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The following question is related to the families of high rank elliptic curves with torsion subgroup $\mathbb{Z}/6\mathbb{Z}$.

The SageMath/Python code below produces a list of small fractions $a$ for which $e=\sqrt{a(a+1)}$ is a multiple of $\sqrt{r}=\sqrt{2}$.

from time import time
t0 = time()
r = 2
listA = []
top = 100
for n in range(-top, top + 1):
    for d in range(1, top + 1):
        if gcd(n,d) == 1:
            a = QQ(n) / QQ(d)
            e = sqrt(a * (a + 1) / r)
            if (e in QQ) and a != -1 and a != 0:
                listA.append(a)
print(listA)
print(len(listA))
print(time() - t0)

[-98/17, -98/73, -98/89, -98/97, -81/31, -81/49, -81/73, -81/79, -72/23, -72/47, -72/71, -50, -50/41, -50/49, -49/17, -49/31, -49/41, -49/47, -32/7, -32/23, -32/31, -25/7, -25/17, -25/23, -18/17, -9, -9/7, -8/7, -2, 1, 1/7, 1/17, 1/31, 1/49, 1/71, 1/97, 2/7, 2/23, 2/47, 2/79, 8, 8/17, 8/41, 8/73, 9/23, 9/41, 9/89, 18/7, 18/31, 25/7, 25/47, 25/73, 32/17, 32/49, 32/89, 49, 49/23, 49/79, 50/31, 50/71, 72/49, 72/97, 81/17, 81/47, 98/23, 98/71]
66
1.3610780239105225

Given some $a$, I would like to derive a simple formula to produce a different value, $b$, in the same list. I would measure the complexity of the formula by the sum $s$ of degrees of its numerator and denominator.

By analyzing the high rank $\mathbb{Z}/6\mathbb{Z}$ families, three simple formulas were derived to date. Note that these formulas are true for any value of $r$, not just $r=2$.

(1) $b=-\dfrac{(a+1)}{(1-3a)}$; $s=2$

(2) $b=\dfrac{a(a+1)}{(1-3a)}$; $s=3$

(3) $b=-\dfrac{(5a+1)^2}{(3a-1)^2}$; $s=4$

Applying (2) and (2) consecutively:

(4) $b=\dfrac{a(a+1)(a^2-2a+1)}{(-1+6a+3a^2)(-1+3a)}$; $s=7$

For a higher value of $s$:

(5) $b=-\dfrac{(1+2a+5a^2)^2}{(-1+6a+3a^2)^2}$; $s=8$

My questions:

  1. Is it possible to derive some new formula(s) for $s\le6$?
  2. Is there a way to write a smarter/faster code, e.g., by skipping some values of $n$ and/or $d$ for given values of $r$ and $top$?
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    $\begingroup$ You use the symbol $b$, but never define it. Anyway, you want $\sqrt{a(a+1)}=c\sqrt2$, This equation can be rewritten in the form $u^2-2d^2=1$, where $u=2a+1$, and this gives you $\sqrt{a(a+1)}=c\sqrt2$ with $d=2c$. Taking a point on $u^2-2d^2=1$, such as $u=1$, $d=0$, drawing a line through it with rational slope $t$, and finding the other intersection of this line with the hyperbola, should give all rational solutions, parametrized by $t$. $\endgroup$
    – Gerry Myerson
    Commented Jan 23, 2021 at 23:20
  • $\begingroup$ @Gerry Myerson: $b$ is a different value in the same list. E.g., if $r=2$, $top=289$, $a=-2$, then formulas $(1)-(5)$ produce $b_1=\frac{1}{7}$, $b_2=\frac{2}{7}$, $b_3=-\frac{81}{49}$, $b_4=\frac{18}{7}$, $b_5=-289$, all of which are in the same list. $\endgroup$
    – Maksym Voznyy
    Commented Jan 24, 2021 at 1:35
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    $\begingroup$ I am not sure what you are really asking. What it seems to me, you are asking is for rational functions $b=b(a)$ such that, if $a(a+1)=2e^2$ for some $e$, then $b(b+1)=2d^2$ for some $d$ (in terms of $a$ and $e$). It is not explained at all what it is the relation with elliptic curves, and it is probably misleading. About the code, you are only searching solutions of a "Pell equation", easily parametrizable (as explained by @GerryMyerson). $\endgroup$
    – Xarles
    Commented Jan 24, 2021 at 10:47
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    $\begingroup$ From one value of $a$, your five formulas give you five new values of $a$. From one value of $a$, my construction gives you an exhaustive infinity of new values of $a$. $\endgroup$
    – Gerry Myerson
    Commented Jan 24, 2021 at 11:39
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    $\begingroup$ I converted some \fracs into \dfracs, which I think enhances readability. However, you say "Applying (2) and (2) consecutively", which probably is not what you mean. $\endgroup$
    – LSpice
    Commented Jan 24, 2021 at 14:50

2 Answers 2

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I found $$b = 16 \frac{a^2+a}{8a+9}$$

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  • $\begingroup$ Thank you so much! This is exactly an example of a new formula I was looking for! The formula is true for any $a$ and any $r$, and is simple ($s=3$). Could there be more of them with $s\le6$? Did you use brute-force or any other approach? In what math package? $\endgroup$
    – Maksym Voznyy
    Commented Jan 24, 2021 at 16:04
  • $\begingroup$ Applying the new formula (6) and (1) consecutively: (7) $b=\dfrac{(4a+3)^2}{48a^2+40a-9}$, $s=4$. Applying (1) and (6) consecutively: (8) $b=\dfrac{64a(a+1)}{(3a-1)(35a-1)}$, $s=4$. $\endgroup$
    – Maksym Voznyy
    Commented Jan 24, 2021 at 16:41
  • $\begingroup$ Six more formulas with $s=4$ derived from (6) due to Nulhomologous, all true for any $r$: (9) $b=\dfrac{(4a+3)^2}{240a^2+232a-9}$, (10) $b=\dfrac{(23a-1)^2}{495a^2+1070a-1}$, (11) $b=\dfrac{(5a+6)^2}{231a^2+196a-36}$, (12) $b=\dfrac{256a(a+1)}{273a^2-302a+1}$, (13) $b=\dfrac{(a+3)^2}{255a^2+250a-9}$, (14) $b=\dfrac{64a(a+1)}{465a^2-110a+1}$. $\endgroup$
    – Maksym Voznyy
    Commented Jan 24, 2021 at 18:42
  • $\begingroup$ The formulas with $s=4$ that could have been included in my original question: (15) $b=\dfrac{4a(a+1)}{21a^2+6a+1}$, (16) $b=\dfrac{(a-1)^2}{15a^2+18a-1}$, (17) $b=\dfrac{a(a+1)}{24a^2+9a+1}$. And a new one with $s=2$: (18) $b=-(2a+1)^2$, this seed will produce many more! This discussion opened a new universe for me, thank you so much, everybody! Promise to update here if/when the new high rank $\mathbb{Z}/6\mathbb{Z}$ curves are produced by any of these formulas. Back to searching for rank 8, rank 9 and above! $\endgroup$
    – Maksym Voznyy
    Commented Jan 25, 2021 at 3:33
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  1. The following is a much more productive SageMath/Python code due to Gerry Myerson's idea to convert to a hyperbola (when $r > 0$) or an ellipse (when $r < 0$).
from time import time
t0 = time()
r = 2
listA = []
top = 100
for n in range(1, top + 1):
    for d in range(1, top + 1):
        if gcd(n,d) == 1:
            t = QQ(n) / QQ(d)
            a = t^2 / (r - t^2)
            e = sqrt(a * (a + 1) / r)
            if (e in QQ) and a != -1 and a != 0:
                listA.append(a)
print(listA)
print(len(listA))
print(time() - t0)

[<6087 values of a, instead of original 66>]
6087
0.41802382469177246
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