Let $p_n$ denote the $n$th prime. Recall Andrica's conjecture, which states that $$\sqrt{p_{n+1}}-\sqrt{p_n}<1\quad\text{ for all }\,n.$$ I think Andrica's conjecture implies Bertrand's postulate. I have found that it actually implies $\frac{p_{n+1}}{p_n} < \frac32$ (for $p_n>11$). Is this true?

Assume $$(\sqrt{p_{n+1}}-\sqrt{p_n})(\sqrt{p_{n+1}}+\sqrt{p_{n}}) > \frac{p_{n}}2$$ ( in other words $p_{n+1}/p_{n} > 3/2$). Assume also $$1> \sqrt{p_{n+1}}-\sqrt{p_{n}}$$ (this is Andrica's conjecture) then $\sqrt{p_{n+1}}+ \sqrt{p_{n}} > p_{n}/2$ therefore $$\frac{\sqrt{p_{n+1}}}{\sqrt{p_{n}}} + 1 > \frac{\sqrt{p_{n}}}2.$$ So $\sqrt2+ 1 > \sqrt{p_{n}}/2$, this implies $24 > p_{n}$. Note, $p_{n}$ means the $n$-th prime.

So if Andrica's conjecture is true for any $p_{n} > 24$ then $$\frac{p_{n+1}}{p_{n}} <\frac32,$$ for any primes greater than 29.

If $p_{n+1}-p_n> p_n/2$. $$(\sqrt{p_{n+1}}-\sqrt{p_n})(\sqrt{p_{n+1}}+\sqrt{p_n})>\frac{p_n}2$$ If $\sqrt{p_{n+1}}-\sqrt{p_n} <1$ therefore $\sqrt{p_{n+1}}+\sqrt{p_n}>p_n/2$.

Therefore $\sqrt{p_{n+1}/p_n}+1 > \sqrt{p_n}/2$, therefore $\sqrt2 + 1 > \sqrt{p_n}/2$ hence $24>p_n$.

So if $\frac{p_{n+1}}{p_n} > \frac32$ and $\sqrt{p_{n+1}} - \sqrt{p_n} <1$ then $p_n < 24$.

Therefore if $\sqrt{p_{n+1}} - \sqrt{p_n} <1$ and $P_n > 24$ then $\frac{p_{n+1}}{p_n} < \frac32$. (Note if $p_n >2 p_{n+1}/p_n$ does not equal $3/2$.)

  • $\begingroup$ Is this a bad question? $\endgroup$ – user128932 Jul 27 '14 at 6:42
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    $\begingroup$ It would be a good thing to include (a sketch of) your proof in the question. Ohtherwise you are asking users to reproduce the work you did. $\endgroup$ – Bart Michels Aug 5 '14 at 11:02
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    $\begingroup$ Assume (SQRT(P(n+1))-SQRT(P(n)))(SQRT(P(n+1)+SQRT(P(n)) > P(n)/2 ( in other words P(n+1)/P(n) > 3/2), given SQRT means square root. Assume also 1> (SQRT(P(n+1)-SQRT(P(n))) (this is Andrica's conjecture) then (SQRT(P(n+1))+ SQRT(P(n))) > P(n)/2 THERFORE (SQRT(P(n+1)/SQRT(P(n)) + 1 > SQRT(P(n))/2. So SQRT(2)+ 1 > SQRT(P(n))/2 , this implies 24 > P(n). Note ,P(n) means the n-th prime. So if Andrica's conjecture is true for any P(n) > 24 then P(n+1)/P(n) <3/2 , for any primes greater than 29. $\endgroup$ – user128932 Aug 9 '14 at 4:28
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    $\begingroup$ I have copied your comment from the deleted answer. (I did not notice your comment above.) I hope it can help you with basics of the typesetting math. And you can edit the post to the version which you like (and which is also readable for others). I think that expression like $\sqrt{p_{n+1}}-\sqrt{p_n}$ is easier to read than (SQRT(P(n+1))-SQRT(P(n))). This is true especially if such things appear in longer expressions or derivations. $\endgroup$ – Martin Sleziak Aug 16 '14 at 13:39
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    $\begingroup$ I have also added LaTeXed version of your first comment. Please if you have time, edit your question further. (Leave only the version which you prefer. Add further comments or clarification if necessary. If I unintentionally made some mistakes when transcribing, please, correct them.) $\endgroup$ – Martin Sleziak Aug 17 '14 at 7:12

Several conjectures (and results) on prime numbers can be expressed using gap size $$g_n=p_{n+1}-p_n.$$

Bertrand's postulate is equivalent to $g_n<p_n$.

Oppermann's conjecture is equivalent to $g_n<\sqrt{p_n}$

Andrica's conjecture says that $\sqrt{p_{n+1}}-\sqrt{p_n}<1$. If we multiply both sides of the inequality by $\sqrt{p_{n+1}}+\sqrt{p_n}$ we get $$g_n=p_{n+1}-p_n<\sqrt{p_{n+1}}+\sqrt{p_n}.$$ Using $\sqrt{p_{n+1}}-\sqrt{p_n}<1$ once again we have $\sqrt{p_{n+1}}<\sqrt{p_n}+1$ and $$g_n < 2\sqrt{p_n}+1.$$

It is clear that Opperman's conjecture implies both Bertrand's postulate and Andrica's conjecture

Since $2\sqrt{p_n}+1<p_n$ for $p_n\ge11$, Andrica's conjecture implies Bertrand's postulate.

And also for any positive constant $c$ we have $$2\sqrt{p_n}+1< c p_n$$ for all large enough $n$'s.

So this implies $g_n < cp_n$, i.e. $$p_{n+1}<(1+c)p_n$$ for large enough $n$. (Which means that there is only finitely many $n$'s left to check.)

  • $\begingroup$ this is a bit unorthodox, but I am wondering given your knowledge of Opperman's and Andrica's conjectures if you had any thoughts on this question: math.stackexchange.com/q/1422179/252983 $\endgroup$ – Colm Bhandal Sep 4 '15 at 19:15
  • $\begingroup$ I think that calling this knowledge of Opperman's and Andrica's conjectures is a bit of an overstatement. I basically copied here what I read in the linked Wikipedia article and made a few simple algebraic operations. (I am afraid I can't help you much there.) $\endgroup$ – Martin Sleziak Sep 7 '15 at 14:21
  • $\begingroup$ Haha OK cheers for the reply anwyay. $\endgroup$ – Colm Bhandal Sep 7 '15 at 14:47

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