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Oct 17, 2020 at 21:38 history edited YCor CC BY-SA 4.0
removed capitals from title (the question was bumped anyway)
Oct 17, 2020 at 17:30 history edited Pietro Paparella CC BY-SA 4.0
Improved exposition
Feb 9, 2018 at 4:03 vote accept Pietro Paparella
Mar 15, 2017 at 7:47 answer added Włodzimierz Holsztyński timeline score: 4
Mar 15, 2017 at 5:13 answer added Zurab Silagadze timeline score: 11
Mar 15, 2017 at 4:44 comment added individ The parameterization equation $x^2+y^3=z^2$ can record such. artofproblemsolving.com/community/c3046h1217955_the_cube_view For $x^2+y^3=z^4$ artofproblemsolving.com/community/… As already said - it is better for each individual degree to solve directly the equation.
Mar 15, 2017 at 3:57 comment added Pietro Paparella @NoamD.Elkies: there are references but they concern the cases when $3 \leq n \leq 5$.
Mar 15, 2017 at 3:21 comment added Noam D. Elkies The last, and by far the hardest, case is $n=5$, which is solved in: J. Edwards: A Complete Solution to $X^2 + Y^3 + Z^5 = 0$, Journal f. d. reine und angew. Math. (Crelle's Journal) 571 (2004), 213-236. (Change $(X,Y,Z)$ to $(x,y,-z)$ for your form of the equation.) That paper should also give references for $n<5$.
Mar 15, 2017 at 2:54 history edited Pietro Paparella CC BY-SA 3.0
Reference added
Mar 15, 2017 at 2:39 history edited Pietro Paparella CC BY-SA 3.0
edited body
Mar 15, 2017 at 2:33 comment added Gerhard Paseman You can treat this as a special case of factoring, e.g. $Y=(z+x)(z-x)$, where you can have one of the factors (say $z-x$) be $y$ to a power. If you want everything integral, make sure the factors have the same parity. Gerhard "Difference Of Squares Is Easy" Paseman, 2017.03.14.
Mar 15, 2017 at 2:21 history edited Pietro Paparella CC BY-SA 3.0
title edited
Mar 15, 2017 at 2:16 history asked Pietro Paparella CC BY-SA 3.0