Why does $x^2+47y^2 = z^5$ involve solvable quintics? This is related to the post on $x^2+ny^2=z^k$. In response to my answer on,
$$x^2+47y^2 = z^3\tag1$$
where $z$ is not of form $p^2+nq^2$, Will Jagy provided one for,
$$x^2+47y^2 = z^5\tag2$$
$$ (14p^5 + 405p^4q + 3780p^3q^2 + 13410p^2q^3 + 11550pq^4 - 14647q^5)^2 + 47  ( p^5 - 270p^3q^2 - 2520p^2q^3 - 8115pq^4 - 8344q^5)^2 = (3p^2 + 28pq  + 81q^2)^5\tag3 $$
As noticed by Elaqqad, the cubic polynomials I used for $(1)$ involve the discriminant $d=-47$ and I assumed it would be same with the (irreducible) quintic polynomials used by Jagy for $(3)$. Then I wondered if they were solvable in radicals as well. (I knew Ramanujan played with a solvable quintic with $d=-47$.) It turns out they are.
This Magma calculator computes the Galois group and the command is:

Z := Integers(); P < x > := PolynomialRing(Z); f := 14*x^5 + 405*x^4 + 3780*x^3 + 13410*x^2 + 11550*x - 14647; G, R := GaloisGroup(f); G;

Testing both polynomials, it shows the group has order 20 and hence is solvable.

Q: Given $x^2+dy^2=z^k$ where $z\neq p^2+dq^2$, is it true that if
$$\big(P_1(x)\big)^2+d\big(P_2(x)\big)^2=\big(P_3(x)\big)^k$$
then the equations $P_1(x) = P_2(x) = P_3(x) = 0$ are solvable in radicals?

P.S. Or is this $5$th parameterization special only because the class number $h(-47) = 5$? One way to check would be to solve $x^2+47y^2 = z^\color{red}7$ analogous to $(3)$ (Will, care to oblige?) and see if it involves solvable septics.
 A: (A partial answer.) Thanks to Jagy's two parameterizations for degs $5$ and $7$, a general identity has been found. The clue was to transform $(3)$ using $p=u-14v,\,q=3v$ to the form,
$$(14 u^5 + 235 u^4 v - 6580 u^3 v^2 - 22090 u^2 v^3 + 154630 u v^4 + 47^3 v^5)^2 + 47(u^5 - 70 u^4 v - 470 u^3 v^2 + 6580 u^2 v^3 + 11045 u v^4 - 14\cdot47^2 v^5)^2\\=3^5(u^2+47v^2)^5$$
(and similarly for the deg $7$). Inspecting the coefficients, patterns were found. Thus, 

If $a^2+db^2=c^5$, then,

$$(\color{blue}+a u^5 \color{blue}+ 5 b d u^4 v \color{red}- 10 a d u^3 v^2 \color{red}- 10 b d^2 u^2 v^3 \color{blue}+ 5 a d^2 u v^4 \color{blue}+ b d^3 v^5)^2 
+ d (\color{blue}+ b u^5 \color{red}- 5 a u^4 v \color{red}- 10 b d u^3 v^2 \color{blue}+ 10 a d u^2 v^3 \color{blue}+ 5 b d^2 u v^4 \color{red}- a d^2 v^5)^2 \\= c^5 (u^2 + d v^2)^5$$
where the above used $14^2+47\times1^2=3^5$. 
(Edited later.) In general, if,

If $a^2+db^2=c^k$, then,

$$\Big(a\,\phi_1-b\sqrt{-d}\,\phi_2\Big)^2+d \Big(b\,\phi_1-\frac{a}{\sqrt{-d}}\,\phi_2\Big)^2 =c^k(u^2+dv^2)^k$$
where,
$$\phi_1 = \frac{(u+\sqrt{-d}\,v)^k+(u-\sqrt{-d}\,v)^k}{2}$$
$$\phi_2 = \frac{(u+\sqrt{-d}\,v)^k-(u-\sqrt{-d}\,v)^k}{2}$$ 
Since an initial solution to $a^2+db^2=c^k$ is easily found for any $d$ (such as using the obvious $a,b,c = 1,0,1$), then the class number $h(-d)$ need not matter, answering part of my original question. For example, using $a^2+47b^2=c^7$ where $a,b,c = 866458,55861,51$, then one can find a $k=7$ parameterization similar to the one found by Jagy for $d=71$, even though $h(-47) = 5$.
The discriminants $D$ of the first polynomials $P_1(u,v)$ for $k=5,7$ are,
$$D_5 = 2^{12}\cdot5^5c^{20}d^{10}$$
$$D_7 = 2^{30}\cdot7^7c^{42}d^{21}$$ 
Their form and discriminants highly suggest that, equated to zero, then $P_i(u,v)=0$ is solvable in radicals for any $a,b,d$. But I do not (yet) have a rigorous proof that this is the case. 
A: There is a standard calculation procedure.  Suitable for any degree. 
We make the conversion.
$$x^2+ay^2=(p^2+as^2)(k^2+an^2)(t^2+ar^2)=z^5$$
And then solve the system of equations.
$$\left\{\begin{aligned}&z^2=k^2+an^2=t^2+ar^2\\&z=p^2+as^2\end{aligned}\right.$$
In the first equation to parameterize the substitution in the second.  And then, we decide and we obtain the parameterization of all solutions.  
Not clearly only one. Need to find other solutions?
