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Please help or hints me to solve this question:

Suppose that $p>2$ be a prime and $q=p^e$, for some integer $e$. Suppose $E: y^2= x^3+ax+b$ be an elliptic curve over $F_q$, ($a,b \in \Bbb F_q$). Let $N_q$ be the number of $\Bbb F_q$-rational points on $E$.

Show that $|N_q - q|\leq‎ \dfrac{q+3}{2}‎$.

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  • $\begingroup$ Do you know that $|N_q - (q+1)| \leq 2\sqrt q$ ? $\endgroup$ – Watson Feb 9 '17 at 12:23
  • $\begingroup$ @Watson, Yes, hasse theorem. $\endgroup$ – Masoud Feb 9 '17 at 12:28
  • $\begingroup$ It seems to be true if $q>13$ : $$|N_q - q| = |N_q - (q + 1) + 1| \leq 2\sqrt q + 1 \leq \dfrac{q+3}2$$ since $$ 0 \leq q^2-14q+1 \iff 16q \leq q^2+2q+1 \iff 4\sqrt q \leq q+1 \iff 4\sqrt q + 2 \leq q+3 $$ and $0 \leq q^2-14q+1$ if $q>13$ is an integer (so $q \geq 14$). $\endgroup$ – Watson Feb 9 '17 at 12:32
  • $\begingroup$ @Watson, If we consider $f(x) \in \Bbb F_q[x]$, then we know that every solutions of equation $ 1‎\pm‎ f^\frac{q-1}{2} =0$ in $\Bbb F_q$ are a multiple root of $ R(x) = 2f(x)(1 \pm f^\frac{q-1}{2})+ f^\prime (x) (x^q-x)$. With this we can conclusion ? $\endgroup$ – Masoud Feb 9 '17 at 12:50

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