How to find all rational solutions of $\ x^2 + 3y^2 = 7 $? I knew that for $ x^2 + y^2 = 1$ the x and y can be expressed by introducing one more variable where $\ m=y/(x+1) $, then $\ x= 2m/(1+m^2) $ and $\ y= (1-m^2)/(1+m^2) $. What about $\ x^2 + 3y^2 = 7 $, should I divide the equation by 7 in order to get the 1 at the right hand side ? Then how to deal with the $\ 3y^2 $ ? Thank you!

Using the method of pg 7 of this paper on this related equation
$$x^2+3y^2=7z^2 \quad \text{with initial solution} \quad (x,y,z)=(2,1,1)$$
A line $y=t(x-2)+1$, which will cut through the ellipse $x^2 + 3y^2 = 7$ at rational points if $t$ is rational.... when substituted into the ellipse yields:
$$\begin{align}
x^2+3\left[t(x-2)+1\right]^2&=7 \\
x^2+3\left[t^2(x^2-4x+4)+2t(x-2)+1\right]&=7 \\
(1+3t^2)x^2+(-12t^2+6t)x+(12-12t+3-7)&=0 \\
\text{vieta:  the two roots,} \quad x_1, x_2 \quad \text{are such that} 
 \quad -(x_1+x_2)&=\frac{-12t^2+6t}{1+3t^2} \\
\text{since} \quad x_1=2, \quad \text{we have} \quad x_2(t)=\frac{12t^2-6t}{1+3t^2}-2 &=\frac{6t^2-6t-2}{1+3t^2} \\
\text{substituting this} \quad x(t) \quad \text {into the line:} \quad  y&=t\left(\frac{6t^2-6t-2}{1+3t^2}-2\right)+1 \\
y(t)=t\left(\frac{-6t-4}{1+3t^2}\right) + \frac{1+3t^2}{1+3t^2}&=\frac{-3t^2-4t+1}{1+3t^2}
\end{align}$$
Letting $y(t)\to |y(t)|$, Solution set with one parameter: 
$$\begin{cases}
x(t)&=\frac{6t^2-6t-2}{1+3t^2} \\
y(t)&=\frac{3t^2+4t-1}{1+3t^2} 
\end{cases}$$
$t$ was rational, so let $t=\frac{m}{n}$
Solution set now with two parameters:
$$\begin{align}
x(m,n)&=\frac{6m^2-6mn-2n^2}{n^2+3m^2} \\
y(m,n)&=\frac{3m^2+4mn-n^2}{n^2+3m^2} 
\end{align}$$
