### 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:

if our textbook authors hadn't

wanted at least *some* details,

they presumably would not have

asked that part of the question

at all.)

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