Does there exist a prime number within the interval?

Question

Conjecture

$\forall p_{n}\in \mathbb{P} : n\geq3, \: \exists p_{m}\in \mathbb{P} : 3p_{n} - 4 \geq p_{m} > \sqrt{2(p^2_{n+1} - 1)} $

How would you go about proving/disproving this?

Answer 1

This is true indeed. First consider $n$ s.t. $p_n \geq 29$. It has been proven that for $m \geq 3$, there is a prime in the interval $(m, \frac{4(m+2)}{3})$, see here, Corollary 2.2.

Since $p_n \geq 3$, there is a prime in $(p_n, \frac{4(p_n+2)}{3})$, so $p_{n+1}<\frac{4(p_n+2)}{3}$. Now $$\sqrt{2(p_{n+1}^2-1)}<\sqrt{2}p_{n+1}<\frac{4\sqrt{2}(p_n+2)}{3}<\left\lceil \frac{4\sqrt{2}(p_n+2)}{3} \right\rceil$$

Since $\left\lceil \frac{4\sqrt{2}(p_n+2)}{3} \right\rceil>\frac{8\sqrt{2}}{3} \geq 3$, there is a prime in $\left(\left\lceil \frac{4\sqrt{2}(p_n+2)}{3} \right\rceil, \frac{4(\left\lceil \frac{4\sqrt{2}(p_n+2)}{3} \right\rceil+2)}{3}\right)$.

Thus $\exists p_m$ s.t.

$$\sqrt{2(p_{n+1}^2-1)}<\left\lceil \frac{4\sqrt{2}(p_n+2)}{3} \right\rceil<p_m<\frac{4(\left\lceil \frac{4\sqrt{2}(p_n+2)}{3} \right\rceil+2)}{3}<\frac{4(\frac{4\sqrt{2}(p_n+2)}{3}+3)}{3}$$

Now since $p_n \geq 29$, we have

$$(3p_n-4)-(\frac{4(\frac{4\sqrt{2}(p_n+2)}{3}+3)}{3})=(3-\frac{16 \sqrt{2}}{9})p_n-(8+\frac{32\sqrt{2}}{9})>0$$

Thus

$$\sqrt{2(p_{n+1}^2-1)}<p_m<\frac{4(\frac{4\sqrt{2}(p_n+2)}{3}+3)}{3}<3p_n-4$$

It remains to check $5 \leq p_n<29$.

Checking small cases:

$p_n=5$: The interval is $(\sqrt{96}, 11]$ so $p_m=11$ works.

$p_n=7$: The interval is $(\sqrt{240}, 17]$ so $p_m=17$ works.

$p_n=11$: The interval is $(\sqrt{336}, 29]$ so $p_m=29$ works.

$p_n=13$: The interval is $(\sqrt{576}, 35]$ so $p_m=31$ works.

$p_n=17$: The interval is $(\sqrt{720}, 47]$ so $p_m=47$ works.

$p_n=19$: The interval is $(\sqrt{1056}, 53]$ so $p_m=53$ works.

$p_n=23$: The interval is $(\sqrt{1680}, 65]$ so $p_m=61$ works.