Proof
(related to Proposition: Number of Quadratic Residues in Reduced Residue Systems Modulo a Prime)
 By hypothesis, $p > 2$ is a prime number and we consider a reduced residue system modulo $p$. Let's call it $R.$
 $R$ can be respresented by the $p1$ numbers $n=1,\ldots,p1.$
 Let $n\in R,$ and assume that $n$ is a quadratic residue modulo $p.$
 Since $p\not\mid n,$ there is at least one solution of the congruence $x^2(p)=n(p)$ in $R,$ and, by counting the roots of a diophantine polynomial modulo $p$, there are at most two solutions in $R.$
 Because $(px)^2(p)\equiv (x)^2(p)\equiv x^2(p)$, if $x$ solves the polynomial, so does $px.$
 Therefore, $n$ must be congruent to exactly one of the $\frac{p1}{2}$ numbers in $R$ from the interval $1\le x \le \frac{p1}{2}.$
 Hence, there are exactly $\frac{p1}{2}$ quadratic residues modulo $p,$ since the $\frac{p1}{2}$ numbers $1^2,2^2,\ldots,\left(\frac{p1}{2}\right)^2$ are all incongruent.
 Since $R$ has $p1$ elements, $R$ contains $p1\frac{p1}{2}=\frac{p1}{2}$ quadratic nonresidues modulo $p.$
∎
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References
Bibliography
 Landau, Edmund: "Vorlesungen über Zahlentheorie, Aus der Elementaren Zahlentheorie", S. Hirzel, Leipzig, 1927