### if they give you drooled paper, writhe the other way

Let p be an odd prime.
Let L = (1^2)(3^2)…([p-1]^2) and
let R = (-1)^{(p+1}/2}.
We will show that L = R (mod p).

Note that for 1 \le k < p
("\le" denotes is-less-than-or-equal-to)
we have k = -(p-k) (mod p).
Rearranging the factors of L gives
L=1(-1)(p-1)2(-1)(p-2)…(p-2)(-1)(2).
L=(-1)^{(p-1)/1}1*2*3*…*(p-1).
Since 1*2*…*(p-1) = (p-1)!, we can
apply Wilson's Theorem:
L\equiv (-1)^{(p-1)/2}*(-1) (mod p).
But then L = R as we were to show.

One may now derive
(2^2)(4^2)…([p-1]^2) \equiv R
by squaring "Wilson's Equation"
[(p-1)!]^2 \equiv (-1)^2 (mod p).
Note that (-1)^2 = 1 = (-1)^{p-1}:
[(p-1)!]^2 \equiv(-1)^{p-1} (mod p).

Divide on the left by L;
divide on the right by R.
The result follows.
(Division "works" because p
is a prime… we can always
"cancel" factors less than p
in this case.)

(One should *not*… if instructed
to prove *two* equivalences…
simply remark that the second
"is similar to" the first:
wanted at least *some* details,
they presumably would not have
asked that part of the question
at all.)