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I was reading a Mathematical paper from the '40s, and came across the following formula:

$$q = \theta\sqrt{\frac{n}{\nu}}$$

Where $q$ is a prime number, $\nu$ is a given integer, and $\theta \in (1, 1+\vartheta)$, where $\vartheta$ is a given real. $n$ is an integer.

For the argument in the paper to work, I believe this formula may only fail for finitely many $n$. The value of $\theta$ may of course depend on $n$. The only reference why this should be the case, is that: "According to the law of prime numbers, for sufficiently large $n$ such a prime $q$ exists." I was not able to find any such law, and this is were I need help.

Why should this be the case?

Link to the original paper, which is in Russian: http://www.mathnet.ru/php/archive.phtml?wshow=paper&jrnid=sm&paperid=5985&option_lang=eng

Xander L
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1 Answers1

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It seems the full claim is:

Let $\vartheta >0$ and $\nu\in\Bbb N$. Then for almost all $n\in\Bbb N$, there exists $\theta\in(1,1+\vartheta)$ such that $\theta\sqrt{\frac n\nu}$ is prime.

or equivalently

Let $\vartheta >0$ and $\nu\in\Bbb N$. Then for almost all $n$, there exists a prime $q$ such that $$\sqrt{\frac n\nu}<q<(1+\vartheta)\sqrt{\frac n\nu}.$$

For this it is sufficient to show

Let $\vartheta>0$. Then for all $x\gg0$, $$ \pi((1+\vartheta) x)>\pi(x).$$

What the Prime Number Theorem gives us is: For $\alpha <1<\beta$ of our choice, we have $$ \frac{\alpha x}{\ln x}<\pi(x)<\frac{\beta x}{\ln x}$$ for all $x\gg 0$. So we can certainly achieve $\frac \beta\alpha < 1+\vartheta$. Then $$\begin{align} \pi(x+\vartheta x)-\pi(x) &>\frac{\alpha x(1+\vartheta)}{\ln x+\ln(1+\vartheta)}-\frac{\beta x}{\ln x}\\ &>\frac{\alpha x(1+\vartheta)}{\ln x+\vartheta}-\frac{\beta x}{\ln x}\\ &=\frac{\alpha x(1+\vartheta)\ln x - \beta x\ln x - \beta\vartheta x}{(\ln x+\vartheta)\ln x}\\ &=\frac{x}{(\ln x+\vartheta)\ln x}\cdot\bigl(\overbrace{(\alpha(1+\vartheta)-\beta)}^{>0} \ln x-\beta\vartheta\bigr ). \end{align}$$ This is positive for $x\gg0$.

Hagen von Eitzen
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