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:

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

Note that $(n+1)!$ is divisible by $k$ because $k \leq n+1$ 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 $(n+1)! \geq (n+1) \cdot n \cdot \dots \cdot 2 \cdot 1$, which is obviously larger than $k$ for all $n \geq 1$).
• 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. $\blacksquare$