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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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    **Hint:** Lagrange's theorem and Hasse's theorem. – 2017-01-03
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    @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 $k=1$? – 2017-01-04

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By Lagrange $\#E(\Bbb F_p)=kp$ for some integer $k$. By Hasse's theorem $$kp7)$$ 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 $p$.

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    It is not necessary for $1+2\sqrt{p}7$. I am confounded. – 2017-01-05
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    Could you explain why it is not? – 2017-01-05
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    excuse me, your answer is correct!!!! – 2017-01-05