In my cryptography course I found the following problem:

Find $|E(\mathbb{F}_{7^{100}})|$ where $E$ is given by $y^2=x^3+1$.

I know how to do it for small numbers, using quadratic residues, but this doesn't work with $7^{100}$. My question is if there is a general method or algorithm that works in general and can be done by hand (no computer) or if there is a clever solution in this particular case.

Many thanks.

  • 2
    $\begingroup$ Did you see thngs like $\# E(\Bbb{F}_{7^n})=N(\phi^{7^n}-1) = (\phi^{7^n}-1) ^*(\phi^{7^n}-1)$ already? (Frobenius endomorphism and dual endomorphism) This is detailed in Silverman's AEC. $\endgroup$
    – reuns
    Dec 4 '21 at 18:49
  • $\begingroup$ @reuns thanks, do you know if we can solve it without strong results? If possible I would like to solve it directlly with elementary methods $\endgroup$
    – Marcos
    Dec 4 '21 at 20:00
  • $\begingroup$ I don't think there is any alternative method. $\endgroup$
    – reuns
    Dec 4 '21 at 20:33
  • $\begingroup$ Related. I have gotten used to doing it that way because it generalizes to higher genus curves. Knowing the answer for the fields $\Bbb{F}_{p^\ell}$, $\ell=1,2,\ldots,g$, gives you everything. In the case of elliptic curves $\ell=1$ suffices. It is equivalent to using the zeta function as in reuns's comment and WhatsUp's answer. $\endgroup$ Dec 5 '21 at 5:20
  • $\begingroup$ Anyway, the theory of zeta functions gives the recipe $$\#E(\Bbb{F}_{p^n})=p^n+1-\omega_1^n-\omega_2^n$$ for a pair of complex conjugate numbers $\omega_1,\omega_2$ satisfying $|\omega_{1,2}|=\sqrt p$. Counting also the point at infinity we have $\#E(\Bbb{F}_7)=12$, so $\omega_1\omega_2=7$ and $\omega_1+\omega_2=-4$. This yields $\omega_{1,2}=-2\pm i\sqrt{3}$. And gives the answer $$\#E(\Bbb{F}_{7^{100}})=7^{100}+1-\omega_1^{100}-\omega_2^{100}=3234476509624757991344647769100216810857205479048020383989725994314227 883581889277376.$$ $\endgroup$ Dec 5 '21 at 5:47

The zeta function for $E/\Bbb F_p$ is, by definition, the formal power series $$Z(T) = \exp\left(\sum_{r = 1}^\infty \frac{|E(\Bbb F_{p^r})|}rT^r\right).$$ It turns out that $Z(T)$ is a rational function: there exists $a \in \Bbb Z$ such that $$Z(T) = \frac{1 - aT + pT^2}{(1 - T)(1 - pT)}.$$ This result appears e.g. in the book of Silverman, Arithmetic of Elliptic Curves, Chapter V, Theorem 2.4 (Page 136).

By calculating $|E(\Bbb F_p)|$, you can determine the value of $a$, which then gives you all the values of $|E(\Bbb F_{p^r})|$.

  • $\begingroup$ One question, is It posible to solve this particular exercise without this general result? Because if It is possible I want to avoid to study all this theory for just one exercise $\endgroup$
    – Marcos
    Dec 4 '21 at 19:55
  • 2
    $\begingroup$ I can't see an easier way to do it. I think the number $7^{100}$ is chosen in purpose. For some other choices, there are easier methods. E.g. if it were $p^r$ with $p^r \not\equiv 1\pmod 3$, then we could use the fact that $x \mapsto x^3$ is a bijection from $\Bbb F_{p^r}$ to itself. But this doesn't work for $7^{100}$. $\endgroup$
    – WhatsUp
    Dec 4 '21 at 20:05
  • $\begingroup$ Ok, thanks for your help :) $\endgroup$
    – Marcos
    Dec 4 '21 at 20:09

Let $\,a_n\,$ be the number of solutions to $\, y^2 \equiv x^3 + 1 \,$ in $\,\mathbb{F}_{7^n}\,$ and also the point at infinity.

A search for $\,n=1\,$ yields the $11$ solutions for $\,(x,y)\,$

$$ (0,\pm1),\, (1,\pm3),\, (2,\pm3),\, (3,0),\, (4,\pm3),\, (5,0),\, (6,0) $$

in $\,\mathbb{F}_{7}\,$ and thus $\,a_1=12.\,$

The general result is

$$ \sum_{n=1}^\infty \frac{a_n}n T^n = \log{Z(T)}\quad \text{ where } \quad Z(T) = \frac{1 + t_pT + pT^2}{(1 - T)(1 - pT)} $$

as mentioned in another answer. Here $\,t_7=4\,$ and the power series is

$$ Z(T) = 1 + 12T + 96T^2 + 684T^3 + 4800T^4 + 33612T^5 + \cdots. $$

The generating function for $\,a_n\,$ is

$$ A(T) := \sum_{n=1}^\infty a_nT^n = 1 + 12T + 48T^2 + 324T^3 + 2496T^4 + \cdots. $$

Let $\,b_n := a_n -1\,$ be the number of solutions not including the point at infinity. Then

$$ B(T) := \sum_{n=1}^\infty b_nT^n = 11T + 47T^2 + 323T^3 + 2495T^4 + \cdots.$$

Note that $\,b_n\,$ satisfies a linear recursion and has a rational generating function which is

$$ B(T) = \frac{T (11 + 14 T - 49 T^2)}{(1 - 7 T) (1 + 4 T + 7 T^2)}. $$

The general results are that

$$ B(T) = \frac{T ((p+t_p) + 2pT -p^2T^2)}{(1 - pT)(1 + t_pT +pT^2)} $$


$$ b_n = p^n - (\alpha^n + \beta^n) \quad \text{ where } \quad \alpha\beta = p \text{ and } \alpha+\beta = -t_p. $$

For $\,p=7\,$ the conjugate root constants are $\,\alpha = -2+i\sqrt{3},\; \beta = -2-i\sqrt{3}.$

  • $\begingroup$ That was exactly what I was looking for, many thanks $\endgroup$
    – Marcos
    Dec 4 '21 at 22:46
  • 4
    $\begingroup$ I haven't gone through the argument, but this answer cannot be correct. In the book of Silverman, Arithmetic of Elliptic Curves, Chapter V, Theorem 1.1 (Page 131), it is proved that $|\#E(\Bbb F_q) - q - 1| \leq 2\sqrt q$ (this is in accordance with Weil's conjecture, now a theorem of Deligne). Thus for large $q$, the number of points of $E(\Bbb F_q)$ should be close to $q$. It is impossible that $a_n = 2(7^n - 1)$ for all $n > 0$. $\endgroup$
    – WhatsUp
    Dec 4 '21 at 23:35
  • $\begingroup$ @WhatsUp I agree with you, but where is the error? $\endgroup$
    – Somos
    Dec 4 '21 at 23:41
  • 2
    $\begingroup$ Your formula produces $a_2 = 2(7^2 - 1) = 98$. The value of $a_2$ should be $48$ (including infinity point). $\endgroup$
    – WhatsUp
    Dec 4 '21 at 23:53
  • 1
    $\begingroup$ @WhatsUp Of course, Again, Thanks for your correction. $\endgroup$
    – Somos
    Dec 4 '21 at 23:54

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