Proof that, ∀n∈ℕ, there are n consecutive composite numbers

Prove that, for every natural number n, there are n consecutive composite numbers.

Proof: For every natural number $n$, there are $n$ consecutive composite numbers.

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Step 1: Construct the sequence.

Consider the number:

$$N=(n+1)!$$

where $(n+1)!$ is the factorial of $n+1$, meaning:

$$(n+1)!=(n+1)\cdot n\cdot (n-1)\cdots 2\cdot 1$$

Step 2: Build the sequence of $n$ consecutive numbers:

$$N+2,N+3,N+4,\dots ,N+(n+1)$$

This is a sequence of $n$ consecutive numbers (from $N+2$ to $N+(n+1)$).

Step 3: Prove each number is composite.

Consider any number in the sequence:

$$N+k\text{where}k=2,3,4,\dots ,n+1$$

Note that $(n+1)!$ is divisible by $k$ because $k\le n+1$ and the factorial contains every integer up to $n+1$ as factors.

Therefore:

$$N+k=(n+1)!+k$$

is divisible by $k$ (since $(n+1)!$ is divisible by $k$).

Check whether this number is composite:

• It is greater than $k$ (because $(n+1)!\ge (n+1)\cdot n\cdot \cdots \cdot 2\cdot 1$, which is obviously larger than $k$ for all $n\ge 1$).
• It has $k$ as a divisor other than 1 and itself.

Thus, every number in the sequence is composite.

Step 4: Conclusion.

For every natural number $n$, there exists a sequence of $n$ consecutive composite numbers — specifically the numbers:

$$(n+1)!+2,(n+1)!+3,\dots ,(n+1)!+(n+1)$$

This proves the statement. $◼$