How to prove that the curve $x^2+y^2-3=0$ has no rational points? 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.
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.
$ 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 $1&lt;|x|&lt;\sqrt 3, y\notin\Bbb{Q}$. Again for $x=\pm1, y=\sqrt 2\notin \Bbb{Q}$. Also $x$ cannot be $\pm\sqrt 3$.  Hence proved.