To: rusin@math.niu.edu (Dave Rusin)
From: [Permission pending]
Date:         9 Feb 95 13:35:41 MET
Subject:      Re: pythagorean triples

> I haven't been following the thread too closely, but it's easy to see
> what is happening. Given a point on a quadratic curve, you can translate
> it to the origin and assume the point is  (0,0), the curve is
> 0=(ax+by)+(cx^2+dxy+ey^2), and a line through it is y=mx. The x coordinates
> of points of intersection between the curve and the line satisfy
> 0=x[(a+bm)+x(c+dm+em^2)].

Yes. That's smart. Thank you.

> So "how do you know there is another point" is an unclear question.

My concern was whether (there is a simple proof that) 'the'
intersection-point is *rational*, given that one already
has good grasp on the geometry over IR.
Thank you very much for your post,

[sig deleted -- djr]