Prime number between $\sqrt{n}-n^{1/3}$ and $\sqrt{n}$

Solution 1:

Some thoughts:

Put $n=t^4$. You are basically saying that the is a prime between $t^2-t^\frac43$ and $t^2$. For large enough values of $t$:

$$t^2-t^{\frac43}<t^2-2t<t^2-2t+1=(t-1)^2$$

So if there is always a prime between $(t-1)^2$ and $t^2$ (Legendre's conjecture), the statement in your book is definitely correct. But Legendre's conjecture has not been either proved or disproved yet.

On the other side, if you put $n=t^6$ you are basically saying that there is always a prime between $t^3-t^2$ and $t^3$. But:

$$t^3-t^2>t^3-3t^2+3t+1=(t-1)^3$$

It is proved that there is a prime between two consecutive cubes, like $(t-1)^3$ and $t^3$ for large enough values of $t$ ($t\ge e^{e^{33.217}}$) and the proof is fairly complex, not the easiest piece of math to read. But that fact is useless in this case. The statement in your book seems to be narrowing the gap even further because for large enough $t$:

$$(t-1)^3<t^3-t^2<t^3$$

Proving that there is a prime between $t^3$ and $t^3-t^2$ is even more difficult because the gap between numbers is now just 1/3 of the gap that we had before. It is also possible that such prime is not guaranteed to exist.

Solution 2:

As suggested by Barry Cipra and myself , putting $n=x^6$ gives the interval $[x^3, x^3-x^2]$. Consider: $$π(x^3) - π(x^3-x^2)$$ ,where $π$ is the prime counting function. As $n$ is large, $x$ is also large and for sufficiently large values of $x$ , $π$ can be approximated by the logarithmic integral $li (z) := \int^z_0 \frac {dt}{\ln t}$. The series definition is: $$li (e^z) := \gamma + \ln |z| + \sum_{n=1}^∞ \frac {z^n}{n.n!}$$ Now taking $e^u = x^3$ and $e^v = x^3 - x^2$, using the above mentioned definitions and approximations , and simplifying we have: $$π(x^3) - π(x^3-x^2) ≈ \ln |\frac {u}{v}| + \sum \frac {u^n - v^n}{n.n!}$$ For large enough values of $u$ and $v$ the sum converges to values much larger than $1$. Hence, $$π(x^3) - π(x^3-x^2) ≥1$$ As there is atleast one prime between $x^3$ and $x^3 - x^2$ our result is proved.

Solution 3:

I think this gets you in the ball park.

Number of primes $<x\approx \frac{x}{ln(x)}$

$\frac{\sqrt{x}}{ln(\sqrt{x})}=\frac{2}{\sqrt{x}}\frac{x}{ln(x)}$

$\frac{x^{1/3}}{ln(x^{1/3})}=\frac{3}{x^{2/3}}\frac{x}{ln(x)}$

So it suffices to prove $(\frac{2}{\sqrt{x}}-\frac{3}{x^{2/3}})\frac{x}{ln(x)}>1$ for sufficient large x.

Perhaps the ratios could help mitigate relative error issues.

We want: $$\pi(\sqrt{x})-\pi(x^{\frac{1}{3}})>1$$ $$\frac{\pi(\sqrt{x})}{\pi(x^{\frac{1}{3}})}>1+\frac{1}{\pi(x^{\frac{1}{3}})}$$ $$\pi(\sqrt{x})>2\pi(x^{\frac{1}{3}})$$ Taking ratios of the approximations above,

$\frac{\pi(\sqrt{x})}{\pi(x^{\frac{1}{3}})}\approx \frac{2}{3}x^{\frac{1}{6}}$

$\frac{2}{3}x^{\frac{1}{6}}\pi(x^{\frac{1}{3}})>\pi(x^{\frac{1}{3}})+1$

$\frac{2}{3}x^{\frac{1}{6}}>1+\frac{1}{\pi(x^{\frac{1}{3}})}$

$\frac{2}{3}x^{\frac{1}{6}}>2$

$x>729$