There exists n such that number of primes between $(n+1)^{2021}$ and $n^{2021}$ is greater than 1122021140.

elementary-number-theorynumber theoryprime numbers

To show there exists a natural number $n$ such that there are at least 1122021140 primes between $(n+1)^{2021}$ and $n^{2021}$.

Here is my attempt :

$\Rightarrow$
We define $\pi(x)=\#\{p \leq x \mid p$ is prime $\}$.

Then by prime number theoremn $\pi(x) \sim \frac{x}{\log x}$.

So I need to show $\pi(y)-\pi(x) > 1122021140$ for $y=(n+1)^{2021}$ and $x=n^{2021}$

For large $x$ we have (saw it here)
$$
\pi(y)-\pi(x) \sim \frac{y-x}{\log x} \tag1
$$

$$
\begin{aligned}
&=\frac{(n+1)^{2021}-n^{2021}}{\log (n)^{2021}} \\
&=\frac{(n+1)^{2021}-n^{2021}}{2021 \log n}
\end{aligned}
$$

I have two questions :

  1. How to prove equation $(1)$ i.e., $\pi(y)-\pi(x) \sim \frac{y-x}{\log x}$ for large $x$.

  2. How to show there exists a $n$ such that $\frac{(n+1)^{2021}-n^{2021}}{2021 \log n} > 1122021140$.

Best Answer

Let $\epsilon$, $M$ be positive numbers. Then there exists $n$ such that between $n^{1+\epsilon}$ and $(n+1)^{1+\epsilon}$ there are more than $M$ prime numbers. Indeed, otherwise a simple estimate would show that $$\sum_{p \textrm{ prime}} \frac{1}{p} \le M \cdot \sum_{n=1}^{\infty} \frac{1}{n^{1+\epsilon}}< \infty$$ contradiction

$\bf{Added:}$ A similar argument shows that for every increasing sequence $a_n$ with $\sum \frac{1}{a_n} < \infty$ we have $$\lim \sup_{n\to \infty} (\pi(a_{n+1}) -\pi(a_n)) = +\infty$$

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