This question comes from Rational Points on Elliptic Curves (Silverman & Tate), exercise $1.6$ (b).
I want to calculate all rational solutions for $3x^2+5y^2 =4$. However, I think that there are no rational solutions because if we homogenize we get $3X^2+5Y^2 =4Z^2$ and mod $3$ the only solution is $Z=Y=0$. Is this sufficient?
Best Answer
This is a prime example of a proof by descent, I believe that this method was made famous by Fermat who used it in Diophantine equations.
The idea is that if you are looking for rational solutions to $3x^2+5y^2 =4$, you can instead look at integer solutions to $3X^2+5Y^2=4Z^2$.
If you now suppose that there is a minimal triple $(X,Y,Z)$ that solves this Diophantine equation. If $(X,Y,Z)$ is minimal, it follows that $X,Y,Z$ have no common factor, $p$. Otherwise, $(X/p,Y/p,Z/p)$ would be a solution which is smaller than our minimal solution.
If you reduce the equation $(\mathrm{mod}\;3)$, you get $Z^2=2Y^2$, which is only possible if $Z=Y=0$ $\pmod{3}$.
But, $3X^2=4Z^2-5Y^2$, which means that the $RHS$ has 2 copies of 3, which in turn implies that $3|X$.
This is a contradiction, as we have a solution which is "smaller" than our minimal solution.