I already posted this question on [MSE][1].

Using theorem $IV$ from [this article][2], it is possible to prove that when $p$ is a prime $p ≡ 3\bmod4$, $x ≢ y\bmod{p}$ and $\gcd(x,y) = 1$, then the equation $ \dfrac{x^p - y^p}{x - y} = a^2+pb^2 $ ever admits solutions for some integers $\{a, b\}$ (see proof below).

I want to know how $\{a, b\}$ can be expressed as a function of $\{x, y\}$, the case $p=3$ is easy

$ a = \dfrac{x-y}{2}$ , $ b = \dfrac{x+y}{2}$
(of course not all the solutions have this form). 
With the great help of Will Jagy on MSE, the closed form for the following cases are

$p = 7$

$ a = \frac{ (x-y)  (2x^2 + 3xy + 2y^2 )}{  2}$ , $ b = \frac{xy(x+y)}{2}$

$p = 11$

$ a = \frac{( x-y) (2x^4 + 3x^3  y  + x^2  y^2 + 3  x  y^3 + 2  y^4)}{  2}$ , $ b = \frac{ xy(x+y)(x^2 - xy+y^2)}{  2}$

$p = 19$

$ a = \frac{ (x-y)  (2x^8 + 3x^7y - x^6y^2 + 2x^5y^3 + 7x^4y^4 + 2x^3y^5 - x^2y^6 + 3xy^7 + 2y^8 )}{ 2}$, $ b = \frac{  x  y  (x +y) ( x^2 -xy +y^2)(x^4 - x^2 y^2 + y^4   )}{  2}
$

**What can be said in general?** 

I guess it is possible to restate the problem in terms of cyclotomic binary forms: if we define $ f(k) = y^{\phi(k)} \Phi_k(x/y) $ where $ \phi(k)$ is Euler's phi function and $ \Phi_k(x/y) $ is the $k$-th cyclotomic polynomial, so $ f(p) = \dfrac{x^p - y^p}{x - y} $ and

$ f(7) = a^2 + 7\left (\dfrac{xyf(2)}{2} \right )^2 $

$ f(11) = a^2 + 11\left (\dfrac{xyf(2)f(6)}{2} \right )^2 $

$ f(19) = a^2 + 19\left (\dfrac{xyf(2)f(6)f(12)}{2} \right )^2 $




*Proof:*

When $(x - y) ≢ {0}\bmod{p}$

$ \dfrac{x^p - y^p}{x - y} = P(p) $

where $P(p)$ is the [arithmetic primitive factor][3] of $ (x^p - y^p) $ defined in that article (p. $175$). With other considerations we can say that each one of those primitive prime factors is of the form $(2k_i p+1)$ where $k_i$ $∈ N$.


Now $q\mid{a^2+pb^2} $, with $q\nmid p$, iff $ \left(\frac{-p}{q}\right) = 1$ and equivalently $q ≡ \{q_1, q_2, q_3,..\}\bmod{4p}$
(see corollary $1.19$ in David Cox's book *Primes of the Form $x^2+ny^2$*), so being $q=2k_i p+1$, we have just $q_1=1$ and $q_2 = 2p+1$. Using that $p ≡ 3\bmod4$, we have $ \left(\frac{-p}{2p+1}\right) = 1$ and so $ \dfrac{x^p - y^p}{x - y} = a^2+pb^2 $ ever admits solutions for some integers $\{a, b\}$.


  [1]: https://math.stackexchange.com/questions/4936038/about-the-solutions-of-dfracxp-ypx-y-a2pb2
  [2]: http://www.jstor.org/stable/pdf/2007263.pdf
  [3]: https://mathworld.wolfram.com/PrimitivePrimeFactor.html