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Two different circles can have an integer point in common (for example, $P=(1,1)$ belongs to both $x^2+y^2-2=0$ and $x^2+y^2-4(x+y)+6=0$) but any pair of distinct elliptic curves on the class defined over $\mathbb {Q}$ by $X^3+Y^3=A$ where $A$ is a cube-free integer, excepting for the trivial $0=(1,-1)$, can not. One wonders on a family of elliptic curves defined over $\mathbb {Q}$ in which there are couples of them that -as with the circles- may have an integer point in common. How to express this family as with the class $X^3+Y^3=A$, I mean analitically?

Piquito
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1 Answers1

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I'm not quite sure what you mean. Given two curves $f(X,Y) = 0$ and $g(X,Y) = 0$ where $f$ and $g$ are polynomials, let $r_X(Y)$ and $r_Y(X)$ be the resultants of $f(X,Y)$ and $g(X,Y)$ with respect to $X$ and $Y$ respectively. A point $(x,y)$ common to both curves must satisfy $r_X(y) = r_Y(x) = 0$. In particular, if you want a common integer solution, a necessary condition is that $r_X$ and $r_Y$ must both have integer roots.

EDIT (incorporating last comment): If you want two continuous parametric families of curves that always intersect in at least one integer point, then by continuity that point is constant. Choose the intersection point, and construct the curves to pass through that point. Thus the curves $Y^2+X^3+aX+b=0$ and $X^2+Y^3+cY+d=0$ intersect at $(1,1)$ if $a+b+2=c+d+2=0$. In almost all cases, these are elliptic.

Robert Israel
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  • Right but the resultants roots ¨live¨in general out of $\mathbb {Q}$. What I want to is write like $X^3+Y^3=A$ but for other class of elliptic curves defined over $\mathbb {Q}$ and admiting an integer point in common. Thank for your comment. Regards. – Piquito Jul 06 '15 at 22:19
  • Well, that's just the point: in general the intersections will involve non-integers, only in special cases will you have integer roots in common. – Robert Israel Jul 07 '15 at 00:36
  • The point is to build two curves which are not currents equations, like the given circles, but provided with a group law that makes it very special (elliptic curves) and from this deduce a family like the given one where figure a parameter $A$ (Selmer Curves). For other family the most probable is with more than one parameter. – Piquito Jul 07 '15 at 01:46
  • If you want two continuous parametric families of curves that always intersect in at least one integer point, then by continuity that point is constant. Choose the intersection point, and construct the curves to pass through that point. Thus the elliptic curves $Y^2 + X^3 + a X + b=0$ and $X^2 + Y^3 + c Y + d=0$ intersect at $(1,1)$ if $a+b+2=c+d+2=0$. – Robert Israel Jul 07 '15 at 06:36
  • Let me suggest change (X,Y) by (Z,W) or (U,V) in your second equation. – Piquito Jul 07 '15 at 19:15