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:

$N := \paren {n + 1}!$

where $!$ denotes the factorial.

Let $i \in I$ where:

$I = \set {i \in \N: 2 \le i \le n + 1}$

We have that $\size I = n$.

Then for all $i \in I$:

\(\ds i\) \(\divides\) \(\ds N\) where $\divides$ denotes divisibility
\(\ds \leadsto \ \ \) \(\ds i\) \(\divides\) \(\ds N + i\)

and so $N + i$ is composite.

That is:

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

are all composite.

$\blacksquare$


Sources