5
$\begingroup$

Let $p_1,p_2,...,p_k$ be the first $k$ primes and $k>1$. Does there exist a prime $p_{k+1}$ such that $p_{k+1} = p_1p_2...p_k + 1$?

$\endgroup$

closed as off-topic by Saad, Don Thousand, José Carlos Santos, Gibbs, GNUSupporter 8964民主女神 地下教會 Oct 12 '18 at 20:50

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Saad, Don Thousand, José Carlos Santos, Gibbs, GNUSupporter 8964民主女神 地下教會
If this question can be reworded to fit the rules in the help center, please edit the question.

  • 1
    $\begingroup$ Are you asking whether $p_1 p_2 \dots p_k + 1$ is ever a prime, always a prime, or ever / always the next prime after $p_k$? The answers to these questions, respectively, are 1) yes, 2) no, 3) almost never, it's too big. $\endgroup$ – Qiaochu Yuan Oct 9 '18 at 20:49
  • $\begingroup$ @Qiaochu I was asking whether there exists a other primes $p$ other than $3$ such that is a product of all the preceding primes plus one $\endgroup$ – Tom Carter Oct 9 '18 at 20:53
  • 1
    $\begingroup$ Your phrasing of the question is confusing because you ask "does there exist a prime" and then call it $p_{k+1}$, which leaves it ambiguous whether you want this to be some arbitrary prime or specifically the next prime after $p_k$. If the latter, this is really a question about $k$, and once you've picked $k$ then $p_{k+1}$ is already uniquely determined, so it doesn't make grammatical sense to ask the question about $p_{k+1}$. It's like saying "let $k$ be an integer. Does there exist an integer $k+1$ such that..." $\endgroup$ – Qiaochu Yuan Oct 9 '18 at 20:54
5
$\begingroup$

Bertrand's postulate states that there is always a prime between $n$ and $2n$, whenever $n > 3$. Since $p_1 = 2$, you are asking for instances when $p_{k+1} = 2 p_2 \dots p_k + 1$; but we know $p_{k+1}$ lies between $p_k$ and $2 p_k$, so in particular it is less than $2 p_k$ and hence less than $2 p_2 \dots p_k + 1$.

$\endgroup$
  • 1
    $\begingroup$ To add an example to this question that I think might be useful to future readers: If $k=2$, then $p_1p_2+1=2\cdot 3+1=7$. $7$ is prime, but it is not the next prime after $3$ -- that prime is $5$. Using our standard nomenclature, $p_1p_2+1=p_4\neq p_3$. $\endgroup$ – Carl Schildkraut Oct 10 '18 at 16:21
3
$\begingroup$

There only exists one such prime, and that's 3, because by Bertrand's postulate we see that... oh, too slow, Patrick already mentioned Bertrand.

I'm going to try another way. The prime number theorem tells us that $$\pi(n) \sim \frac{n}{\log n}.$$ It's not very precise for small numbers, but it will do for our purposes here.

So, if $$\pi\left(\prod_{i = 1}^{k - 1} p_i\right) \approx \frac{\prod_{i = 1}^{k - 1} p_i}{\log \prod_{i = 1}^{k - 1} p_i}$$ and $$\pi(p_k) \approx \frac{p_k}{\log p_k},$$ then $$\pi\left(\prod_{i = 1}^{k} p_i\right) \approx \frac{\prod_{i = 1}^{k} p_i}{(\log p_k) + \log \prod_{i = 1}^{k - 1} p_i}.$$

Remember the basic properties of logarithms: we have a multiplication for the numerator and an addition for the denominator.

And since $\log p_k > 1$ for $k > 1$, the denominator does increase but not as much as the numerator. This storngly suggests that there are at least a couple of primes between $$\prod_{i = 1}^{k - 1} p_i$$ and $$\prod_{i = 1}^{k} p_i,$$ so even if $$1 + \prod_{i = 1}^{k} p_i$$ is indeed prime, it can't be the next prime.

Try it out with, say, the first four primes and a scientific calculator.

$\endgroup$

Not the answer you're looking for? Browse other questions tagged or ask your own question.