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I tried to calculated the integral points for the following curve by Sage, but after a few hours I didn't receive any answer .

[0,0,0,-1609983754533564186692237854003906250000,0]

How can I calculate integral points for this curve?

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    $\begingroup$ Care to explain your notation and motivation for what you were trying to compute? $\endgroup$ Mar 7, 2012 at 16:31
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    $\begingroup$ $[a_1,a_2,a_3,a_4,a_6]$ is standard notation (in this subculture) for the elliptic curve $y^2 + a_1 x y + a_3 y = x^3 + a_2 x^2 + a_4 x + a_6$. So he's asking for integer solutions of $y^2 = x^3 - 1609983754533564186692237854003906250000 x$. Why this particular curve, I don't know. mwrank says it has rank 8. $\endgroup$ Mar 7, 2012 at 19:04
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    $\begingroup$ There are at least four pairs of integral points, since the mwrank basis contains points with $x = -1577320410156250000$, $43474893804931640625$, $-20492969555664062500$, and $19818207721836914062500$. There may be others. $\endgroup$ Mar 7, 2012 at 19:06
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    $\begingroup$ It seem that the curve comes by taking $a=25$, $b=5$ in the family of elliptic curves from this paper arxiv.org/abs/1202.5676 $\endgroup$
    – duje
    Mar 7, 2012 at 19:31
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    $\begingroup$ mw basis is generated by the points with $x$-coordinates: -695070141601562500, -1577320410156250000, -39433010253906250000, 43474893804931640625, 378951228540039062500/9, -5623634204101562500, -101469119551562500000000/994009, 415752567622704536133789062500/472932009 $\endgroup$
    – duje
    Mar 7, 2012 at 19:35

1 Answer 1

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This elliptic curve $E: x^3 - ax$, where $$ a = 1609983754533564186692237854003906250000 = 2^4 5^{28} 73 \cdot 97 \cdot 2281 \cdot 390001 \cdot 428801, $$ has the integral $2$-torsion point $T = (0,0)$, plus at least $17$ pairs $\pm P = (x,\pm y)$ of nontorsion integral points, with $(x,y)$ as follows:

(-695070141601562500, 33447209632705688476562500000)
(2316289620532226562500, 111461304221023574829101562500000)
(-1577320410156250000, 50354105671882629394531250000)
(1020708122564697265625, 32584910670546829700469970703125) *
(-39433010253906250000, 46579627221641540527343750000)
(40828324902587890625, 48227820848605342864990234375) *
(43474893804931640625, 110347464366896152496337890625) *
(-20492969555664062500, 156163643848510742187500000000)
(-5623634204101562500, 94213113773239135742187500000)
(-154178930664062500, 15755059779357910156250000000)
(-39975982055664062500, 21815100375976562500000000000)
(-3450306250000000000, 74255386668395996093750000000)
(-40124306640625000000, 976627163925170898437500000)
(211550113970947265625, 3021093403700607776641845703125) *
(2520419030914306640625, 126518514595970671176910400390625) *
(318603088420664062500, 5641612672625483917968750000000) *
(19818207721836914062500, 2789945206846541022216796875000000) *

These are listed in order of increasing canonical height; the first six points are in pairs $P,P'$ of equal height, with each $P' = T - P$ where $T=(0,0)$ as above. The list is probably complete: the heights ranged from $12.3$ to $30.3$, and none of the other $15000$ or so point pairs of height at most $100$ is integral.

As suggested in the commentary, proving that there are no other integral points is feasible but requires rather more effort than one would normally expend without having quite a good reason. Thanks to the presence of the $2$-torsion point (and some luck with the $2$-descent, as implemented in Cremona's mwrank), one can compute generators for the group of rational points on $E$. The rank of this group is $8$, which is large enough that the proof that there are no more integral points is not routine.

In case somebody does want to take on this computation, note that the equation for $E$ is not minimal due to the factors $2^4 5^{28} = 156250^4$ of $a$. For a minimal equation, write $(x,y) = (156250^2 x_0,156250^3 y_0)$ and divide through by $156250^4$ to get the minimal form $E_0: y_0^2 = x_0^3 - a_0^{\phantom0} x_0^{\phantom0}$ where $$a_0 = a / 156250^4 = 2701104520630058561.$$ Seven of the pairs $(x,\pm y)$ of integral points on $E$ become non-integral on $E_0$; those are marked with asterisks in the above list.

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  • $\begingroup$ FYI found all of your integral points and none new so far. $\endgroup$
    – joro
    Mar 8, 2012 at 13:52
  • $\begingroup$ @joro: using the same kind of method or something else? $\endgroup$ Mar 8, 2012 at 15:23
  • $\begingroup$ Elkies, probably not your method, i didn't care about the height. Just tried an opportunistic search. If G is the set of the generators + the torsion point tried subset sums of {P,-P,2P, -2P} P in G. The known points were found fast (the subsets are still running). $\endgroup$
    – joro
    Mar 8, 2012 at 18:32
  • $\begingroup$ I've tried similar things in the past, recursively trying all 2- and 3-point sums from the known points. Using the canonical height seems more efficient in this range: it took only a minute or so to try all candidates of height $200$ or less (yes, double what I posted -- I'll update this in the next edit). As expected no new integral points appeared. $\endgroup$ Mar 8, 2012 at 21:38

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