# Mills' Theorem

## Theorem

There exists a real number $A$ such that $\left\lfloor{A^{3^n} }\right\rfloor$ is a prime number for all $n \in \N_{>0}$, where:

$\left\lfloor{x}\right\rfloor$ denotes the floor function of $x$
$\N$ denotes the set of all natural numbers.

## Proof

We define $f \left({x}\right)$ as a prime-representing function if and only if:

$\forall x \in \N: f \left({x}\right) \in \Bbb P$

where:

$\N$ denotes the set of all natural numbers
$\Bbb P$ denotes the set of all prime numbers.

Let $p_n$ be the $n$th prime number.

$p_{n+1} - p_n < K p_n^{5/8}$

where $K$ is an unknown but fixed positive integer.

### Lemma 1

$\forall N > K^8 \in \Z: \exists p \in \Bbb P: N^3 < p < \left({N + 1}\right)^3 - 1$

#### Proof

Let $p_n$ be the greatest prime less than $N^3$.

 $\displaystyle N^3$ $<$ $\displaystyle p_{n+1}$ because $p_n$ is the greatest prime less than $N^3$ $\displaystyle$ $<$ $\displaystyle p_n + K {p_n}^{\frac 5 8}$ Difference between Consecutive Primes $\displaystyle$ $<$ $\displaystyle N^3 + K N^{\frac {15} 8}$ because $p_n < N^3$ $\displaystyle$ $<$ $\displaystyle N^3 + N^2$ because $N > K^8$ $\displaystyle$ $<$ $\displaystyle N^3 + 3N^2 + 3N$ $\displaystyle$ $=$ $\displaystyle \left({N + 1}\right)^3 - 1$

Therefore:

$N^3 < p_{n+1} < \left({N + 1}\right)^3 - 1$

$\Box$

Let $P_0 > K^8$ be a prime number.

By Lemma 1, there exists an infinite sequence of primes:

$P_0, P_1, P_2, \ldots$

such that:

$\forall n \in \N_{>0}: {P_n}^3 < P_{n+1} < \left({P_n + 1}\right)^3 - 1$

Then we define two functions $u, v: \N \to \Bbb P$:

$\forall n \in \N: u \left({n}\right) = {P_n}^{3^{-n} }$
$\forall n \in \N: v \left({n}\right) = \left({P_n + 1}\right)^{3^{-n} }$

It is trivial that $v \left({n}\right) > u \left({n}\right)$.

### Lemma 2

$\forall n \in \N_{>0}: u \left({n + 1}\right) > u \left({n}\right)$

#### Proof

 $\displaystyle u \left({n + 1}\right)$ $=$ $\displaystyle {P_{n+1} }^{3^{-\left({n+1}\right)} }$ $\displaystyle$ $>$ $\displaystyle \left({P_n^3}\right)^{3^{-n-1} }$ because $P_{n+1} > {P_n}^3$ $\displaystyle$ $=$ $\displaystyle {P_n}^{3 \times 3^{-n-1} }$ $\displaystyle$ $=$ $\displaystyle {P_n}^{3^{-n} }$ $\displaystyle$ $=$ $\displaystyle u \left({n}\right)$

$\Box$

### Lemma 3

$\forall n \in \N_{>0}: v \left({n + 1}\right) < v \left({n}\right)$

#### Proof

 $\displaystyle v \left({n + 1}\right)$ $=$ $\displaystyle \left({P_{n+1} + 1}\right)^{3^{-\left({n+1}\right)} }$ $\displaystyle$ $<$ $\displaystyle \left({\left({\left({P_n + 1}\right)^3 - 1}\right) + 1}\right)^{3^{-n-1} }$ because $P_{n+1} < \left({P_n + 1}\right)^3 - 1$ $\displaystyle$ $=$ $\displaystyle \left({\left({P_n + 1}\right)^3}\right)^{3^{-n-1} }$ $\displaystyle$ $=$ $\displaystyle \left({P_n + 1}\right)^ {3^{-n} }$ $\displaystyle$ $=$ $\displaystyle v \left({n}\right)$

$\Box$

It follows trivially that $u \left({n}\right)$ is bounded and strictly monotone.

Therefore, there exists a number $A$ which is defined as:

$A := \lim_{n \mathop \to \infty} u \left({n}\right)$

From Lemma 2 and Lemma 3, we have:

$u \left({n}\right) < A < v \left({n}\right)$
 $\displaystyle u \left({n}\right)$ $<$ $\displaystyle A$ $\displaystyle <$ $\displaystyle \left({n}\right)$ $\displaystyle \leadsto \ \$ $\displaystyle {P_n}^{3^{-n} }$ $<$ $\displaystyle A$ $\displaystyle <$ $\displaystyle \left({P_n + 1}\right)^{3^{-n} }$ $\displaystyle \leadsto \ \$ $\displaystyle P_n$ $<$ $\displaystyle A^{3^n}$ $\displaystyle <$ $\displaystyle P_n + 1$

The result follows.

$\blacksquare$

## Source of Name

This entry was named for William H. Mills.