A conjecture for primes $p\equiv\pm1\pmod5$ For any prime $p\equiv\pm1\pmod5$, we can write $p$ uniquely in the form $x_p^2+3x_py_p+y_p^2$ with $x_p,y_p\in\mathbb Z$ and $x_p>y_p>0$. 
I have the following conjecture.
Conjecture. We have
$$\lim_{N\to+\infty}\frac{\sum_{p\le N\atop p\equiv\pm1\pmod5}(3x_p^2+2x_py_p)}
{\sum_{p\le N\atop p\equiv\pm1\pmod5}(3y_p^2+2x_py_p)}=4.\tag{$*$}$$
I have checked this conjecture via computation. For example,
\begin{align}\sum_{p\le 2\times10^9\atop p\equiv\pm1\pmod5}(3x_p^2+2x_py_p)=&88916125007243531,
\\\sum_{p\le 2\times10^9\atop p\equiv\pm1\pmod5}(3y_p^2+2x_py_p)=&22228898519387861,
\\\frac{88916125007243531}{22228898519387861}\approx&4.000023885.\end{align}
Similarly, any prime $p\equiv1\pmod3$ can be written uniquely as $u_p^2+u_pv_p+v_p^2$ with $u_p,v_p\in\mathbb Z$ and $u_p>v_p>0$, and I observe  that
$$\lim_{N\to+\infty}\frac{\sum_{p\le N\atop p\equiv1\pmod3}(u_p^2+2u_pv_p)}
{\sum_{p\le N\atop p\equiv1\pmod3}(v_p^2+2u_pv_p)}=2.$$
This is essentially provable in view of Hecke's equidistribution theorem.
QUESTION. How to prove the equality $(*)$?
 A: Let us consider $\mathbb{Q}(\sqrt{5})$ as a subfield of $\mathbb{R}$. Let us also consider the positive fundamental unit $\epsilon:=(1+\sqrt{5})/2$, whose square generates the group of totally positive units. The conditions on $x_p$, $y_p$ can be rewritten as
$$p=x_p^2+3x_py_p+y_p^2\qquad\Longleftrightarrow\qquad p=(x_p+\epsilon^2 y_p)(x_p+\epsilon^{-2} y_p),$$
$$x_p+\epsilon^{-2} y_p<x_p+\epsilon^2 y_p<\epsilon^2(x_p+\epsilon^{-2} y_p)\qquad\Longleftrightarrow\qquad p^{1/2}<x_p+\epsilon^2 y_p<p^{1/2}\epsilon.$$
By Hecke's theorem, $\log(x_p+\epsilon^2 y_p)$ is equidistributed in $[\log(p^{1/2}),\log(p^{1/2}\epsilon)]$, cf. Example 3 in Ch. XV, §5 of Lang: Algebraic number theory. From here we get by a similar argument as in my response here that
$$\lim_{N\to+\infty}\frac{\sum_{p\le N\atop p\equiv\pm1\pmod5}(3x_p^2+2x_py_p)}
{\sum_{p\le N\atop p\equiv\pm1\pmod5}(3y_p^2+2x_py_p)}
=\frac{\int_0^1 (3x(t)^2+2x(t)y(t))\,dt}{\int_0^1 (3y(t)^2+2x(t)y(t))\,dt},$$
where the real functions $x,y:[0,1]\to\mathbb{R}$ are defined by the equations
$$\epsilon^{-1}x(t)+\epsilon y(t)=\epsilon^{-t}\qquad\text{and}\qquad
\epsilon x(t)+\epsilon^{-1}y(t)=\epsilon^t.$$
Solving for $x(t)$ and $y(t)$, the ratio of the two integrals turns out to be exactly $4$. This proves $(\ast)$.
