Show that the curve $x^2+y^2-3=0$ has no rational points 
Show that the curve $x^2+y^2-3=0$ has no rational points, that is, no points $(x,y)$ with $x,y\in \mathbb{Q}$.

Update: Thanks for all the input! I've done my best to incorporate your suggestions and write up the proof. My explanation of why $\gcd(a,b,q)=1$ is a bit verbose, but I couldn't figure out how to put it more concisely with clear notation.
Proof: Suppose for the sake of contradiction that there exists a point $P=(x,y)$, such that $x^2+y^2-3=0$, with $x,y\in\mathbb{Q}$. Then we can express $x$ and $y$ as irreducible fractions and write $(\frac{n_x}{d_x})^2+(\frac{n_y}{d_y})^2-3=0$, with $n_x, d_x, n_y, d_y\in\mathbb{Z}$, and $\gcd(n_x,d_x)=\gcd(n_y,d_y)=1$.
Let $q$ equal the lowest common multiple of $d_x$ and $d_y$. So $q=d_xc_x$ and $q=d_yc_y$ for the mutually prime integers $c_x$ and $c_y$ (if they weren't mutually prime, then $q$ wouldn't be the lowest common multiple). If we set $a=n_xc_x$ and $b=n_yc_y$, we can write the original equation as $(a/q)^2+(b/q^2)-3=0$, and equivalently, $a^2+b^2=3q^2$.
In order to determine the greatest common divisor shared by $a$, $b$, and $q$, we first consider the prime factors of $a$. Since $a=n_xc_x$, we can group them into the factors of $n_x$ and those of $c_x$. Similarly, $b$'s prime factors can be separated into those of $n_y$ and those of $c_y$. We know that $c_x$ and $c_y$ don't share any factors, as they're mutually prime, so any shared factor of $a$ and $b$ must be a factor of $n_x$ and $n_y$.
Furthermore, $q=d_xc_x=d_yc_y$, so it's prime factors can either be grouped  into those of $d_x$ and those of $c_x$, or those of $d_y$ and those of $c_y$. As we've already eliminated $c_x$ and $c_y$ as sources of shared factors, we know that any shared factor of $a$, $b$, and $q$ must be a factor of $n_x$, $n_y$, and either $d_x$ or $d_y$. But since $n_x/d_x$ is an irreducible fraction, $n_x$ and $d_x$ share no prime factors. Similarly, $n_y$ and $d_y$ share no prime factors. Thus $a$, $b$, and $q$ share no prime factors, and their greatest common divisor must be $1$.
Now consider an integer $m$ such that $3\nmid m$. Then, either $m\equiv 1\pmod{3}$, or $m\equiv 2\pmod{3}$. If $m\equiv 1\pmod{3}$, then $m=3k+1$ for some integer $k$, and $m^2=9k^2+6k+1=3(3k^2+2k)+1\equiv 1\pmod{3}$. Similarly, if $m\equiv 2\pmod{3}$, then $m^2=3(3k^2+4k+1)+1\equiv 1\pmod{3}$. Since that exhausts all cases, we see that $3\nmid m \implies m^2\equiv 1\pmod{3}$ for $m\in\mathbb{Z}$.
Notice that $a^2+b^2=3q^2$ implies that $3\mid (a^2+b^2)$. If $3$ doesn't divide both of $a$ and $b$, then $(a^2+b^2)$ will be either $1\pmod{3}$ or $2\pmod{3}$, and thus not divisible by $3$. So we can deduce that both $a$ and $b$ must be divisible by $3$.
We can therefore write $a=3u$ $\land$ $b=3v$ for some integers $u$ and $v$. Thus, $9u^2+9v^2=3q^2$, and equivalently, $3(u^2+v^2)=q^2$. So $3$ divides $q^2$, and must therefore divide $q$ as well. Thus, $3$ is a factor of $a,b,$ and $q$, but this contradicts the fact that $\gcd(a,b,q)=1$, and falsifies our supposition that such a point $P=(x,y)$ exists.
 A: Suppose to the contrary that there is a rational solution of the equation. Then there exist integers $a$, $b$, and $q$, with $q\ne 0$, such that $a^2+b^2=3q^2$, and $a$, $b$, and $q$ have no common factor greater than $1$.
Note that $a$ and $b$ must both be divisible by $3$. For if an integer $m$ is not divisible by $3$, then $m^2$ has remainder $1$ on division by $3$. So if one or both of $a$ and $b$ is not divisible by $3$, then $a^2+b^2$ has remainder $1$ or $2$ on division by $3$, and therefore cannot be of the shape $3q^2$.
Thus both $a$ and $b$ are divisible by $3$. It follows that $q$ is divisible by $3$, contradicting our assumption that $a$, $b$, and $q$ have no common divisor greater than $1$. 
A: Suppose $a^2 + b^2 = 3 c^2$
Write $a = 3^p u$, $b = 3^q v$, and $c = 3^r w$,
where $u, v, w$ are all relatively prime to 3.
This does not assume that $a, b, c$ have no common divisor greater than 1.
Assume $p \le q$ (if $p > q$, switch their roles in what follows).
$a^2+b^2 = (3^p u)^2 + (3^q v)^2
= 3^{2p}(u^2+ 3^{2(q-p)}v^2)
$, so an even power of 3 divides $a^2+b^2$.
($u^2+ 3^{2(q-p)}v^2$ has a remainder of 1 or 2 mod 3
depending on if $p < q$ or $p = q$.)
But $3 c^2 = 3 (3^r w)^2
= 3^{2r+1} w^2$,
so an odd power of 3 divides $c^2$.
By unique prime factorization, this is a contradiction.
Note: I wrote this because the assumption
of $a, b, c$ having no common factor is,
to me, either an implicit use of unique factorization
or an infinite descent contradiction based on
the powers of 3 dividing them.
A: Here is a more geometric flavored proof:
$x^2+y^2-3=0 \iff x^2+y^2 = \sqrt{3}^2$ is a circle with radius $\sqrt{3}$ centered at the origin. Think of the points along the circle as polar coordinates $(r, \theta)$, i.e. 
\begin{equation}(\sqrt{3}, \theta) \text{ where } 0 \leq \theta \leq 2\pi \end{equation}
The formulas for converting a polar coordinate into a cartesian coordinate is just right triangle trigonometry:
\begin{align*}
  x &= r\cos\theta \\
  y &= r\sin\theta
\end{align*}
Since $r=\sqrt{3}$, we have that
\begin{align*}
  x &= \sqrt{3}\cos\theta \\
  y &= \sqrt{3}\sin\theta
\end{align*}
Then you can say that $\sqrt{3}$ multiplied by any number between $-1$ and $1$ is irrational. 
A: It is actually sufficient to analyze the curve modulo 3 (or 2, even), where the question becomes whether $-1 \equiv 2$ is a square modulo 3. You can reduce the proof to this step directly by clearing denominators/divisibility arguments, but this curve is related to the quaternion algebra generated by the square roots of -1 and 3 which ramifies at the prime 3. Here, I'm making use of the Albert-Brauer-Noether-Hasse Theorem. 
