This particular curve, which I'll call $E$, may be quite challenging. The analytic rank is probably 2, but as far as I know, the only way to prove this is to show that the algebraic rank is not 0. Assuming the analytic rank is 2 and the full BSD formula, the product of the Regulator and the order of the Tate-Shafarevich group is approximately 1435241.110225344. (Using the command ConjecturalRegulator(E); in MAGMA).

Since the the rank is probably 2, I would guess that the Shafarevich-Tate group is trivial, so that the regulator is quite large.

In this case a 4-Descent was feasible. Below are the genus 1 curves in $\Bbb P^3$ that represent the elements of the $4$-Selmer group modulo torsion. If we can find points on these curves, they will map to points of infinite order on $E$.

```
> E := EllipticCurve([0,-86069^5]);
> ConjecturalRegulator(E);
1435241.11022534407264592039437 2
> TD := TwoDescent(E);
> SetClassGroupBounds("GRH");
> time FD := [FourDescent(C) : C in TD];
Time: 3.890
> FD;
[
[
Curve over Rational Field defined by
15*x1^2 + 111*x1*x2 + 4*x1*x3 + 10*x1*x4 - 38*x2^2 - 95*x2*x3 + 67*x2*x4
- 54*x3^2 - 14*x3*x4 - 71*x4^2,
31*x1^2 - 71*x1*x2 - 69*x1*x3 - 23*x1*x4 + 9*x2^2 - 92*x2*x3 - 24*x2*x4
+ 35*x3^2 - 148*x3*x4 + 168*x4^2,
Curve over Rational Field defined by
35*x1^2 + 26*x1*x2 + 41*x1*x3 + 61*x1*x4 + 54*x2^2 + 11*x2*x3 + 25*x2*x4
+ 68*x3^2 + 3*x3*x4 - 78*x4^2,
36*x1^2 - 138*x1*x2 - 38*x1*x3 + 127*x1*x4 + 25*x2^2 + 42*x2*x3 +
47*x2*x4 + 81*x3^2 - 12*x3*x4 + 60*x4^2
],
[
Curve over Rational Field defined by
21*x1^2 + 13*x1*x2 + 13*x1*x3 + 44*x1*x4 - 32*x2^2 - x2*x3 - 18*x2*x4 -
45*x3^2 + 24*x3*x4 - 238*x4^2,
5*x1^2 - 14*x1*x2 + 122*x1*x3 + 268*x1*x4 + 26*x2^2 + 6*x2*x3 +
149*x2*x4 - 57*x3^2 - 23*x3*x4 - 78*x4^2,
Curve over Rational Field defined by
4*x1^2 + 49*x1*x2 + 34*x1*x3 + 26*x1*x4 + 26*x2^2 - 33*x2*x3 - 74*x2*x4
+ 53*x3^2 - 74*x3*x4 + 111*x4^2,
38*x1^2 - 84*x1*x2 + 3*x1*x3 - 88*x1*x4 - 29*x2^2 + 27*x2*x3 - 154*x2*x4
+ 5*x3^2 - 234*x3*x4 - 120*x4^2
],
[
Curve over Rational Field defined by
7*x1^2 + 78*x1*x2 + 106*x1*x3 + 62*x1*x4 - 21*x2^2 - 26*x2*x3 + 22*x2*x4
+ 34*x3^2 - 25*x3*x4 - 118*x4^2,
33*x1^2 + 2*x1*x2 - 14*x1*x3 + 106*x1*x4 + 48*x2^2 - 33*x2*x3 +
165*x2*x4 + 69*x3^2 + 31*x3*x4 - 26*x4^2,
Curve over Rational Field defined by
7*x1^2 + 46*x1*x2 + 33*x1*x3 + 23*x1*x4 + 13*x2^2 + 36*x2*x3 - 108*x2*x4
- 69*x3^2 - 88*x3*x4 + 145*x4^2,
19*x1^2 - 28*x1*x2 - 14*x1*x3 + 8*x1*x4 + 150*x2^2 - 52*x2*x3 +
190*x2*x4 - 46*x3^2 + 33*x3*x4 + 248*x4^2
]
]
```

I searched for points on these curves up to height $10^9$ and didn't find any, which isn't that surprising given that the regulator is probably so large. Maybe an $8$- $9$- or $12$-descent would help, but I'm not sure since the regulator looks to be so large.

I've come across a lot of curves like this one, and honestly I don't know what to do other than spend a great deal of time implementing higher and higher descents, which will require better and better architecture for working with algebraic number fields.

Finding points on curves with rank at least 2 is much harder than the rank 1 case, where a non-torsion Heegner point can be constructed. I've often wondered if the notion of visibility of Mordell-Weil groups could be useful here to prove that the rank is 2. There might be some hope since this curve is just a sextic twist of a very simple elliptic curve, but I have no idea what other abelian variety one would try to use to "visualize" the Mordell-Weil group.

EDIT: Found a rational point!

Since this is a Mordell curve, it has a 3-isogeny. So ThreeDescentByIsogeny(E); works in magma. This, together with the 4-Descent above, can be patched together with Tom Fisher's 12-Descent code. Running PointSearch(C, 10^15) on all the 12-covers, I found one rational point on a 12-cover. This maps back to the HUGE point $P = (r/t^2, s/t^3) \in E(\Bbb Q)$ where

$$t = 1064315237527062416197829497356636659645269584461099593669088031745089037672135800872679057089531240249644479256790731579010363138838881840905955995370777995601208434305913644468575606,$$

$$r = 280330358182289626756155234063598323321079390462341827366518987565228009421332474964681906370265513552399459534408574388270067641670339702167066942312699829200035334349465315027488431491497462250163493458119988220147933027219004129214673758569410719992144923411319943555067857245550021843686621974171705862665984304126540657354963027685841117202627482318291749321065520239017,$$

and

$$s = -3894169495021322706664690332034015776766200723677334488117565017206832575610394605401029654682981124772659459298812689312208671041841435640288530899795853801784491292694434680924857836480106291120205903428248497270005655966231499766892812223437056078000735443855989892506051021805114412318184420256429821942583648074219767873044937972437645324252633026228735728785740079828040374108087622426272823877167554852034435686184476368398239806496596492675365799418315372050294056476779196364271993469777717240193414071588739725730054647942389914059081838403974737198837$$

Magma (and sage) says the canonical height of $P$ is 862.589016739449. $P$ has infinite order and is not a multiple of another point.

I'll keep looking for more points on 12-covers, but there should be a smarter way to look for a second generator since we know conjecturally what the regulator is supposed to be, at least up to division by an integer square (the order of Sha).

Second Edit: Second generator found by Tom Fisher!

I received email from Tom Fisher who got wind of this example thanks to Kevin Acres and Mark Watkins. He used some more careful minimization techniques for 12-covers which made it much easier for him to find the second generator $Q = (u/w^2, v/w^3)$ where

$$w = 182114807484521200807106885195433046250677033531559359037472093313104122418454868285361195199297225042175078422885282555317033956772998359062866824462985656413093380882725866443379879289096885109176415930832786071510257551923931848041894889001647111552913336376307990237676516910188887260245394301062683134535223719019999940520792408074037957510825778941886955164545102$$

$$u = 6409818948420000148009253515371033320765674658334995509069337486116156301741657451140832523716129416946553724548364793698626405777843956622076380997023235142487299844326278568312577933328949209325079504842164930964499614652180484249884310193722360960206856049327696413372088438499522988880226096525828673972002212550418445757426791463589511241682169507191857404060127705996714164059762587639453970388047710151991644286292989545241436751637355865317439825495449344823977758797766025325026099986880721224779341838599953004314077314458400203457162779096001177667625960285991372461677348744634118572760595736047817251613300845834732694598202906328385669708044981270667425075403152129513088853355196780998723571290752400352784381236905647783762282233$$

$$v = 9541721267275526440336656075081968921795973155189044266523956895286489666750185996900611973660561530367870689843326743006933458752307224811373451969681421106849987021781748740069623362388923316032840282833074591791294185661671902447470420594105932470690611184875968605655279379274234568020070054315280341961704988604471445169462057158108739193383808091705876229256447451432215635271166605138776098284834213663487614216411244364417456782497135614069447487775516105502956819954881548171764933506408613533330700718054166278531808088754093650937927404096948372079871885150489168059803000137727292308698902319300888120656605108547686297443217509881949442132801485412503095358513515604711600740777338910658971774981045705755311582771691405245163713860189075823123694400673272454942433819801867842344237200886745700934179591344744409814700762619336056693409131813436072830610006822421568399180878828865624130779460438617259828853398661927306856293129558704629256045989464076110208592569978237366668166654822438426674566555490387538893656476099175185704061248678944708728574308512767720849000403950411614279621166869036373101$$

This height of $Q$ is 1715.46805605884712533431816634 and the regulator is actually 1435241.11022534407264592039437 as BSD predicts if Sha(E/Q) is trivial.

magmacomputation) here the ConjecturalRegulator is only 2612.64 (those $k^{\pm\epsilon}$ factors can make a big difference in practice), so you can expect to get at least one point. Meanwhile I ran FourDescent and PointsQI, and got $(46160353/42^2, 438188195663/42^3)$, whichmwrankmay have found by now (height $18.29$), and ... – Noam D. Elkies Sep 2 '12 at 3:09mwranksurely could not find in reasonable time. – Noam D. Elkies Sep 2 '12 at 3:11