Timeline for On $\sum_{k=1}^nk^3 = x^3 + y^3$ with $x,y \ge 1$
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Aug 8, 2020 at 10:17 | history | edited | Alkan | CC BY-SA 4.0 |
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| Aug 7, 2020 at 8:56 | history | edited | Alkan | CC BY-SA 4.0 |
added 3 characters in body
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| Aug 7, 2020 at 8:42 | comment | added | Alkan | Sorry. I have a typo, what I mean is $293^{3}$. Best regards. | |
| Aug 7, 2020 at 8:42 | history | edited | Alkan | CC BY-SA 4.0 |
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| Aug 7, 2020 at 5:20 | comment | added | Alkan | @Tomita: Thanks, I understand. Your example is also confirmation of last term of OEIS entry. Best regards. | |
| Aug 7, 2020 at 4:44 | comment | added | Alkan | @Tomita: Many thanks for your comment. Yes, I agree with your transformation. Is this larger solution that you found yet? May you look further ranges and can you confirm $291^{3}$ ? I am also not sure that there are other solution(s) between $2112278$ and $291^{3}$. Best regards. | |
| Aug 7, 2020 at 0:15 | comment | added | Tomita | @Alkan: $\sum_{k=1}^nk^3 = x^3 + y^3$ can be transformed to $v^2 = -3s^4+(3n^4+6n^3+3n^2)s$. We get $(n,s,x,y)=(2112278, 237469827, 70918018, 166551809).$ | |
| Aug 6, 2020 at 16:52 | comment | added | Alkan | @zeraouliarafik: Yes, we can rewrite the equation based on $\sum_{k=1}^nk^3 = T_{n}^{2}$ as you did, $n^2 + 2n^3 + n^4 = 4(x^3+y^3)$. Best regards. | |
| Aug 6, 2020 at 13:02 | comment | added | zeraoulia rafik | I think you are wolcome to study solutions of $ 2(n^4+n^2)+4n^3= 8(x^3 + y^3)$ | |
| Aug 6, 2020 at 8:41 | history | asked | Alkan | CC BY-SA 4.0 |