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Nope! Amazingly enough, no elementary proof of this fact is yet known (Edit: See KConrad's answer). The difficulty is tied up in some pretty fantastic algebraic/analytic number theory, namely the analytic class number formula. But, without getting into that, here's the short form of the story: let $L(s)$ be the $L$-function attached to the character arising from the Kronecker symbol mod $q$. Then one can compute analytically via the Euler product formula for this $L$-function that (for an explicit positive constant $C$), $$L(1)=C\left[\sum_{m=0}^{q/2}\left(\frac{m}{q}\right)-\sum_{m=q/2}^{q}\left(\frac{m}{q}\right)\right]=\frac{\pi}{\left(2-\left(\frac{2}{q}\right)\right)q^{1/2}}\sum_{m=0}^{q/2}\left(\frac{m}{q}\right),$$ the positivity of which gives the desired statement about the distribution of quadratic residues.

For a reference (from whence I pulled this out of), Davenport's "Multiplicative Number Theory" is pretty fantastic. This is all done in the first 4-5 pages.

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Nope! Amazingly enough, no elementary proof of this fact is yet known. The difficulty is tied up in some pretty fantastic algebraic/analytic number theory, namely the analytic class number formula. But, without getting into that, here's the short form of the story: let $L(s)$ be the $L$-function attached to the character arising from the Kronecker symbol mod $q$. Then one can compute analytically via the Euler product formula for this $L$-function that (for an explicit positive constant $C$), $$L(1)=\frac{\pi}{\left(2-\left(\frac{2}{q}\right)\right)q^{1/2}}\sum_{m=0}^{q/2}\left(\frac{m}{q}\right)>0L(1)=C\left[\sum_{m=0}^{q/2}\left(\frac{m}{q}\right)-\sum_{m=q/2}^{q}\left(\frac{m}{q}\right)\right]=\frac{\pi}{\left(2-\left(\frac{2}{q}\right)\right)q^{1/2}}\sum_{m=0}^{q/2}\left(\frac{m}{q}\right),$$ the positivity of which combines with a separate calculation that $L(1)=-\frac{\pi}{q^{3/2}}\sum_{m=1}^{q-1}m\left(\frac{m}{q}\right)$ is negative to give gives the desired statement about the distribution of quadratic residues.

For a reference (from whence I pulled this out of), Davenport's "Multiplicative Number Theory" is pretty fantastic. This is all done in the first 4-5 pages.

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Nope! Amazingly enough, no elementary proof of this fact is yet known. The difficulty is tied up in some pretty fantastic algebraic/analytic number theory, namely the analytic class number formula. But, without getting into that, here's the short story form of the story: let $L(s)$ be the $L$-function attached to the character arising from the Kronecker symbol mod $q$. Then one can compute analytically that $$L(1)=\frac{\pi}{\left(2-\left(\frac{2}{q}\right)\right)q^{1/2}}\sum_{m=0}^{q/2}\left(\frac{m}{q}\right)>0,$$ which combines with a separate calculation that $L(1)=-\frac{\pi}{q^{3/2}}\sum_{m=1}^{q-1}m\left(\frac{m}{q}\right)$ is negative to give the statement about the distribution of quadratic residues.

For a reference (from whence I pulled this out of), I've found Davenport's "Multiplicative Number Theory" is pretty fantastic. This is all done in the first 4-5 pages.

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