If $p - a \equiv -a \pmod{p}$ then what is $\frac{p-1}{2} \equiv ? \pmod{p}$? If $p - a \equiv -a \pmod{p}$ then what is $\frac{p-1}{2} \equiv ? \pmod{p}$? Where $p$ is an odd prime.
I read in the book, they claimed:
$$p - 1 \equiv -1 \pmod{p}$$
$$p - 2 \equiv -2 \pmod{p}$$
$$p - 3 \equiv -3 \pmod{p}$$
$$ ... $$
$$\frac{p - 1}{2} \equiv -(\frac{p - 1}{2}) \pmod{p}$$
However, I realized that the last statement was wrong because if it's true then:
$$\frac{p - 1}{2} + \frac{p - 1}{2}\equiv 0 \pmod{p}$$
$$p - 1 \equiv 0 \pmod{p}$$

Any idea?   
Thanks, 

In the 5th edition, they have got the signs right, at least thats what the scanned version of the ebook says.

EDIT
We have $p-k \equiv -k \pmod{p}$ and hence $$ \prod_{k=1}^{\frac{p-1}{2}} (p-k) \equiv \prod_{k=1}^{\frac{p-1}{2}} (-k) \pmod{p}$$
Note that $$(p-1)! = \left(\prod_{k=1}^{\frac{p-1}{2}} k \right) \left(\prod_{k=1}^{\frac{p-1}{2}} (p-k) \right)$$
Hence, $$(p-1)! \equiv \left(\prod_{k=1}^{\frac{p-1}{2}} k \right) \left(\prod_{k=1}^{\frac{p-1}{2}} (p-k) \right) \pmod{p} \equiv \left(\prod_{k=1}^{\frac{p-1}{2}} k \right) \left(\prod_{k=1}^{\frac{p-1}{2}} (-k) \right) \pmod{p}$$
Hence, $$(p-1)! \equiv (-1)^{\frac{p-1}{2}} \left(\prod_{k=1}^{\frac{p-1}{2}} k \right)^2 \pmod{p}$$
$$\left(\prod_{k=1}^{\frac{p-1}{2}} k \right) = \left(\frac{p-1}{2}\right)!$$
From Wilson's theorem, $$(p-1)! \equiv -1 \pmod{p}$$
Hence, $$-1 \equiv (-1)^{\frac{p-1}{2}} \left[ \left(\frac{p-1}{2}\right)! \right]^2 \pmod{p}$$
Hence, $$\left[ \left(\frac{p-1}{2}\right)! \right]^2 \equiv (-1)^{\frac{p+1}{2}}  \pmod{p}$$
Hence, if $p = 4k+3$, then $$(-1)^{\frac{p+1}{2}} = 1$$
Hence, if $p = 4k+3$, $$\left[ \left(\frac{p-1}{2}\right)! \right]^2 \equiv 1  \pmod{p}$$
