If $p\mid4n!-1$ then $p>n$

Question

Let $n$ be a positive integer and define $a_{n}=4(1\cdot 2\cdot3 \cdots n)-1.$ Let $p$ be a prime number such that $p\mid a_{n}.$ Prove that $$p>n.$$

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HInt: $p≤n\implies p;|;n!$ – lulu Apr 07 '16 at 00:19

score 1 · Answer 1 · answered Apr 07 '16 at 01:56

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If $p\leq n$ then $p | n!$ since $n!=1 \cdot 2 \cdot... \cdot p \cdot... \cdot n$ Then you will have $p|4n!-1$ and $p|n!$. which means $p|1$. That cannot be true.

answered Apr 07 '16 at 01:56

christina_g

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I think you may have taken a wrong turn at Albuquerque. – Bob Happ Apr 07 '16 at 21:39
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@BobHapp I'm sorry but I dont understand what that means... – christina_g Apr 08 '16 at 21:39
As I reread your answer, I think now it's a matter of clarity, that your answer could be clearer. – Bob Happ Apr 11 '16 at 20:41

score 1 · Answer 2 · answered Apr 07 '16 at 21:38

Given a prime number $p \geq 2$, if $p \leq n$, then $p \mid n!$ as well as $4n!$ but not $4n! - 1$. Therefore we need $p > n$ such that $p \nmid n!$ nor $4n!$ in order to allow the possibility that $p \mid 4n! - 1$.

To give a concrete example: consider $n = 4$ and $p = 5$ or $19$. Then we have $4n! - 1 = 95 = 5 \times 19$. The primes less than $4$ are $2$ and $3$, but obviously $96$ is quite divisible by both of those.

If $p\mid4n!-1$ then $p>n$

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