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Diophantine equation. $X^2+Y^2=qZ^3$

I wonder at what values ​​of the coefficient $q$ equation has a solution.

And of course I wonder how she looks like a formula describing their solutions.

For the special case when $X^2+Y^2=Z^3$ You can get a basic formula.

Has the solutions:

$X=2k^6+8tk^5+2(7t^2+8qt-9q^2)k^4+16(t^3+2qt^2-tq^2-2q^3)k^3+$

$+2(7t^4+12qt^3+6q^2t^2-28tq^3-9q^4)k^2+8(t^5+2qt^4-2q^3t^2-5tq^4)k+$

$+2(q^6-4tq^5-5q^4t^2-5q^2t^4+4qt^5+t^6)$

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$Y=2k^6+4(3q+t)k^5+2(9q^2+16qt+t^2)k^4+32qt(2q+t)k^3+$

$+2(-9q^4+20tq^3+30q^2t^2+12qt^3-t^4)k^2+4(-3q^5-tq^4+10q^3t^2+6q^2t^3+5qt^4-t^5)k-$

$-2(q^6+4tq^5-5q^4t^2-5q^2t^4-4qt^5+t^6)$

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$Z=2k^4+4(q+t)k^3+4(q+t)^2k^2+4(q^3+tq^2+qt^2+t^3)k+2(q^2+t^2)^2$

$q,t,k$ - What are some integers any sign. After substituting the numbers and get a result it will be necessary to divide by the greatest common divisor. This is to obtain the primitive solutions.

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  • $\begingroup$ I find the method you used for this and the answer below yields an $X,Y,Z$ with the form $$(ax)^2+n(ay)^2=(az)^3$$ hence have a common factor. $\endgroup$
    – Tito Piezas III
    Aug 7, 2015 at 14:52
  • $\begingroup$ @TitoPiezasIII You know how to solve a system of nonlinear Diophantine equations? This is unlikely! $\endgroup$
    – individ
    Aug 7, 2015 at 14:57
  • $\begingroup$ I've solved up to 10th degree (see this post) so 3rd degree is not that difficult. The common factor of your $X,Y,Z$ above is $a=(k+q)^2+t^2$. Don't you have Mathematica? $\endgroup$
    – Tito Piezas III
    Aug 7, 2015 at 15:31
  • $\begingroup$ @TitoPiezasIII I have a pen and paper. Why do I need a computer? Although I agree that some decisions are just. Then you should choose a more complex equation. $\endgroup$
    – individ
    Aug 7, 2015 at 16:35

3 Answers 3

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There are solutions with $(x,y)\ne (0,0)$ for every $q\ne 0$. To see this, recall that the equation $x^2+y^2=w$ has a non-trivial solution if the prime power factorization of $w$ has all primes of the form $4k+3$ appearing to an even power. So if some prime $p$ of the form $4k+3$ appears to an odd power in the prime power factorization of $q$, we just pick $z$ with $p$ appearing to an odd power in the prime power factorization of $z$.

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  • $\begingroup$ I is not welcome and do not think so. Prefer to write exactly the formula opisyvayushie solutions. $\endgroup$
    – individ
    Apr 28, 2014 at 7:13
  • $\begingroup$ You did not read what I wrote? I painted the decision not as tsiferek and formula describing his decision. $\endgroup$
    – individ
    Apr 28, 2014 at 7:24
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    $\begingroup$ I did not understand the assertion in a comment that "$1$ is not welcome" and for a minute thought it meant there is no solution for $q=1$. But then I realized you were aware of the solutions when $q=1$ so must have meant something else. But the answer to your first question remains that there is a solution with $(x,y)\ne (0,0)$ for any $q\ne 0$. $\endgroup$
    – André Nicolas
    Apr 28, 2014 at 7:26
  • $\begingroup$ We must solve the equation - and not talk. As an example, look at my decision, and then we'll talk. $\endgroup$
    – individ
    Apr 28, 2014 at 7:30
  • $\begingroup$ Let's translate A. Nicolas' answer into an identity. Define $$c = e^3 - 3e f^2,\quad d = 3e^2f - f^3$$ Then, $$(a d + b c)^2 + (a c - b d)^2 = (a^2+b^2)(e^2+f^2)^3$$ Thus, any number $q$ expressible as the sum of two squares $a^2+b^2$ will do. $\endgroup$
    – Tito Piezas III
    Nov 18, 2014 at 1:56
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In the equation: $X^2+Y^2=qZ^3$

If the ratio is such that the root of an integer: $c=\sqrt{q-1}$

Then the solution is:

$X=-2(c+1)p^6+4(2c(q-2)-3q)p^5s+2(c(5q^2-2q-8)-q^2-22q+8)p^4s^2+$

$+8q(5q^2-14q+4)p^3s^3+2(c(5q^2-2q-8)+q^2+22q-8)q(q-2)p^2s^4-$

$-4(2c(q-2)+3q)q^2(q-2)^2ps^5-2(c-1)q^3(q-2)^3s^6$

...........................................................................................................................................................

$Y=2(c-1)p^6+4(2qc+q-4)p^5s+2(c(-5q^2+18q-8)+15q^2-22q-8)p^4s^2+$

$+8q(q^2+6q-12)p^3s^3-2(c(5q^2-18q+8)+15q^2-22q-8)q(q-2)p^2s^4-$

$-4(2qc-q+4)q^2(q-2)^2ps^5+2(c+1)q^3(q-2)^3s^6$

............................................................................................................................................................

$Z=2p^4+8p^3s+4(q^2-2q+4)p^2s^2-8q(q-2)ps^3+2q^2(q-2)^2s^4$

And more.

$X=-2(c-1)p^6-4(2c(q-2)+3q)p^5s+2(c(5q^2-2q-8)+q^2+22q-8)p^4s^2+$

$+8q(5q^2-14q+4)p^3s^3+2(c(5q^2-2q-8)-q^2-22q+8)q(q-2)p^2s^4+$

$+4(2c(q-2)-3q)q^2(q-2)^2ps^5-2(c+1)q^3(q-2)^3s^6$

.............................................................................................................................................................

$Y=2(c+1)p^6-4(2qc-q+4)p^5s+2(c(-5q^2+18q-8)-15q^2+22q+8)p^4s^2+$

$+8q(q^2+6q-12)p^3s^3+2(c(-5q^2+18q-8)+15q^2-22q-8)q(q-2)p^2s^4+$

$+4(2qc+q-4)q^2(q-2)^2ps^5+2(c-1)q^3(q-2)^3s^6$

.............................................................................................................................................................

$Z=2p^4-8p^3s+4(q^2-2q+4)p^2s^2+8q(q-2)ps^3+2q^2(q-2)^2s^4$

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For the equation: $$X^2+qY^2=Z^3$$

You can write this simple solution:

$$X=(p^2+qs^2)((p^4-q^2s^4)t^3-3(p^2+qs^2)^2kt^2+3(p^4-q^2s^4)tk^2-(p^4-6qp^2s^2+q^2s^4)k^3)$$

$$Y=2ps(p^2+qs^2)((p^2+qs^2)t^3-3(p^2+qs^2)tk^2+2(p^2-qs^2)k^3)$$

$$Z=(p^2+qs^2)((p^2+qs^2)t^2-2(p^2-qs^2)tk+(p^2+qs^2)k^2)$$

$q - $The ratio is given for the problem.

$p,s,t,k - $ integers asked us.

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