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Numbers k such that Mordell's equation y^2 = x^3 + k has exactly 5 integral solutions.
8

%I #19 Jun 02 2023 01:56:53

%S 1,64,729,1000,2744,4096,15625,21952,35937,46656,50653,64000,117649,

%T 262144,343000,531441,592704,681472,729000,753571,1000000,1124864,

%U 1771561,2000376,2197000,2299968,2744000,2985984,3652264,4096000,4826809,5451776,6229504,7189057,7529536

%N Numbers k such that Mordell's equation y^2 = x^3 + k has exactly 5 integral solutions.

%C Contains all sixth powers: suppose that y^2 = x^3 + t^6, then (y/t^3)^2 = (x/t^2)^3 + 1. The elliptic curve Y^2 = X^3 + 1 has rank 0 and the only rational points on it are (-1,0), (0,+-1), and (2,+-3), so y^2 = x^3 + t^6 has 5 solutions (-t^2,0), (0,+-t^3), and (2*t^2,+-3*t^3). - _Jianing Song_, Aug 24 2022

%H J. Gebel, <a href="/A001014/a001014.txt">Integer points on Mordell curves</a> [Cached copy, after the original web site tnt.math.se.tmu.ac.jp was shut down in 2017]

%F a(n) = A356711(n)^3.

%Y Cf. A054504, A081119, A179145-A179162, A356711.

%K nonn

%O 1,2

%A _Artur Jasinski_, Jun 30 2010

%E Edited and extended by _Ray Chandler_, Jul 11 2010

%E a(31)-a(35) from _Max Alekseyev_, Jun 01 2023