Prove that if $4t + 1$ is prime, it divides $(4t)! - [(2t)!]^2$ How do I show the following:
$4t+1$ divides $(4t)! - [(2t)!]^2$
$4t+1$ is a prime number in this case, though I think the above should hold for all positive integer values of $t$.
 A: You might find it worth reading about Wilson's theorem first for the case where $4t+1$ is prime, but you're right that we can do this without assuming that in this case. We have:
\begin{align*}
  [(2t)!]^2 &= (1\times 2 \times \ldots \times 2t)\times (1\times 2 \times \ldots \times 2t)\\
&= (1\times 2 \times \ldots \times 2t)\times(-1\times -2 \times \ldots \times -2t)\times (-1)^{2t}\\
&\equiv (1\times 2 \times \ldots \times 2t)\times(((4t+1)-1)\times ((4t+1)-2) \times \ldots \times ((4t+1)-2t))\mod (4t+1)\\
&=(1\times 2 \times \ldots \times 2t) \times (4t \times(4t-1)\times \ldots \times (2t+1))\\
&= 1\times 2 \times\ldots \times 2t \times (2t+1)\times \ldots \times 4t\\
&= (4t)!
\end{align*}
So their difference is congruent to $0 \mod (4t+1)$, and is divisible by $(4t+1)$.
A: We can pair off factors of $(4t)!$ and $[(2t)!]^2$ as follows
for $1,...2t$, we can match items exactly, so they have the same residue mod $4t+1$
for $2t+1,..., 4t$ from $(4t)!$ we can match with $2t,...,1$ from $[(2t)!]^2$ mod $4t+1$ with a change of sign,
e.g. $2t+1= -2t$ mod $4t+1$,  ....., $4t = -1$ mod $4t+1$
As there are an even number of the items that are paired off with a sign change, we conclude that $(4t)!= [(2t)!]^2$ mod $4t+1$
