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Are there any cubes of the form $3x^2\pm xy+5y^2$, with x, y coprime ?

Partly inspired by this question. I tried various computer searches of the form $|x|\le10^a$, $|y|\le10^b$ with $a+b=6$, all of which failed to return any solutions.

Davood
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Lucian
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  • Lucian, notice that the form in the previous problem was $3 x^2 + xy - 5 y^2,$ an indefinite form of discriminant $61$ and class number one. Your form, $3 x^2 + xy + 5 y^2$ is positive, discriminant $-59,$ and class number three. Completely different behavior. – Will Jagy Feb 09 '15 at 03:34
  • @WillJagy: I am not surprised. A small computer routine had no troubles whatsoever in finding countless coprime solutions to $z^3=3x^2\pm xy{\color{red}-}5y^2$. – Lucian Feb 09 '15 at 07:10

2 Answers2

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Lucian, I think you can do this yourself (for primes) with some hints and some formalism, from Leonard Eugene Dickson, Introduction to the Theory of Numbers. If a prime $q$ is represented as $q = x^2 + xy+ 15 y^2,$ then $3 x^2 + xy + 5 y^2$ does not represent it or its square or cube. If a prime $p$ is represented as $p = 3x^2 + xy+ 5 y^2,$ then $ x^2 + xy + 15 y^2$ does not primitively represent $p^2.$ Meanwhile, $ 3x^2 + xy+ 5 y^2$ does primitively represent $p^2,$ and does represent $p^3$ but not primitively. For primitively, Dickson says properly.

Start with problem 4 at the bottom of page 93, continues onto page 94.

Oh, for $2$ and any prime $r$ such that $(-59|r) = -1,$ no form of discriminant $-59$ represents $r$ or $r^3.$

Not needed: the primes represented by $x^2 + xy + 15 y^2$ are $59$ along with those $p$ for which $$ z^3 + 2 z + 1 \equiv 0 \pmod p $$ has three distinct solutions. The form $3 x^2 + xy + 5 y^2$ represents all other primes with $(-59|p)=1.$

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Will Jagy
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For the equation:

$$3x^2+xy+5y^2=z^3$$

write the formula so that it was easier to go through. To facilitate calculations will make the replacement.

$$p=15k^2+3q^2-21n^2-11qk$$

$$s=25n^2+15k^2+3q^2+10qn-11qk-38kn$$

$$a=42n^2-30k^2-46qn+60qk-16q^2$$

$$t=42n^2-6q^2+8k^2-46nk+12qk$$

$$b=3p^2+6ps-17s^2$$

$$j=p^2+6ps-3s^2$$

Then decisions can be recorded and they are.

$$x=ab-5tj$$

$$y=(a-3t)b+3(2t-a)j$$

$$z=3p^2+4ps+21s^2$$

$q,k,n$ - integers, which we ask.

individ
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  • I tested your expressions with Mathematica. Your $x,y$ have the common factor $(15 k^2 - 27 k n + 21 n^2 - 11 k q + 4 n q + 3 q^2)$. Lucian wants them be co-prime, meaning to have no common factor. – Tito Piezas III Feb 08 '15 at 14:51
  • @TitoPiezasIII all solutions of the equation give not one simple solution. Perhaps they do not exist. – individ Feb 08 '15 at 14:55
  • Check your answer before posting. If you knew your $x,y$ had a common factor, then no need to post. – Tito Piezas III Feb 09 '15 at 15:50