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Not sure how to show $p_{n+1} \leq 4p_1...p_n - 1$, where $p$ are primes of the form $4n + 3$.

Only way that leaps out to me right now is thinking about it intuitively, the right side being a larger number than the next prime but I'm not sure how to show that.

amWhy
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SS'
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  • Your question as currently written is very difficult to understand. Please fix your indices $n$ and $k$ -- which currently seem to be placed in random ways -- to create a meaningful question. – zipirovich Jan 19 '18 at 05:43
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    You have to adapt Euclid's proof of the infinitude of primes, using the extra consideration that every prime is congruent to $\pm 1$ mod 4. – fred goodman Jan 19 '18 at 05:43
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    ...and that every number of the form 4n+3 must have at least one prime divisor of the same form. – Ivan Neretin Jan 19 '18 at 05:44
  • Relevant. –  Jan 19 '18 at 05:47
  • Gotcha, makes sense adapting Euclid's argument. Thanks! – SS' Jan 19 '18 at 05:56
  • Do you mean every prime in the expression $p_{n+1} \leq 4p_1 p_2 ... p_n -1$ is of the form $4n+3$? Or you take all consecutive primes like Bonse's inequality? – rtybase Jan 19 '18 at 20:42
  • You know what, it's not very polite to ask a question and then abandon it. Your question statement was somewhat unclear. You have a specific question addressed to you asking for clarification. You have two different answers based on two different interpretations. And you are not paying any attention to any of this, presumably because you have already understood the matter, or no longer care. But you actually have a responsibility to curate your question. – fred goodman Jan 25 '18 at 21:13
  • My apologies, I'm not always on here and didn't get an email regarding the question. Saw the comment, understood what to do from there but didn't see that there were answers. – SS' Jan 26 '18 at 01:16

2 Answers2

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Problem statement: I think the intended problem statement is the following.

Let $p_k$ denote the $k$-th prime congruent to $-1$ mod $4$, i.e. the $k$-th prime of the form $4n + 3$ for some $n$. Show that for all $k \ge 1$, $$ p_{k+1} \le 4 p_1 p_2 \cdots p_k - 1. $$

Solution: the integer $n(k) = 4 p_1 p_2 \cdots p_k - 1$ is not divisible by $p_j$ for $1 \le j \le k$, and is congruent to $-1$ mod $4$. Now we have to use that every integer congruent to $-1$ mod $4$ must have at least one prime factor congruent to $-1$ mod $4$. Let $p$ be a prime factor of $n(k)$ that is congruent to $-1$ mod $4$. Then $p \le n(k)$ and $p$ is not equal to any of $p_1, \dots, p_k$. It follows that $p_{k+1} \le p \le n(k)$.

fred goodman
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We always have $$ p_{n+1}\le p_1p_2\cdots p_n+1\le 4p_1p_2\cdots p_n-1. $$

The second estimate is clear because of $3p_1p_2\cdots p_n\ge 2$. The first estimate is shown here:

$(n+1)th$ prime $p_{n+1}$ less than or equal to $p_1p_2\dots p_n+1$

So you do not need the assumption that the primes are of the form $4n+3$.

Dietrich Burde
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