Show that $x^2 + y^2 = 3$ has no rational points Are there rational numbers such that $x^2 + y^2 = 3$ ? 

If I want to find a rational paramterizatio of $x^2 + y^2 = 1$  could start with the point $(1,0)$ and find lines $\ell$ of slope $m \in \mathbb{Q}$ and the intersection points $[\ell] \cdot [circle] = 2 [pt] $.
However, if I use the circle $x^2 + y^2 = 2$ there's no rational point on the axes.  Instead we should use $(x,y) = (1,1)$.  
In the case of $x^2 + y^2 = 3$ there's no obvious rational point that comes to mind.  I'm concerned there might be no rational point at all.  In integers we'd have $a^2 + b^2 = 3c^2$ with $a,b,c \in \mathbb{Z}$.  We'd have $c \equiv 0 \pmod 4$.  Then $a \equiv b \equiv 0 \pmod 4$.  This could lead to an infinite descent argument.  

As a bonus could there exist a small rationqal $\epsilon > 0$ with $\epsilon \ll 1$ and $\epsilon \in \mathbb{Q}$ such that $x^2 + y^2 = 3 + \epsilon$ has a solution (and therefore infinitely many solutions)?
 A: Hint: Show that if $a^2+b^2=3c^2$ for some integers $a,b,c$ with $c\neq 0.,$ then it has a solution with neither $a$ nor $b$ divisible by $3$. 
Then show that means there must be an integer $n$ such that $n^2+1$ is divisible by $3$. 
Is that possible?
A: Any integer square is of form $3k$ or $3k+1$ for some integer $k$. Then $3|a^2+b^2$ if both $a$ and $b$ are divisible by $3$ which turns out $a=3a_1$ and $b=3b_1$ and by substitution $$9a_1^2+9b_1^2=3c^2\\3a_1^2+3b_1^2=c^2$$which leads to $c=3c_1$ and we have $$a_1^2+b_1^2=3c_1^2$$ and this procedure continues infinitely many times, then such integers don't exist
A: Your concerns are well justified. 
We will prove that there cannot be any non-trivial (i.e. a non zero) integer solution to the diophantine equation $ x^2 +y^2 = pz^2$ where $ p \equiv 3$ (mod $4$). As a corollary, there cannot be any rational points that satisfy the equation $ x^2 +y^2 = 3z^2$ (why?). 
Let us assume we CAN find a non-trivial solution to $ x^2 +y^2 = pz^2$. We will derive a contradiction.
Without loss of generality, we may assume that $x,y $ and $z$ have no common divisors i.e. gcd$(x,y,z) = 1$. Otherwise, we may always divide through by the common divisor to obtain another non-trivial solution. 
Next, note that a square of an integer is either congruent to $0$ or $1$ (mod$4$). This may be seen by squaring the expressions $2k$ and $2k+1$ respectively. 
Hence, 
$ x^2 \equiv 0,1 $ (mod$4$),
$ y^2 \equiv 0,1 $ (mod$4$), 
$ pz^2 \equiv 0,3 $ (mod$4$). 
(listing all posible options).
Thus, 
$x^2 +y^2 \equiv 0,1,2 $ (mod$4$).
Therefore, as equality implies equality in congruence, it follows the only possible way for $x^2 +y^2 \equiv pz^2$ is for $x, y$ and $z$ to be even. 
This contradicts the fact that we assumed gcd$(x,y,z) = 1$. 
