1. ## Modulo

$

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

Show that:

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

2. For the hypothesis it'd be enough to say that $n$ is a positive integer such that $n+1$ is not a prime number.

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

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

(*) $2p iff $2 and this holds since we were considering $p>2$

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

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

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