Induction - Prime Numbers

Question

Prove that, for every natural number $n > 2$, there is a prime number between $n$ and $n!$. [Hint: There is a prime number that divides $n! - 1$.]

Let $p$ be that prime number. If $p\leq n$ then $p$ divides $n!$. But then it must also divide $n!-(n!-1)=1$. — Pp.., Jan 17 '15 at 16:39

score 2 · Answer 1 · answered Jan 17 '15 at 16:40

2

If induction is not mandatory,

using the hint, either $n!-1$ is prime(then we are done)

Otherwise it is divisible a prime $>n$

as $n!-1\equiv-1\pmod{n!}\equiv-1\pmod r\implies(n!-1,r)=1$ for $2\le r\le n$

answered Jan 17 '15 at 16:40

lab bhattacharjee

2

You don't need to do the separate case when $n!-1$ is prime. – Thomas Andrews Jan 17 '15 at 17:00

1 Answers1