Wednesday, May 28, 2025

Proof that ∀n∈ℕ(n>2⟹∃p∈ℙ(n<p<n!))

Prove that, for every natural number n greater than 2, there is a prime number between n and n!.

Proof: For every natural number $n > 2$, there is a prime number between $n$ and $n!$.


Step 1: Understand the problem.

We are asked to show that for any integer $n > 2$, there exists at least one prime $p$ such that:

$$ n < p < n! $$


Step 2: Use a classical approach based on divisor arguments.

Consider the number:

$$ N = n! - 1 $$

Let's analyze the prime factors of $N$.


Step 3: Key observation.

  • For any integer $k$ with $2 \leq k \leq n$, $k$ divides $n!$ (by the definition of factorial).
  • But $k$ does not divide $n! - 1$ because:

$$ n! \equiv 0 \pmod{k} \implies n! - 1 \equiv -1 \not\equiv 0 \pmod{k} $$

This means none of the integers $2, 3, \dots, n$ divides $N = n! - 1$.


Step 4: Consequence.

  • Any prime divisor of $N$ must be greater than $n$.

Step 5: Conclusion.

  • $N = n! - 1$ is either prime itself (in which case it's a prime greater than $n$) or composite.
  • If it's composite, it must have a prime factor greater than $n$, since none of the primes $\leq n$ divide it.

Thus, there exists a prime $p > n$ such that $p$ divides $n! - 1$.

Since $n! - 1 < n!$, this prime $p$ satisfies:

$$ n < p < n! $$


Therefore, for every integer $n > 2$, there is a prime number between $n$ and $n!$.


This proof uses an idea similar to Euclid's proof of the infinitude of primes but applied to the interval $(n, n!)$.

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