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a) use the lines through the point (1 1) to describe all the points on the circle $x^2 + y^2 = 2$ whose coordiates are rational numbers.

b) what goes wrong if you try to apply the same procedure to find all the points on the circle $x^2 + y^2 = 3$

Now I figured out a solution to this, but it was not very elegant and based upon what I'm asked in the second part of the question I feel Imam missing something.

With that said my solution:

using $x^2 + y^2 = 2$ I also have the equation of the line at (1,1) : $y = m(x-1) + 1$ , where $m$ is a rational number. So to describe all the points on the circle whose coordinates are rational, I shoul analyze the point where th line and circle intersect. Doing that I end up having to use the quadratic formula and obtain:

$$x = \frac{-(2m-2m^2) \pm [(2m-m^2)^2 - 4(1+m^2)(m^2-2m-1)]^{\frac{1}{2}}}{2(1+m^2)}$$ and then to solve for $y$, i would plug this back into the equation of the line.

As you can see not easy to draw any conclusions for the second portion of the question. Is this what has to be done, or is there something else I am missing?

Adola
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D.C. the III
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  • With $x^2+y^2=2$ you started with a rational point $(1,1)$ on the circle, and then considered lines with rational slopes through this point. For $x^2+y^2=3$, you will not be able to find a rational point to start with. The proof that there is no rational point is not hard, but involves some number theory. The argument for $x^2+y^2=2$ can be made nicer than yours, we do not need the discriminant. – André Nicolas Jan 18 '16 at 16:18
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    There's a $2$ missing in your formula for $x$: where you wrote $(2m-m^2)^2$ should be $(2m-2m^2)^2$, and then the whole thing inside the square root simplifies to $4(m+1)^2$, so $x$ is either $1$ (the point you started from) or $(2m^2-4m-2)/(2m^2+2)$, and in the latter case $y$ is $(-2m^2-4m+2)/(2m^2+2)$. – Gro-Tsen Jan 18 '16 at 16:25
  • Also, you should remember when solving a quadratic equation that if you already know one solution (here $1$), you easily get the other without extracting square roots by using the fact that the sum of the two roots of $ax^2+bx+c$ is $-b/a$. This is the reason why we get rational points here. – Gro-Tsen Jan 18 '16 at 16:28
  • For the case when the coefficient is the sum of squares. http://www.artofproblemsolving.com/community/c3046h1051335_pythagorean_triple Knowing the first solution or the General formula to get there, you see. http://math.stackexchange.com/questions/1513733/solving-a-diophantine-equation-of-the-form-x2-ay2-byz-cz2-with-the-co/1514030#1514030 http://math.stackexchange.com/questions/985733/can-legendres-theorem-really-help-solve-this-equation-ax2by2-cz2/988813#988813 – individ Jan 18 '16 at 17:20

1 Answers1

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The general procedure you used for $x^2+y^2=2$ works for $x^2+y^2=c$, where $c\gt 0$ is rational, if you can find one rational point on the circle $x^2+y^2=c$. With $c=2$, you used the rational point $(1,1)$. This procedure will not work for $x^2+y^2=3$, for there are no rational points on this circle.

Suppose to the contrary there is a rational point. We can assume it has shape $(a/q,b/q)$ where $a$, $b$, and $q$ are integers, with $q\ne 0$.

Clearing denominators, we get the equation $a^2+b^2=3q^2$.

We show that the equation $s^2+t^2=3w^2$ has no non-zero integer solutions. If it does, there is a solution with $w$ minimal. It is easy to verify that $s^2+t^2$ cannot be divisible by $3$ unless both $s$ and $t$ are divisible by $3$. So let $s=3s_1$, $t=3t_1$. Then we get $9s_1^2+9t_1^2=3w^2$. That forces $w$ to be divisible by $3$, say $w=3w_1$.

Then $s_1^2+t_1^2=3w_1^2$, contradicting the minimality of $w$.

André Nicolas
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  • I am only a week in to my number theory course, we are using Silverman's book. Should I already know about this minimality concept? The only thing with regards to unit circles we have done is primitive pythagorean triples. – D.C. the III Jan 18 '16 at 17:21
  • Nice book. I proved there was no rational solution to start with, using little number theory. Do you already have congruences? – André Nicolas Jan 18 '16 at 17:28
  • just got to the fundamental theorem of arithmetic. probably by next lecture. But I am familiar with them – D.C. the III Jan 18 '16 at 17:33
  • We could reword as follows. Let $3^k$ be the highest power of $3$ that divides both $s$ and $t$. Let $s=3^ks_1$, $t=3^kt_1$. Then $3^{2k}(s_1^2+t_1^2)=3w^2$. So $w=3^kw_1$ for some $w_1$. Substituting we get $s_1^2+t_1^2=3w_1^2$. Now at least one of $s_1$ and $t_1$ is not divisible by $3$. But this is impossible, since the square of something not divisible by $3$ is congruent to $1$ modulo $3$. I consider this basically the same proof as the one given in the answer, just messier. – André Nicolas Jan 18 '16 at 17:47