Kindly asking for any hints about the following question:

Let $E$ be an elliptic curve over $F_p$ where $p>7$ is a prime. Suppose $E(F_p)$ had a point of order $p$, then if #$E(F_p)=p$?


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    $\begingroup$ Hint: Lagrange's theorem and Hasse's theorem. $\endgroup$ – yyyyyyy Jan 3 '17 at 20:38
  • $\begingroup$ @yyyyyyy: With Lagrange's theorem we know that #$E(Fp)=kp$, for some $k \in \mathbb{‎N}$‎. By Hasse's theorem we know that $kp \in [p+1−2‎\sqrt{p}‎,p+1+2‎\sqrt{p}‎]$. But why $k=1$? $\endgroup$ – Masoud Jan 4 '17 at 16:16

By Lagrange $\#E(\Bbb F_p)=kp$ for some integer $k$. By Hasse's theorem $$kp<p+1+2\sqrt{p}<2p\quad (\text{since }1+2\sqrt{p}<p\text{ for }p>7)$$ and $$kp>p+1-2\sqrt{p}>0\quad (\text{since }1-2\sqrt{p}>-p\text{ for }p>7).$$ Hence $kp$ is a multiple of $p$ such which is strictly larger than 0, but strictly smaller than $2p$. The only remaining option is $p$.

  • $\begingroup$ It is not necessary for $1+2\sqrt{p}<p$ that $p>7$. I am confounded. $\endgroup$ – Masoud Jan 5 '17 at 17:22
  • $\begingroup$ Could you explain why it is not? $\endgroup$ – CurveEnthusiast Jan 5 '17 at 17:31
  • $\begingroup$ excuse me, your answer is correct!!!! $\endgroup$ – Masoud Jan 5 '17 at 18:56

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