# Making an intuitive guess for performing a projection

I was reading Rational Points On Elliptic Curves and after the first chapter, this is a question I found-

Describe all rational points on the circle $$x^2 + y^2 = 2$$ by projecting from the point $$(1, 1)$$ onto an appropriate rational line (Your formulas will be simpler if you are clever in your choice of the line).

Now, I know how to do this projection and find out the rational points. However, I was stuck at

Your formulas will be simpler if you are clever in your choice of the line

My question is, is there any way to understand beforehand without doing any calculations (or doing minimum calculations) which line is the most suitable one to perform this projection?

Also, the question specifies the point $$(1,1)$$. What if it was not specified? Do we have any way to make a guess for the easiest choice of this point?

And, I am not just talking about this circle. Is there any general way to guess the best choice for a general circle $$x^2+y^2=r^2$$?

Please note that I am aware of some other methods that can lead us to the solution of this problem. But, I am specifically asking for the clever choice of the line and the point for a general circle. In other words, is there any kind of a hack to know from beforehand whether one choice of a point and a line is better than the other?

I don't know if this is easier than your solution: The line is $$y-1=t(x-1)$$ with rational slope $$t$$. We can rewrite this as $$(x,y)=(a+1,at+1)$$.

Plugging this into the equation, we get $$a = \frac{-2(t+1)}{t^2+1}$$ for $$a \not = 0$$, which immediately gives us $$(x,y) = \left( \frac{t^2-2t-1}{t^2+1}, \frac{t^2+2t-1}{t^2+1} \right)$$

• I know this method, but the question specifically asks for performing that projection. Also, the question I am asking is how to make the most intuitive guess for the point and the line to make the projection. So, your answer is irrelevant to my question. Jul 15, 2021 at 7:16

$$a^2+b^2=qc^2$$

There are solutions when the coefficient can be represented as the sum of squares. $$q=t^2+k^2$$

$$a=-tp^2+2kps+ts^2$$

$$b=kp^2+2tps-ks^2$$

$$c=p^2+s^2$$

• I know this method, but the question specifically asks for performing that projection. Also, the question I am asking is how to make the most intuitive guess for the point and the line to make the projection. So, your answer is irrelevant to my question. Jul 15, 2021 at 7:17
• Intuition is not needed here. Decompose the coefficient into the sum of squares. This will be the first point on which you will build a tangent. Jul 15, 2021 at 7:32

The problem is about parametrizing rational solutions on a curve. Generally, this is a very difficult problem. Indeed, even finding one rational solution to a given equation can be extraordinarily hard. But in some cases, for instance the case of a circle, this is not so hard. The idea, as the problem walks you through, is to find a rational point on the circle. Then you can draw a line through that point with some rational slope. The line will intersect the circle at another point (except for where the line would be tangent, in which case we say we get the original point), and this point is rational (because you already have a point of intersection which is rational and the slope is rational). Varying this rational slope, we get all the rational points on the circle.

In this case, you are trying to parametrize all the rational solutions on the circle $$x^2 + y^2= 2$$. So we follow the plan above. But notice the first step - finding a rational point at all - is really the "tough" step. So why the point $$(1,1)$$? Simple. This is because $$(x,y)= (1,1)$$ is an "obvious" rational point on the circle. But you could have used any rational point, e.g. $$(-1,1)$$, $$(1,-1)$$, $$(-1,1)$$, $$(-1/5,7/5)$$, etc. Each may yield a different parametrization - but a parametrization nevertheless. Why did they choose $$(1,1)$$? Probably because it seemed the simplest. But again, any rational point will do.

What makes for the "best" choice of rational point? You would first need to make clear what one means by "best." Again, any rational point on the circle will yield a parametrization, so what makes some "better" than another? I am not really sure, other than it would be nice to choose points which make the rational parametrization "nice". But that is in the eye of the beholder and what makes some "nicer" than others really depends, in the end, on your method. You could work out assuming you have a rational point $$(a,b)$$ and seeing what the parametrization would be, i.e. is the parametrization "simpler" for some $$a, b$$ than others? If so, check to see if that pair actually gives a rational point on the circle.

• There are no difficulties for the mentioned equation. The solutions will be when it is possible to decompose the coefficient into the sum of squares. Decomposing it into squares , we will have the first point for parametrization. Jul 15, 2021 at 7:35
• @individ You have misunderstood the question. The question was not "how do I solve this problem." The questions given were "given this solution approach, how does one choose a 'nice' line to project on" and "Also, the question specifies the point (1,1). What if it was not specified? Do we have any way to make a guess for the easiest choice of this point?" and for general circles. Jul 15, 2021 at 7:41
• @individ You answered how to find a rational point in a specific case. But for a general circle, this will not always be possible - indeed, not all circles have any rational points at all. Moreover, for a general polynomial equation $p(x,y)= 0$, you'd be famous indeed if you could find a way of finding a rational point on the curve (assuming there is one). This was the more general question side I addressed. We have different solutions, addressing different question issues. There is nothing wrong with either. Jul 15, 2021 at 7:43
• That's what I'm talking about. For this equation $x^2+y^2=q$ It is necessary to decompose the coefficient into the sum of squares. This will be the first point. Jul 15, 2021 at 7:44
• @individ Well then, try this with $x^2 + y^2= 3$......... Jul 15, 2021 at 7:45