1. ## Greatest common divisor

Hi.
Can anybody prove that?

Let $\displaystyle n>4$

$\displaystyle GCD(n, (n-1)!)=1 \Longleftrightarrow{} n$ its prime.

and

if $\displaystyle n$ its not prime $\displaystyle \Longrightarrow{} GCD(n, (n-1)!)=n$

Thanks

2. Originally Posted by asub1
$\displaystyle GCD(n, (n-1)!)=1 \Longleftrightarrow{} n$ its prime.
This should be clear. If $\displaystyle n$ is prime then $\displaystyle 1,2,3...,n-1$ show no non-trivial factors with $\displaystyle n$. Thus, $\displaystyle (n,(n-1)!)=1$.

if $\displaystyle n$ its not prime $\displaystyle \Longrightarrow{} GCD(n, (n-1)!)=n$
If $\displaystyle n$ is not prime then $\displaystyle n=ab$ where $\displaystyle 1<a,b<n$. Therefore, we are able to find $\displaystyle a,b$ among $\displaystyle 1,2,...,n-1$.
Thus, $\displaystyle n|(n-1)!$ and so $\displaystyle (n,(n-1)!)=n$.

3. Hi!

If n is not prime then $\displaystyle n=ab$ where $\displaystyle 1< a,b <n$. Therefore, we are able to find $\displaystyle a, b$ among $\displaystyle 1, 2, ..., n-1$.
Thus, $\displaystyle n|(n-1)!$ and so $\displaystyle (n, (n-1)!)=n$.
What about $\displaystyle a=b$?
$\displaystyle n=aa$ where $\displaystyle 1<a, a<n$. we are able to find $\displaystyle a$ among $\displaystyle 1, 2, ..., n-1$, but... and the other $\displaystyle a$?

Can you prove?

Thanks