Pick fourth power free integer $n$ ($p^4$ doesn't divide $n$).

Represent $n$ as difference of possibly negative integer squares $n=v_i^2-u_i^2$.

The goal is to find quadratic polynomial with integer coefficients (possibly square) $f(x)$ maximizing the number of solutions to $f(x)=u_i$.

Solutions are integral points on the elliptic curve $n=y^2-f(x)^2$.

Unless one can use the group law and scale rationals (related to fourth power free), boundedness is open problem, so I am interested in explicit records.

In terms of rank, my current record is $37$ points (not counting the sign of $x$) on $ y^2-(x^2-565215^2)^2=40224510201185827416900$ of rank $9$.

This was found by expressing $u_i=u_i'^2-v_i'^2$ and checking for the most common $v_i'$.

Also, what do heuristic say about this record(s)?

  • $\begingroup$ $n = 440224510201185827416900$ is equal to the product of the first $11$ primes squared. I tried your idea for product with less than $11$ primes squared, but achieved nothing of interest. So I wondered how you came across $n = 40224510201185827416900$? And how are you able to compute with an $n$ of this order of magnitude. It seems to demand a prohibitively long computation. $\endgroup$ Nov 26, 2014 at 20:36
  • $\begingroup$ Other than that I really like your ideas of producing elliptic curves of high rank. This one, as well as the previous idea, seems quite novel to me. $\endgroup$ Nov 26, 2014 at 20:37
  • $\begingroup$ @JesperPetersen The record was found in less than 15 minutes. Contact me via mail if you want the sage code. There might be some structure in the divisors leading to better records - chose suitable primes. $\endgroup$
    – joro
    Nov 27, 2014 at 8:52
  • $\begingroup$ Actually there was a bug in my own Sage code. I did notice a vague pattern, which I reported in a separate answer. $\endgroup$ Nov 28, 2014 at 22:12

1 Answer 1


From realizing that

$$440224510201185827416900 = \prod_{\text{p prime} \leq 31} p^2$$ I noticed a vague pattern that might be of interest. Let

$$n_P = \prod_{\text{p prime} \leq P} p^2$$ and consider that curves $E_P: n_P = y^2 - f(x)$.

Now, the number of integral points on $E_P$, which we can denote $S_P$, grows monotonically like this. $$ S_3 = 1, S_5 = 3, S_7 = 6, S_{11} = 9, S_{13} = 10, S_{17} = 14, S_{19} = 30.$$ For $S_{23}$ and $S_{29}$ there is no growth compared to $S_{19}$, but as you have noticed $S_{31} = 37$.

I wonder if there is an explanation for this. Maybe your Sage code can extend the computation beyond $S_{31}$?

  • $\begingroup$ For S_17 I get 15 points and for S_19 get 31. $\endgroup$
    – joro
    Nov 29, 2014 at 8:57
  • $\begingroup$ My implementation uses a lot of RAM (on purpose, to save time) and I get out of memory on my machine beyond. Maybe 64 or 128 GB will be enough for beyond. Too lazy to increase the swap or use SQL. $\endgroup$
    – joro
    Nov 29, 2014 at 13:32
  • $\begingroup$ My computation of $S_{19}$ contains these 30 points:[[73269, 5359083114],[60025, 3593741690],[33853, 1136794966],[25095, 620562810], [16611,266829654],[15701, 237447574],[11775,129741690],[10805,107912810],[8617, 65700614],[7035, 41372310], [6543, 34913274], [5425, 22370810], [5255,20749690],[4831, 17086586],[4641,15638406],[4365,13775190],[185,13394810],[273, 13367046],[495,13250310],[4221,12926634],[1069,12655786],[1311,12293754],[1477,12014954],[1875,11279190],[1933,11168086],[2187,10688106],[3465,10077690],[3379,9934166],[2807, 9799174],[2883, 9747114]]. Which one is missing? $\endgroup$ Nov 29, 2014 at 17:30
  • $\begingroup$ What is your f(x)? Is (3045, 9699690) on your quartic, it is on mine (and your points are on mine). $\endgroup$
    – joro
    Nov 29, 2014 at 17:48
  • $\begingroup$ My $f(x)$ is $x^2 - 3045^2$. $(3045, 9699690)$ is indeed on the quartic. I must investigate how I could miss that point. $\endgroup$ Nov 29, 2014 at 18:05

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