<s>[INST] A conjecture about quadratic residues given $p \equiv 5 \pmod 8$ Original Problem
$p$ is a prime that is congruent to $5$ modulo $8$ and $a$ is a quadratic residue modulo $p$.
Prove that excactly one of $x_1=a^{\frac{p+3}{8}},x_2=(2a)(4a)^{\frac{p-5}{8}}$ is the solution to the congruence $x^2 \equiv a \pmod p$.
What I have already proved
$x_1^2 \equiv a \pmod p$ if $a^{\frac{p-1}{4}} \equiv 1 \pmod p$.
$x_2^2 \equiv a \pmod p$ if $a^{\frac{p-1}{4}} \equiv -1 \pmod p$.
Corollary
Assume that $a \equiv r \pmod p,\quad (1\leq r\leq p-1)$. Then $a^{\frac{p-1}{4}} \equiv r^{\frac{p-1}{4}} \pmod p$, and hence
(a). $x_1^2 \equiv a \pmod p$ if $r^{\frac{p-1}{4}} \equiv 1 \pmod p$.
(b). $x_2^2 \equiv a \pmod p$ if $r^{\frac{p-1}{4}} \equiv -1 \pmod p$.
Examining the cases where $p=5$ and $p=13$, I came up with a conjecture.
Conjecture
(I). $r^{\frac{p-1}{4}} \equiv 1 \pmod p$ if $r$ is odd.
(II). $r^{\frac{p-1}{4}} \equiv -1 \pmod p$ if $r$ is even.
Remark
Notice that $b^2 \equiv a \equiv r \pmod p$ for some nonzero integer $b$.
By Euler's Criterion, we have that $r^{\frac{p-1}{4}} \equiv (b^2)^{\frac{p-1}{4}} \equiv b^{\frac{p-1}{2}} \equiv \genfrac(){}{0}{b}{p} \pmod p$.
Equivalent form of the conjecture
(III). $b$ is a quadratic residue modulo $p$ if $r$ is odd.
(IV). $b$ is a quadratic nonresidue modulo $p$ if $r$ is even.
Can you prove or disprove (III) and (IV) using only the law of quadratic reciprocity?
 [/INST] Let $p=8k+5$.  By Euler's Criterion, $a^{4k+2}\equiv 1\pmod{p}$ and therefore $a^{2k+1}\equiv \pm 1\pmod{p}$. 
Suppose that $a^{2k+1}\equiv 1\pmod{p}$. Then $a^{2k+2}\equiv a\pmod{p}$, and $a$ is congruent modulo $p$ to the square of $a^{k+1}$.
Now suppose that $a^{2k+1}\equiv -1\pmod{p}$. Since $p$ is of the form $8k+5$, it follows that $2$ is a quadratic non-residue of $p$, and therefore $2^{4k+2}\equiv -1\pmod{p}$. It follows that $2^{4k+2}a^{2k+1}\equiv 1\pmod{p}$, and therefore
$$2^{4k+2}a^{2k+2}\equiv a\pmod{p}.$$
Thus the square of $2^{2k+1}a^{k+1}$, that is, of $(2a)(4a)^k$, is congruent to $a$ modulo $p$. 
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