$\displaystyle

a,n\in\mathbb{Z},$

Show that:

$\displaystyle n+1=a^2 \Rightarrow n+1|2(n-1)!$

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- Dec 23rd 2008, 12:51 AMSeaModulo
$\displaystyle

a,n\in\mathbb{Z},$

Show that:

$\displaystyle n+1=a^2 \Rightarrow n+1|2(n-1)!$ - Dec 23rd 2008, 04:04 AMPaulRS
For the hypothesis it'd be enough to say that $\displaystyle n$ is a positive integer such that $\displaystyle n+1$ is not a prime number.

Let $\displaystyle p^s$ be the maximum power of a prime $\displaystyle p$ dividing $\displaystyle n+1$.

- If $\displaystyle n+1$ is divisible by at least 2 primes, then $\displaystyle p^s\leq{\tfrac{n}{2}}\leq{n-1}$ and so $\displaystyle p^s$ appears as a factor of $\displaystyle (n-1)!=1\cdot{2}...\cdot{p^s}\cdot{...\cdot{(n-1)}}$, thus $\displaystyle p^s|(n-1)!$
- Otherwise $\displaystyle n+1=p^s$. If $\displaystyle p=2$ then $\displaystyle 2^{s-1}$ appears as a factor of $\displaystyle (n-1)!$ - this can be seen as in the previous part- so $\displaystyle 2^{s-1}|(n-1)!$ and so $\displaystyle 2^{s}|[2\cdot{(n-1)!}]$. Now if $\displaystyle p>2$ we have that $\displaystyle p^{s-1}$ is a factor of $\displaystyle (n-1)!$, but, since n is not prime, it must be that $\displaystyle s>1$ and so $\displaystyle 2p$ appears also as a factor in the product, because $\displaystyle 2p<p^2-1\leq{p^s-1}=n-1$ (*). Clearly $\displaystyle 2p\neq p^{s-1}$ so we have that $\displaystyle p^s|(n-1)!$

(*) $\displaystyle 2p<p^2-1$ iff $\displaystyle 2<p^2-2p+1=(p-1)^2$ and this holds since we were considering $\displaystyle p>2$

From this analysis it follows that $\displaystyle p^s|[2\cdot{(n-1)!}]$. (this holds for all the maximum powers of primes dividing n+1)

And therefore $\displaystyle (n+1)|[2\cdot{(n-1)!}]$

**Remark**: This doesn't hold when $\displaystyle n+1$ is a prime greater than 2. ( it holds for $\displaystyle n+1=2$)