes it $k=14$ -- but the rest is correct, so $+1$ – J. W. Tanner Feb 16 '23 at 18:00
1

@J.W.Tanner, thanks for spotting this! – lhf Feb 16 '23 at 18:02

The product of the first $k$ prime numbers except the prime number $2$ $\pm 2$ is a prime number?

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