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Sketch a proof: If the five points of a conic are rational, then it contains infinitely many rational points.

In class, we learned about Pascal's Theorem for a hexagon inscribed in a conic. The hexagon may be inscribed in a conic iff the three intersections of opposite sides are co-linear.

And I know that there are either no rational points or infinitely many rational points on a rational conic and also that the intersection of two rational lines is rational ... but I am unsure if I can use those in the proof.

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Consider irreducible conics defined over $\mathbb Q$, i.e. curves of degree two with rational coefficients. Such an irreducible conic in the projective plane over $\mathbb Q$, the field of algebraic numbers, is defined by an irreducible homo- geneous polynomial $G \in \mathbb Q[x,y,z]$ of degree two as the set {$(\bar x : \bar y : \bar z) ∈ \mathbb P^2(\mathbb Q) | G(\bar x, \bar y, \bar z) = 0$}. In the sequel we refer to $$G(x,y,z)=ax^2 +bxy+cy^2 +dxz+eyz+fz^2 =0,...(1)$$ or $$g(x,y)=G(x,y,1)=ax^2 +bxy+cy^2 +dx+ey+f =0,...(2)$$ as the General Conic Equation. $(1)$ defines the projective and the corresponding affine conic, respectively. We denote the projective conic by $C^*$ and the affine conic by $C$.
Definition: We call $P = (\bar x :\bar y : \bar z) \in C^*$ a rational point on $C^*$ iff $P \in \mathbb P^2(\mathbb Q)$. Analogously for the corresponding affine curve.

Theorem: An irreducible conic defined over $\mathbb Q$ has no or infinitely many rational points.

Proof: We give only a sketch of the proof. Suppose there is a rational point P on the conic. Then we intersect the conic with a line through this rational point having a rational direction vector. We will usually get two intersection points – the original rational point P and an additional rational point. Varying the slope of the line leads to infinitely many other rational points on the conic.
Hence, the theorem shows that the existence of one rational point on an irreducible conic implies that there are infinitely many rational points on it.

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  • $\begingroup$ The OP seems to know this theorem. remains to show that every conic with 5 rational points is defined over $\Bbb Q$. $\endgroup$ – Hagen von Eitzen Dec 6 '16 at 7:10
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Given five rational points $A,B,C,D,E$, we can find a sixth rational point as follows:

  1. Let $G$ be the intersction of $AB$ and $DE$.
  2. Let $H$ be a rational point on $CD$.
  3. The line $AH$ intersects the conic in another point $F$
  4. Let $K$ be the intersection of $GH$ and $EF$.

Now $G=AB\cap DE$, $H=CD\cap FA$, $K=EF\cap BC$ are on the Pascal line of the hexagon $ABCDEF$. As $A,B,D,E$ are rational, so is $G=AB\cap DE$. As $A,G,B,C$ are rational, $K=AG\cap BC$ is rational. As $A,H,E,K$ are rational, $F=AH\cap EK$ is rational. So in order to show that there are infinitely many rational points $F$, it suffices to show that the above construction is possible in infinitely many ways.

Step 1 is clear. Step 2 is possible in infinitely many ways because $CD$ is a rational line and contains infinitely many rational points. For step 3, first note that $A$ is not on $BC$, hence the infinitely many lines $AH$ for different choices of $H$ are pairwise different.Also, there are at most two lines through $A$ that do not intersect the conic again. Hence step 3 still works for infinitely many $H$. Step 4 is "deterministic" again; note that in case of parallel lines, $K=\infty$ is still counted as rational.

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