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 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. 
