Why does the author mean by "The $(p+1)/2$ numbers to $x^2$, for $0\leq x\leq (p-1)/2$, ..."

Question

I couldn't get the proof for the given theorem. How is the $(p+1)/2$ numbering to $x^2$, how is the following relation valid??

Answer 1

First: Counting the (square) numbers including zero, we find that there are $1+\frac{p-1}{2}=\frac{p+1}{2}$ of them.

Second: $x_1^2-x_2^2=(x_1-x_2)(x_1+x_2)$, so if $p\mid x_1^2-x_2^2$ then $p\mid x_1-x_2$ or $p\mid x_1+x_2$.

Since $0 \leq x_1, x_2 \leq \frac{p-1}{2}$, from $p\mid x_1-x_2$ it follows that $x_1=x_2$.

Also, $x_1+x_2 \leq p-1$, so if $p\mid x_1+x_2$ then $x_1=x_2=0$.

Therefore, for $x_1, x_2$ in the give range, from $x_1^2 \equiv x_2^2 \pmod{p}$ it follows that $x_1 = x_2$.