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Consider the elliptic curve $$E: y^2= x^3 + x$$ over the finite field $\mathbb{F}_p$ with $p \geq 3$. I want to show that $|E(\mathbb{F}_p)| \equiv 0 \mod 4$.

I know that, if $p \equiv 3\mod 4$, then $|E(\mathbb{F}_p)| = p +1$. If needed, I can give the proof. All things aside, I've found the same question here :

$\forall p\geq 3, E:y^2=x^3+x$ satisfies $\#E(\mathbb{F}_p)=0\mod4$

But I don't have the book he's referring to, so I can't work it out (and believe me, I've tried!). In short: what do you do in the case that $p \equiv 1 \mod 4$? Am I missing something obvious here? Thanks for the help!

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  • $\begingroup$ my guess would be that there is an easy to find subgroup of order $4$ in E(Fp) $\endgroup$ – mercio May 25 '15 at 18:31
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In the case $p = 4 n + 1$ the number $-1$ is a square $\mod p$. So we can partition the points of $E(\mathbb{F}_p)$ as follows:

  1. One point $\infty$ at infinity.
  2. One point with $x = 0, y= 0$.
  3. Two points $(\epsilon_1, 0)$, $(\epsilon_2, 0)$ with an $x$ that maps to $0$ under $x^3 + x$. (It is $\epsilon_i^2 + 1 = 0$).
  4. There remain $4 n - 2$ different $x$ values. Let $m'$ be the number of those $x \neq 0, \epsilon_1,\epsilon_2$ for which $x^3 + x$ is a square in $\mathbb{F}_p$. Then $m'$ is an even number as $x \mapsto -x$ maps $x^3+x$ to $-(x^3+x)$ which is different from $x^3 + x$ for $x \neq 0,\epsilon_1,\epsilon_2$ and is in the square possessing class exactly when $x^3 + x$ is ($-1$ is a square). So the remaining points $(x,y)$ are $2 m'$ many, which is $0 \mod 4$ as $m'$ is even.

Adding 1.,2.,3.,4. together we get a number divisible by $4$.

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  • $\begingroup$ You, kind sir, are a a gentleman and a scholar! Thank you! $\endgroup$ – Riley May 25 '15 at 18:46
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Note that $x^3+x=x(x^2+1)$, so the 2-torsion of this elliptic curve will be defined over any field with a $\sqrt{-1}$. $\mathbb{F}_p$ for $p \equiv 1\ (\text{mod}\ 4)$ is such a field, and as the 2-torsion is a subgroup of order 4 for $E(\mathbb{F}_p)$, the claim follows.

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