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$x,y,z$ be three coprime integers, $a \in \mathbb{Z}>0$ and $k$ an odd integer. How do I find all the non-trivial solutions of the diophantine equation? $$x^2+ay^2=z^k$$ Does the method which consists in assuming that: $$x^2+ay^2=(p^2+aq^2)^k$$ (where $p,q$ are two coprime integers) provide the general parametrization?

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well, no. If $k$ divides the order of the class group, there will be others. For example, along with $x^2 + 11 y^2,$ we also have $3x^2 + 2xy+4y^2$ and $3x^2 - 2xy+4y^2.$ The class group has order 3, and any number $n,$ including but not limited to primes, that has a primitive expression $n = 3 x^2 + 2xy+4y^2,$ also has a primitive expression $$ n^3 = x^2 + 11 y^2. $$ The first few primes represented by $3x^2 + 2xy + 4 y^2$ are $$ 3, 5, 23, 31, 37, 59, 67, 71, 89, 97, $$ other numbers primitively represented include $4$ and $9.$

We find $$ 27 = 16 + 11, $$ $$ 125 = 81 + 11 \cdot 4, $$ $$ 12167 = 54^2 + 11 \cdot 29^2, $$ $$ 64 = 16 + 11 \cdot 4, $$ $$ 729 = 25 + 11 \cdot 64, $$

Tuesday afternoon. Finding individual parametrizations can be left to Dirichlet's version of Gauss composition for binary quadratic forms. If I have a binary form $A x^2 + B xy + A^{k-1}y^2,$ it has order $k$ in its class group. For example, using $3 x^2 + 8 y^2 + 9 y^2,$ we construct $$ r = x^3 - 9 x y^2 - 8 y^3, $$ $$ s = 27 x^2 y + 72 x y^2 + 37 y^3, $$ the result (not the most appealing way to write it, I admit) $$ 27 r^2 + 8 rs + s^2 = (3x^2 + 8xy+9 y^2)^3. $$ One may adjust the left hand side by $s = t - 4 r,$ or $t = s+4r,$ to get $t^2 + 11 r^2.$ Not pretty, but deriving it is algorithmic and deterministic.

For $t^2 + 47 r^2 = k^5,$ we may proceed with $3 x^2 + 28 xy + 81 y^2$ of order $5$ in its class group.

Went through it, $$ (14x^5 + 405yx^4 + 3780y^2x^3 + 13410y^3x^2 + 11550y^4x - 14647y^5)^2 + 47 ( x^5 - 270y^2x^3 - 2520y^3x^2 - 8115y^4x - 8344y^5)^2 = (3 x^2 + 28 x y + 81 y^2 )^5 $$

Below as it actually is done in gp-pari:

a=3; a1=3; b=28; c=27; 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=28; c=27; 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 - 27*y^2
? w
%4 = 6*y*x + 28*y^2

a1 = 9; c = 9; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;

? a1 = 9; c = 9; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? z
%6 = x^3 - 81*y^2*x - 252*y^3
? w
%7 = 27*y*x^2 + 252*y^2*x + 541*y^3

a1 = 27; c = 3; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? a1 = 27; c = 3; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? 
? z
%9 = x^4 - 162*y^2*x^2 - 1008*y^3*x - 1623*y^4
? w
%10 = 108*y*x^3 + 1512*y^2*x^2 + 6492*y^3*x + 8344*y^4

a1 = 81; c = 1; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;


? a1 = 81; c = 1; zz = x * z - c * y * w  ;  ww = a * x * w + a1 * y * z + b * y * w; z = zz; w = ww;
? 
? z
%12 = x^5 - 270*y^2*x^3 - 2520*y^3*x^2 - 8115*y^4*x - 8344*y^5
? w
%13 = 405*y*x^4 + 7560*y^2*x^3 + 48690*y^3*x^2 + 125160*y^4*x + 102169*y^5
? 
? 243 * z^2 + 28 * z * w + w^2
%14 = 243*x^10 + 11340*y*x^9 + 244485*y^2*x^8 + 3200400*y^3*x^7 + 28137390*y^4*x^6 + 173498248*y^5*x^5 + 759709530*y^6*x^4 + 2333091600*y^7*x^3 + 4812198255*y^8*x^2 + 6026540940*y^9*x + 3486784401*y^10
? 243 * z^2 + 28 * z * w + w^2 - (3 * x^2 + 28 * x * y + 81 * y^2)^5
%15 = 0
? 

t = w + 14 * z

? t = w + 14 * z
%16 = 14*x^5 + 405*y*x^4 + 3780*y^2*x^3 + 13410*y^3*x^2 + 11550*y^4*x - 14647*y^5
? z
%17 = x^5 - 270*y^2*x^3 - 2520*y^3*x^2 - 8115*y^4*x - 8344*y^5
? 

t^2 + 47 * z^2 
? 
? t^2 + 47 * z^2
%18 = 243*x^10 + 11340*y*x^9 + 244485*y^2*x^8 + 3200400*y^3*x^7 + 28137390*y^4*x^6 + 173498248*y^5*x^5 + 759709530*y^6*x^4 + 2333091600*y^7*x^3 + 4812198255*y^8*x^2 + 6026540940*y^9*x + 3486784401*y^10
? 
? t^2 + 47 * z^2 -  (3 * x^2 + 28 * x * y + 81 * y^2)^5
%19 = 0
? 
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  • $\begingroup$ So there is not a single method to parametrize the solutions of this equation? $\endgroup$ – user97615 Aug 4 '15 at 0:40
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    $\begingroup$ @WillJagy: +1 The Tues edit is very nice! $\endgroup$ – Tito Piezas III Aug 5 '15 at 2:26
  • $\begingroup$ @Tito, thanks... $\endgroup$ – Will Jagy Aug 5 '15 at 2:36
  • $\begingroup$ @WillJagy: Thanks to your Tues edit, I found an interesting observation. Care to look at this question? $\endgroup$ – Tito Piezas III Aug 5 '15 at 8:35
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No, there is no single method. For example, we have,

$$(p^3-3dpq^2)^2 + d(3p^2q-dq^3)^2 = (p^2+dq^2)^3$$

but it is no longer complete. In particular, for $d=47$, Pepin found there is no rational $p,q$ that corresponds to the solution,

$$(13u^3+30u^2v-42uv^2-18v^3)^2 + 47(u^3-6u^2v-6uv^2+2v^3)^2 = 2^3(3u^2+uv+4v^2)^3$$

Edit:

Using Jagy's $x^2+2xy+4y^2$, we have,

$$(4 x^3 + 15 x^2y - 6 x y^2 - 8y^3)^2 + 11(x^3 - 3 x^2 y - 6 x y^2)^2 = (3x^2 + 2x y + 4y^2)^3$$

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  • $\begingroup$ Yes this is interesting, but actually the second form you gave, they all have the some discriminant which is $-47$ is there a theorem affirming that $n^k$ can be written as $P(u)$ if and only if there exists a polynomial $P'$ having the same discriminant as $P$ and $n=P'(u')$ (in the case where $k$ verified some specific conditions ? $\endgroup$ – Elaqqad Aug 4 '15 at 11:55
  • $\begingroup$ @Elaqqad: Funny you mentioned that. Coz I just noticed myself that $d=-47$ thingie and I'm trying to find a similar formula using Jagy's $3x^2+2xy+4y^2$ (which has $d=11$). $\endgroup$ – Tito Piezas III Aug 4 '15 at 12:00
  • $\begingroup$ Yes this is really wired, and as you know, $3x^2+2xy+4y^2=\frac{1}{3}((3x+y)^2+11y^2)$ and the same thing $3x^2+xy+4y^2=\frac{1}{3}((6x+y)^2+47y^2)$ I think there will be a theorem about this, as representing integers by quadratic forms is well studied. $\endgroup$ – Elaqqad Aug 4 '15 at 12:09
  • $\begingroup$ @Elaqqad: You may be interested in this question. $\endgroup$ – Tito Piezas III Aug 5 '15 at 12:31
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    $\begingroup$ Yes i'm very interested, this is an amazing question, I though about about doing it myself . I did some research on the subject maybe later I will leave a summary in a comment. $\endgroup$ – Elaqqad Aug 5 '15 at 14:16
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For $k=5$ and $a=(4mn-t^2)$

${n}^{5}\,{\left( p\,s\,t+m\,{s}^{2}+n\,{p}^{2}\right) }^{5}=\left( 4\,m\,n-{t}^{2}\right)\cdot\,{\left( -{s}^{5}\,{t}^{4}-5\,n\,p\,{s}^{4}\,{t}^{3}+\left( 3\,m\,n\,{s}^{5}-10\,{n}^{2}\,{p}^{2}\,{s}^{3}\right) \,{t}^{2}+\left( 10\,m\,{n}^{2}\,p\,{s}^{4}-10\,{n}^{3}\,{p}^{3}\,{s}^{2}\right) \,t-{m}^{2}\,{n}^{2}\,{s}^{5}+10\,m\,{n}^{3}\,{p}^{2}\,{s}^{3}-5\,{n}^{4}\,{p}^{4}\,s\right) }^{2}/4+{\left( s\,t+2\,n\,p\right)}^{2}\cdot \,{\left( {s}^{4}\,{t}^{4}+3\,n\,p\,{s}^{3}\,{t}^{3}-5\,m\,n\,{s}^{4}\,{t}^{2}+4\,{n}^{2}\,{p}^{2}\,{s}^{2}\,{t}^{2}-10\,m\,{n}^{2}\,p\,{s}^{3}\,t+2\,{n}^{3}\,{p}^{3}\,s\,t+5\,{m}^{2}\,{n}^{2}\,{s}^{4}\\-10\,m\,{n}^{3}\,{p}^{2}\,{s}^{2}+{n}^{4}\,{p}^{4}\right) }^{2}/4$

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  • $\begingroup$ Do not understand what is written and what! $\endgroup$ – individ Dec 19 '15 at 5:47
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One approach is to write the General formula. For the equation.

$$x^2+ay^2=z^{k}$$

The formula looks for simplicity without coprime solutions.

$$x=(p^2-as^2)(p^2+as^2)^{k-1}$$

$$y=2ps(p^2+as^2)^{k-1}$$

$$z=(p^2+as^2)^2$$

After substituting numbers $p,s - $ to reduce common divisor.

$x,y - $ in such a number. $d^k$

$z - $ in such a number. $d^2$

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