Existence of Prime-Free Sequence of Natural Numbers

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Theorem

Let $n$ be a natural number.

Then there exists a sequence of consecutive natural numbers of length $n$ which are all composite.


Proof

Consider the number

$N := \left({n + 1}\right)!$

where $!$ denotes the factorial.

Let:

$ i \in I $ where $I = \{ i \in \mathbb{N} | i \geq 2 \wedge i \leq n + 1 \} $ and $|I| = n $

Then:

$ N + i = (1 \cdot 2 \cdot \dotsc \cdot i \cdot \dotsc \cdot n \cdot n + 1) + i = i[(1 \cdot 2 \cdot \dotsc \cdot n \cdot n + 1) + 1]$

Then:

$N + 2, N + 3, \ldots, N + n, N + n + 1$

are all composite.

$\blacksquare$


Sources