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There is a question 6.20 from the Book of Proof by Hammock that reads as follows: We say that a point $P=(x,y) \in R^2$ is rational if both x and y are rational. More precisely, P is rational if $P=(x,y) \in Q^2$. An equation $F(x,y)=0$ is said to have a rational pointif there exists $x_0, y_0 \in Q$ such that $F(x_0,y_0)= 0$. For example, the cruve $x^2+y^2-1=0$ has rational point $(x_0,y_0)=(1,0)$. Show that the curve $x^2+y^2-3=0$ has no rational points.

I can only take a guess at how to start a proof. Doing algebra on both sides of $x^2+y^2-3=0$ yields $y = \sqrt{3-x^2}$. Now, assume y is rational -- can x be rational?

Thus, let y = $a\over b$ for some integers a and b, $b \ne 0$,$gcd(a,b)=1$. Then let y = $c\over d$ for some integers c and d, $d \ne 0$,$gcd(c,d)=1$. Hence, ${a \over b} = \sqrt{3 - {{c^2} \over {d^2}}}$.

I think I may want to prove some sort of contradiction in terms of rational and irrational? This as far as I got in my scratchwork for a proof.

mont2223
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  • If both $x,y$ are odd, then $x^2+y^2$ is even, so this will not give you a solution. Hence, either $x^2$ or $y^2$ is $0$ mod $4$. However, $3$ is not a square mod $4$. – Severin Schraven Mar 18 '21 at 16:38
  • Square roots are going to be a losing cause here — or at best, they'll run you around in circles. Instead, I'd look at what a rational solution of the equation would 'look like' — can you go from a rational solution of $x^2+y^2=3$ to an integral solution of a similar equation? Once you've done that, you can prove that there are no integral solutions of the similar equation, and that's your goal. – Steven Stadnicki Mar 18 '21 at 16:38
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    @SeverinSchraven You refer to $x,y$ being even/odd. So are you assuming they are integers? The question is if there might be rational solutions, like maybe $x$ could be $\frac32$ or somthing like that. Which you cannot call "even" or "odd". – 2'5 9'2 Mar 18 '21 at 16:47
  • @alex.jordan That is of course correct. I was thinking about integer solutions (as my parents used to say: People that can read have an advantage in life. Maybe I'll learn it some point). – Severin Schraven Mar 18 '21 at 20:50

2 Answers2

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Suppose $x=\frac{m}{n}$ and $y=\frac{a}{b}$, where both fractions are reduced. Then after a little algebra your equation says:

$$m^2b^2+n^2a^2=3b^2n^2$$

This is an integer equation. Two terms are divisible by $b^2$, so $b^2$ must divide the third term, $n^2a^2$. Since $a$ and $b$ are relatively prime, then actually $b^2$ divides $n^2$.

But similarly, $n^2$ divides $b^2$. So you may conclude $n=b$ and the equation reduces to

$$m^2+a^2=3n^2$$

The residues of squares mod $3$ are just $0$ and $1$, not $2$. So whne you look at that equation mod $3$:

$$m^2+a^2\equiv0$$

the only way it can work out is if $m^2\equiv a^2\equiv0$. So $3$ must divide both $m$ and $a$. But then look again at

$$m^2+a^2=3n^2$$

Now we know $9$ divides the left side. So $3$ must divide $n^2$. This is a contradiction, because now we are saying $3$ divides both $m$ and $n$, but $\frac{m}{n}$ was supposed to be a reduced fraction.

So the assumption that we could have a rational $x$ and $y$ was false.

2'5 9'2
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$ y=\sqrt {3-x^2}$
For $y$ to be rational, we must have $ x^2\le3$ and $y=1\implies |x|\le\sqrt 3$ but there doesn't exist such a rational number $x$ except $|x|=1$ and for any $1<|x|<\sqrt 3, y\notin\Bbb{Q}$. Again for $x=\pm1, y=\sqrt 2\notin \Bbb{Q}$. Also $x$ cannot be $\pm\sqrt 3$. Hence proved.

Manjoy Das
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    Are you confusing "rational" with "integer"? Because for example, $\frac{3}{2}$ is rational, and is less than $\sqrt{3}$. – 2'5 9'2 Mar 18 '21 at 16:46
  • @alex.jordan it would be highly covetable to allow some time between posting an answer and commenting. – Manjoy Das Mar 18 '21 at 16:57