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Proof of Andrica's conjecture by assuming Oppermann's conjecture.

Oppermann's conjecture: $$n\geq2\wedge\pi\left(n^{2}-n\right) < \pi\left(n^{2}\right) < \pi\left(n^{2}+n\right).$$

Andrica's conjecture: $$i\geq2\wedge\sqrt{p_{i+1}}-\sqrt{p_{i}}<1.$$
Insert Gottfried Helms' nice proof here$\ \ \Box$

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Fred Daniel Kline
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  • I don't know why @daniel s answer was deleted? My "hmm" was just a keyboard-reflex, that I was thinking... Just - I would have had expanded the answer by including, that by the truth of the square-root-inequality for $n \ge 1$ the OP's question was completely resolved. – Gottfried Helms Jan 10 '15 at 13:09
  • @GottfriedHelms: Sorry, I told with OP that I would help him with presentation and then delete the answer as it was not really an answer. Thank you for the explanation which is what I finally decided you meant. I think OP is still thinking about changes in the form of the question but for sure must include your suggested expansion. Yes the truth of the inequality I think does it. – daniel Jan 10 '15 at 13:22
  • I have now one concern. Oppermann's conjecture focuses on the interval *around a square-number $n^2$ .* So this conjectures only for the case, that the center $m=n^2$ of the interval [$m-\sqrt m$ .. $m+\sqrt m$] is a square. But what if Oppermann is true for such cases, but untrue for cases, where the center $m$ is not a square-number? (I don't know actually whether this is possibly a non-issue, didn't step into the conceptual part of question too deeply) – Gottfried Helms Jan 10 '15 at 13:57
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    @daniel: not necessarily. Consider there is a modular argument which tells us that between $m=n^2$ and $m-\sqrt m = n^2-n $ there is a prime, but that this does not need to hold if $m$ not a square. Then by Andrica, we need a smaller range than $m+\sqrt m$ for the next prime to occur. But, well, I've definitely not yet stepped into this sufficiently deep... – Gottfried Helms Jan 10 '15 at 14:46

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[update]: Upps, I see that while I'm just writing @Dan has undeleted his earlier answer. So priority (and thus also the bounty) is at Dan's, of course

I) Let's look at some examples due to Oppermann's consideration: $$ \small \begin{array}{} m_{13}=13^2=169 & a_{13} = 169-13+1=157 & b_{13}=169+13-1 = 181 \\ m_{14}=14^2=196 & a_{14} = 196-14+1=183 & b_{14}=196+14-1 = 209 \\ m_{15}=15^2=225 & a_{15} = 225-15+1=211 & b_{15}=225+15-1 = 239 \\ \end{array}$$ or , written more compactly $$ \small \begin{array}{} m_{13}: & 157...168 & 169 &170... 181 \\ m_{14}: & 183...195 & 196 & 197... 209 \\ m_{15}: & 211...224 & 225 & 226...239 \\ \end{array}$$ We can see that the intervals for consecutive $m_k$ cover the integers except the numbers $n^2$, $n^2+n$ etc, which by definition are not prime. Oppermann's conjecture is true, if in each of the intervals is at least one prime.

II) But because the intervals follow each other immediately, we can also look at the two intervals between the two consecutive squares $n^2$ and $(n+1)^2$.

$$ \small \begin{array} {rrrrr} n^2 & \big[(n^2+1) \cdots (n^2+n-1)\big] & (n^2+n) & \big[(n^2+n+1) \cdots ((n+1)^2-1) \big] & (n+1)^2 \\ \end{array}$$

The endpoints $n^2$ , $(n+1)^2$ and the center $n^2+n = (n+1)^2-(n+1)$ are not prime by construction, so the maximal distance between two primes can - assuming Oppermann - at most be $((n+1)^2-1) - (n^2+1)$.

III) Now express the inner neighbours of the two endpoints as $p_0 = n^2+1$ and $p_1=(n+1)^2-1$ which mark the greatest possible distance between two numbers in that interval , then we can refer to the Andrica-formulation $\sqrt{p_1} - \sqrt{p_0}<1$ as $$ \sqrt{((n+1)^2-1)} - \sqrt{(n^2+1)}\lt 1$$ But this is with a small $\small 0 \le \delta \lt 1$ and thus $\small \sqrt{n^2+1} \sim n+\delta $ and $ \small \sqrt{(n+1)^2-1} \sim (n+1)-\delta $
$$ \big((n+1)-\delta \big) - \big(n +\delta \big) = 1 - 2 \delta \lt 1$$ and thus always (for $n$ greater than some small value, perhaps even if greater than 1) true.

IV) From this follows, that Oppermann's conjecture implies Andrica's conjecture

Gottfried Helms
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