# Elliptic curve has points over finite fields

Let $E$ be an elliptic curve with affine equation $$y^2 = x^3+ax+b$$ where $a,b \in \Bbb Z$. Let $P$ be the set of prime numbers not dividing $ab$.

Can we prove directly that if $p \in P$ is at least $5$, then $E(\Bbb F_p) \neq \{ [0:1:0] \}$ ? This should follow from Hasse's bound $$|E(\Bbb F_p)| \geq p+1-2\sqrt p > 1, \quad\forall p \geq 5$$ But my question is about a much weaker statement, so there is probably an easier proof.

• Yes, you can use the formula with the Legendre symbol for the number of points on the curve. This is much easier. – Dietrich Burde Aug 28 '18 at 9:52
• For $p=3$ we cannot assume that our elliptic curve has short Weierstrass form. – Dietrich Burde Aug 28 '18 at 10:42
• The number of points is $$N=1+\sum_{x\in\Bbb{F}_p}\left(1+\left(\frac{x^3+ax+p}p\right)\right).$$ The first $1$ comes from the point at infinity, and the $1$ in the summand comes from the fact that the number of solutions of $y^2=v$ is equal to $1+\left(\dfrac v p\right)$. Of course, it is simpler to write $$N=p+1+\sum_{x\in\Bbb{F}_p}\left(\frac{x^3+ax+b}p\right).$$ – Jyrki Lahtonen Aug 29 '18 at 14:59
• – Alphonse Oct 23 '18 at 12:47
• @JyrkiLahtonen : thank you for your comment. How can I see that $N>1$ in all cases? – Alphonse Oct 23 '18 at 12:50