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?

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.)
