Existence of Prime between Prime and Factorial

Theorem

Let $p$ be a prime number.

Then there exists a prime number $q$ such that:

$p < q \le p! + 1$

where $p!$ denotes the factorial of $p$.

Proof

Let $N = p! + 1$.

If $N$ is prime, then the proof is complete.

Otherwise, from Positive Integer Greater than 1 has Prime Divisor, $N$ has a prime divisor $q$.

From Absolute Value of Integer is not less than Divisors:

$q < N$

Aiming for a contradiction, suppose $q \le p$.

Then by the definition of factorial:

$q \divides p!$

But $q$ was chosen so that :$q \divides N$.

Let $p! = q m_1, N = q m_2$.

Then:

$1 = N - p! = q \paren {m_2 - m_1}$

and so:

$q \divides 1$

From Absolute Value of Integer is not less than Divisors:

$q \le 1$

and so $q$ cannot be a prime number.

It follows by Proof by Contradiction that $q > p$.

$\blacksquare$

Also see

• Euclid's Theorem, of which this is a variant.

Sources

• 1982: Martin Davis: Computability and Unsolvability (2nd ed.) ... (previous) ... (next): Appendix $1$: Some Results from the Elementary Theory of Numbers: Theorem $4$