Are there any rational numbers $x, y, z$ with $xyz \neq 0$ and coprime numerators such that $x^3 +y^3 = z^4$ ?
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5$\begingroup$ $x=1/8$, $y=1/8$ and $z=1/4$? Or $x=1/9$, $y=2/9$ and $z=1/3$? $\endgroup$– NulhomologousCommented Sep 7, 2020 at 13:57
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1$\begingroup$ Also $(x,y,z)=(17/56, 37/56, 3/4)$ looks nice... There are plenty of them... $\endgroup$– NulhomologousCommented Sep 7, 2020 at 14:04
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1$\begingroup$ I believe this is a rational surface and has rational parametrization. $\endgroup$– joroCommented Sep 7, 2020 at 14:09
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$\begingroup$ Parameterization is easy to find. It is sufficient to solve the equation. $x^3+y^3=z^2$ Using these solutions. artofproblemsolving.com/community/c3046h1048734____2 artofproblemsolving.com/community/c3046h1046719___ artofproblemsolving.com/community/c3046h1046717_ Then solve another equation when z is a square. $\endgroup$– individCommented Sep 8, 2020 at 5:02
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2 Answers
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It is a rational surface.
One easy parametrization is $x=s^4 + s\; t^3$, $y=s^3\;t + t^4$ and $z=s^3 + t^3$. From this you should be able to find as many examples as you like.
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1$\begingroup$ Thanks ! Any other parametrization besides yours ? $\endgroup$– user123305Commented Sep 7, 2020 at 14:24
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$\begingroup$ But how did you arrive at your answer ? May you also see mathoverflow.net/q/371113/123305, $\endgroup$– user123305Commented Sep 7, 2020 at 18:08
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8$\begingroup$ Note that this parametrization doesn't only find infinitely many rational solutions, it actually finds all solutions. Here's why. It obviously gives $(0,0,0)$ by taking $s=t=0$. Now suppose that $(x,y,z)$ is a solution with $xyz\ne0$. Then take $s=x/z$ and $t=y/z$, and you'll get an $(s,t)$ pair such that when you plug them into Nullhomologous' formulas, you get back the original $(x,y,z)$. $\endgroup$ Commented Sep 7, 2020 at 19:12
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$x^3+y^3=z^4$ ----$(1)$
In the solution given by @Nulhomologous,
if we put $(s,t)=(2,1)$ we get:
$9^3+18^3=9^4$
Where the integer 'nine' is a common factor.
Eqn.(1) has another parametrization & is shown below:
$(2w(p^4+1))^3+(2w(p^4-1))^3=(2wp)^4$
Where, $w=(p^8+3)$
For, $p=2$, we get:
$8806^3+7770^3=1036^4$
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3$\begingroup$ This is just a special case of nullhomologous' solution, with $s$ and $t$ taken to be $(p^4 + 1)/p$ and $(p^4 - 1)/p$. $\endgroup$ Commented Sep 7, 2020 at 19:05
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$\begingroup$ @Joe Silverman. Your comment about solution given by "Nullomolgous" being a general solution is incorrect because his equation does not produce the numerical solution, (x,y,z)=((17/56),(37/56),(3/4)). Also [z=3/4] in the equation cannot be represented as sum of two rational cubes. $\endgroup$– SamCommented Sep 8, 2020 at 13:40
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$\begingroup$ Regarding the previous comment: for the (x,y,z) solution you have listed, defining s=x/z = 17/42 and t= y/z = 37/42 one can check that s^4+st^3 = 17/56 and t^4+s^3t= 37/56 and s^3+t^3 = 3/4. So it is incorrect to claim that @Nulhomologous's formula does not produce the given numerical solution. $\endgroup$ Commented Sep 8, 2020 at 16:25
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$\begingroup$ I only used explicit numbers to be concrete. It is actually a direct consequence of basic algebra that if x^3+y^3=z^4 with z non-zero, and one then defines s and t as in Joe Silverman's comment, then s(s^3+t^3)=x, t(s^3+t^3)=y and s^3+t^3 = z $\endgroup$ Commented Sep 8, 2020 at 16:26
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1$\begingroup$ @Yemon Choi. You are correct. The numerical solution given by "Nullomolgous" does satisfy the equation. Also his equation look's like a general solution just as "Joe Silverman" commented. I was surprised that the equation is not primitive & has a common factor. Usually a general solutions do not have a common factor like the pythagoras equation of second degree. Namely the general solution, [(m^2+n^2),(m^2-n^2),(2mn)]. $\endgroup$– SamCommented Sep 8, 2020 at 22:09