Finding rational solutions for $3x^2+5y^2 =4$ 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?
 A: 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.
A: In fact you can divide by $4$ and change coordinates so that the problem is equivalent to that of finding rational solutions of
$$
3x^2+5y^2=1,
$$
or integer primitive solutions of
$$
3X^2+5Y^2=Z^2.
$$
The latter do not exist because $3$ is not a square $\bmod5$ or because $5$ is not a square $\bmod3$.
A: Let $x,y,z$ be minimal $x$ nonegative solution.
Writing $$ 3x^2 = 4z^2-5y^2\implies 3\mid z^2+y^2\implies 3\mid z\wedge  3\mid y$$
So $z = 3z'$ and $y= 3y'$ implies $3\mid x$ so $x= 3x'$. So we get an equation 
$$ 3x'^2 +4y'^2 = 5z'^2$$ so we get the same equation with $x'$ smaller than $x$a contradcition, so $x= 0$ and thus $y=0$ and $z=0$.
A: Your argument isn't quite sufficient. This is because the observation that "mod $3$ the only solution is $Z=Y=0$" applies equally well to the equation $9X^2+5Y^2=4Z^2$, which does have solutions, e.g., $(X,Y,Z)=(2,0,3)$.  What you need to say is that "mod $3$ the only solution is $Z=Y=X=0$."  (On a side note, it would be better to use congruence notation, $\equiv$, instead of the equal sign.)
