Timeline for On the equation $x^3 + y^3 = z^4$
Current License: CC BY-SA 4.0
6 events
| when toggle format | what | by | license | comment | |
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| Sep 8, 2020 at 22:09 | comment | added | Sam | @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)]. | |
| Sep 8, 2020 at 16:26 | comment | added | Yemon Choi | 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 | |
| Sep 8, 2020 at 16:25 | comment | added | Yemon Choi | 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. | |
| Sep 8, 2020 at 13:40 | comment | added | Sam | @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. | |
| Sep 7, 2020 at 19:05 | comment | added | David Loeffler | This is just a special case of nullhomologous' solution, with $s$ and $t$ taken to be $(p^4 + 1)/p$ and $(p^4 - 1)/p$. | |
| Sep 7, 2020 at 18:17 | history | answered | Sam | CC BY-SA 4.0 |