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!

There are only six solutions which are also algebraic integers, and they can each be found from whichever one you discover first.
For instance, $$(2 + \sqrt{-3}) \left( \frac{-1 + \sqrt{-3}}{2} \right) = \frac{-5 + \sqrt{-3}}{2}$$ and $$\frac{-5 + \sqrt{-3}}{2} \left( \frac{-1 + \sqrt{-3}}{2} \right) = \frac{1 - 3 \sqrt{-3}}{2}.$$
There are infinitely many solutions, but for each viable denominator $d$ there are only two or four solutions. For example, from the very strategically withheld solution $$\frac{10}{31} + \frac{47 \sqrt{-3}}{31}$$ (which was inexplicably deleted from this page earlier today) we can obtain $$\omega \left( \frac{10}{31} + \frac{47 \sqrt{-3}}{31} \right) = \frac{151 + 37 \sqrt{-3}}{62}$$ and verify that that number does have a norm of $7$.
