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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.

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    $\begingroup$ And solvable septics would be... curable equations? (sorry, couldn't resist) Anyway, you may wish to ping @WillJagy. $\endgroup$ – A.P. Aug 5 '15 at 9:09
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    $\begingroup$ @A.P.: I remember a discussion in the old sci.math where one researcher commented his "sextic" files were often accessed by questionable URLs, if you know what I mean. :) $\endgroup$ – Tito Piezas III Aug 5 '15 at 12:05
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I am not finding a short online summary of the composition of two binary quadratic forms, as done by Dirichlet. Also, the edition of Cox that I have has a typo, corrected in the second edition (2013), here it is.

Given $\gcd(a,a',B) = 1,$ define $$ X = xz-Cyw, $$ $$ Y = axw + a'yz + B yw, $$ then $$ (a x^2 + B xy + a'C y^2) (a' z^2 + B zw + aC w^2) = aa'X^2 + B XY + C Y^2 $$ which you ought to check!

Here are the binary forms (primitive) of discriminant $-284$

Discr  -284 = 2^2 * 71  class  number  7

 all  
     284:  < 1, 0, 71>
     284:  < 3, -2, 24>
     284:  < 3, 2, 24>
     284:  < 5, -4, 15>
     284:  < 5, 4, 15>
     284:  < 8, -2, 9>
     284:  < 8, 2, 9>

The first few primes integrally represented by $3x^2 + 2xy+24y^2$ are $$ 3, 29, 89, 103, 109, 151, 157, 191, $$ and below, we show how to represent each $p^7$ once we have $x,y.$

compared with $-71$ primitive, where this time a form represents the prime $2$

Discr  -71 = 71  class  number  7

 all  
      71:  < 1, 1, 18>
      71:  < 2, -1, 9>
      71:  < 2, 1, 9>
      71:  < 3, -1, 6>
      71:  < 3, 1, 6>
      71:  < 4, -3, 5>
      71:  < 4, 3, 5>

ummmm, $h(-71) = h(-284) = 7.$ Since $4 \cdot 3^7 - 284 = 92^2,$ the principal form is $\langle 1, 92, 2187\rangle.$ The class group is cyclic, everything is a power of $\langle 3, 92, 729 \rangle$ under Dirichlet's version of Gauss composition. All I am doing is repeatedly multiplying by $3 x^2 + 92 xy + 729 y^2,$ the rules for composition eventually give the quadratic form $\langle 2187, 92, 1 \rangle$ with variables which are homogeneous degree seven in the original $x,y.$ Oh, any form that represents $1$ is $SL_2 \mathbb Z$ equivalent to the principal form. At the very end, I show how to write $t^2 + 71 z^2 = (3 x^2 + 92 xy + 729 y^2)^7. $ I am showing the whole gp-pari session, there is nothing difficult once we get that fortunate expression for the coefficients of a generator of the group.

a=3; a1=3; b=92; c=243; z = x; w = y;

zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;

? a=3; a1=3; b=92; c=243; z = x; w = y;
? zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? z
%3 = x^2 - 243*y^2
? w
%4 = 6*y*x + 92*y^2
? 

a1 = 9; c = 81; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? a1 = 9; c = 81; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? z
%6 = x^3 - 729*y^2*x - 7452*y^3
? w
%7 = 27*y*x^2 + 828*y^2*x + 6277*y^3
? 

a1 = 27; c = 27; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;

? a1 = 27; c = 27; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? z
%9 = x^4 - 1458*y^2*x^2 - 29808*y^3*x - 169479*y^4
? w
%10 = 108*y*x^3 + 4968*y^2*x^2 + 75324*y^3*x + 376280*y^4
? 
? 
a1 = 81; c = 9; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;



? a1 = 81; c = 9; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? z
%12 = x^5 - 2430*y^2*x^3 - 74520*y^3*x^2 - 847395*y^4*x - 3386520*y^5
? w
%13 = 405*y*x^4 + 24840*y^2*x^3 + 564930*y^3*x^2 + 5644200*y^4*x + 20889961*y^5
? 



 a1 = 243; c = 3; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;

?  a1 = 243; c = 3; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? z
%15 = x^6 - 3645*y^2*x^4 - 149040*y^3*x^3 - 2542185*y^4*x^2 - 20319120*y^5*x - 62669883*y^6
? w
%16 = 1458*y*x^5 + 111780*y^2*x^4 + 3389580*y^3*x^3 + 50797800*y^4*x^2 + 376019298*y^5*x + 1098952052*y^6
? 
? 

 a1 = 729; c = 1; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;

? 
?  a1 = 729; c = 1; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? 
? z
%18 = x^7 - 5103*y^2*x^5 - 260820*y^3*x^4 - 5931765*y^4*x^3 - 71116920*y^5*x^2 - 438689181*y^6*x - 1098952052*y^7
? w
%19 = 5103*y*x^6 + 469476*y^2*x^5 + 17795295*y^3*x^4 + 355584600*y^4*x^3 + 3948202629*y^5*x^2 + 23077993092*y^6*x + 55417244077*y^7
? 

2187 * z^2 + 92 * z * w + w^2 

( 3 * x^2 + 92 * x * y + 729 * y^2)^7 

? 
? 2187 * z^2 + 92 * z * w + w^2
%20 = 2187*x^14 + 469476*y*x^13 + 46911879*y^2*x^12 + 2892076488*y^3*x^11 + 122889105423*y^4*x^10 + 3807263630268*y^5*x^9 + 88688782583499*y^6*x^8 + 1578039270279536*y^7*x^7 + 21551374167790257*y^8*x^6 + 224815110103695132*y^9*x^5 + 1763324345027822661*y^10*x^4 + 10084047184857263688*y^11*x^3 + 39747900724268273397*y^12*x^2 + 96660945131267433924*y^13*x + 109418989131512359209*y^14
? 
? 
? ( 3 * x^2 + 92 * x * y + 729 * y^2)^7 
%21 = 2187*x^14 + 469476*y*x^13 + 46911879*y^2*x^12 + 2892076488*y^3*x^11 + 122889105423*y^4*x^10 + 3807263630268*y^5*x^9 + 88688782583499*y^6*x^8 + 1578039270279536*y^7*x^7 + 21551374167790257*y^8*x^6 + 224815110103695132*y^9*x^5 + 1763324345027822661*y^10*x^4 + 10084047184857263688*y^11*x^3 + 39747900724268273397*y^12*x^2 + 96660945131267433924*y^13*x + 109418989131512359209*y^14
? 
? 2187 * z^2 + 92 * z * w + w^2 - ( 3 * x^2 + 92 * x * y + 729 * y^2)^7 
%22 = 0
? 

t = w + 46 * z

t^2 + 71 * z^2 

? 
? t = w + 46 * z
%23 = 46*x^7 + 5103*y*x^6 + 234738*y^2*x^5 + 5797575*y^3*x^4 + 82723410*y^4*x^3 + 676824309*y^5*x^2 + 2898290766*y^6*x + 4865449685*y^7
? 
? 
? 
? t^2 + 71 * z^2 
%24 = 2187*x^14 + 469476*y*x^13 + 46911879*y^2*x^12 + 2892076488*y^3*x^11 + 122889105423*y^4*x^10 + 3807263630268*y^5*x^9 + 88688782583499*y^6*x^8 + 1578039270279536*y^7*x^7 + 21551374167790257*y^8*x^6 + 224815110103695132*y^9*x^5 + 1763324345027822661*y^10*x^4 + 10084047184857263688*y^11*x^3 + 39747900724268273397*y^12*x^2 + 96660945131267433924*y^13*x + 109418989131512359209*y^14
? 
? t^2 + 71 * z^2  -  ( 3 * x^2 + 92 * x * y + 729 * y^2)^7
%25 = 0
? 
? 
? t
%26 = 46*x^7 + 5103*y*x^6 + 234738*y^2*x^5 + 5797575*y^3*x^4 + 82723410*y^4*x^3 + 676824309*y^5*x^2 + 2898290766*y^6*x + 4865449685*y^7
? 
? z
%27 = x^7 - 5103*y^2*x^5 - 260820*y^3*x^4 - 5931765*y^4*x^3 - 71116920*y^5*x^2 - 438689181*y^6*x - 1098952052*y^7
? 

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  • $\begingroup$ By the way, I checked that the septic you arrived at does have a solvable Galois group, as expected. $\endgroup$ – Tito Piezas III Aug 6 '15 at 3:51
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(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.

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  • $\begingroup$ Oh, I just realized! If $d=-n$, and $a^2-nb^2=u^2-nv^2=\color{red}1$, then the formulas above can be found from Pell equations. $\endgroup$ – Tito Piezas III Aug 6 '15 at 4:38
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    $\begingroup$ It is likely, given your $a^2 + d b^2 = c^k$ primitive, with $k$ odd (and prime would help) and $c$ an odd prime that does not divide $d,$ that the form of that discriminant that does represent $c$ ( and its "opposite") has order exactly $k$ in the group under composition. The bad news is that there is no guarantee that we can write the principal form as some $\langle 1, f, c^k \rangle,$ although I did get lucky twice. If so, one may get these identities with no $c^k$ coefficient in front; if not, there is existence but no easy cookie cutter way to get the identity. $\endgroup$ – Will Jagy Aug 7 '15 at 3:40
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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?

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  • $\begingroup$ It is easy when $z = p^2+dq^2$. So, yes, we are looking for the other solutions when $z \neq p^2+dq^2$. $\endgroup$ – Tito Piezas III Aug 5 '15 at 11:57

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