Let $p>3$ be a prime, and let $(\frac{\cdot}p)$ be the Legendre symbol. Then $$14\left(\frac 3p\right)-\left(\frac p3\right)-12=\begin{cases}1&\text{if}\ p\equiv1\pmod{12}, \\-25&\text{if}\ p\equiv5\pmod{12},\\-27&\text{if}\ p\equiv 7\pmod{12}, \\3&\text{if}\ p\equiv 11\pmod{12};\end{cases}$$ this is a quadratic residue modulo $p$. It is interesting to find an explicit solution to the congruence $$x^2\equiv 14\left(\frac 3p\right)-\left(\frac p3\right)-12\pmod p.\tag{1}$$
For $n=0,1,2,\ldots$ define $$w_n:=\sum_{k=0}^n\binom nk\binom{n+k}k\binom{2k}k\binom{2(n-k)}{n-k}(-8)^{n-k}.$$ I have the following conjectures related to the congruence equation $(1)$.
Conjecture 1. (i) For every $n=1,2,3,\ldots$, the number $$\frac1n\sum_{k=0}^{n-1}(-1)^k(4k+1)48^{n-1-k}w_k$$ is a positive integer.
(ii) For any prime $p>3$ we have $$\bigg(\frac1p\sum_{k=0}^{p-1}\frac{4k+1}{(-48)^k}w_k\bigg)^2\equiv 14\left(\frac 3p\right)-\left(\frac p3\right)-12\pmod p.\tag{2}$$
Conjecture 2. We have the identity $$\sum_{k=0}^\infty\frac{4k+1}{(-48)^k}w_k=\frac{\sqrt{72+42\sqrt3}}{\pi}.\tag{3}$$
Conjecture 3. For any prime $p>3$, we have $$\sum_{k=0}^{p-1}\frac{w_k}{(-48)^k}\equiv\begin{cases}4x^2-2p\pmod{p^2}&\text{if}\ p=x^2+4y^2\ (x,y\in\mathbb Z),\\0\pmod{p^2}&\text{if}\ p\equiv3\pmod 4.\end{cases}\tag{4}$$
QUESTION. How to solve the above three conjectures? In particular, how to prove the congruence $(2)$?
Your comments are welcome!
