TITLE: Show that for $n!/2 - p$, $p$ is the least prime factor if $n > p$
QUESTION [5 upvotes]: Apparently this isn't clear, so here are some examples:
$$3!/2 - 3$$
$$4!/2 - 3$$
$$5!/2 - 3$$
and 
$$5!/2 - 5$$
$$6!/2 - 5$$
...
How would I go about proving this?
I found it from a matrix I generated from looking at error functions.

REPLY [5 votes]: Not that this question requires any more answers, I figured I would practice writing cancellations in TeX.
Okay, given $n > 5$, $n! = 2 \times 3 \times 4 \times \ldots \times n$. So far nothing you didn't already know, nothing you haven't already been told. Then
$$\require{cancel} \frac{n!}{2} = \frac{\cancel{2} \times 3 \times 4 \times \ldots \times n}{\cancel{2}} = 3 \times 4 \times \ldots \times n.$$
If $p$ is some prime greater than $3$ but less than $n$, that means it appears somewhere in those continuation dots. So then $$\frac{n!}{2} = p \times \frac{3 \times 4 \times \ldots \times \cancel{p} \times \ldots \times n}{\cancel{p}}$$ and therefore $$\frac{n!}{2} - p = p \times \left(\frac{3 \times 4 \times \ldots \times \cancel{p} \times \ldots \times n}{\cancel{p}} - 1 \right).$$ Clearly $$\gcd \left( \frac{3 \times 4 \times \ldots \times \cancel{p} \times \ldots \times n}{\cancel{p}}, \frac{3 \times 4 \times \ldots \times \cancel{p} \times \ldots \times n}{\cancel{p}} - 1 \right) = 1$$ and consequently $$\gcd \left( \frac{n!}{2}, \frac{n!}{2} - p \right) = p$$ and $$\textrm{lpf} \left( \frac{n!}{2} - p \right) = p.$$