Find a non-trivial solution to $5x^2+7y^2=3z^2$. Find a non-trivial solution to $5x^2+7y^2=3z^2$. Note: $x,y,z$ are integers.
My attempt: By a theorem I have $x^2 \equiv 1 \pmod 5$, $x^2 \equiv 1 \pmod 7$ and $x^2 \equiv 1 \pmod 3$. I just guessed $x=1$. So I have $5+7y^2=3z^2$. I now guessed $y=1$ and $z=2$.
Is there a better way to find non-trivial solutions than guessing?

The existence of integer solutions of $ax^2+by^2+cz^2=0$ or other ternary quadratic forms can be decided by using Legendre's theorem. 
Theorem (Legendre): Let $a,b,c$ coprime positive integers, then $ax^2 + by^2 = cz^2$ has a nontrivial solution in integers (or rational numbers) $x,y,z$ if and only if $$\left(\frac{-bc}{a}\right)=\left(\frac{-ac}{b}\right)=\left(\frac{ab}{c}\right)=1.$$
If such solutions exists one can use a theorem of Holzer, that $x,y,z$ must be relatively small (provided $a,b,c$ are squarefree), i.e., $|x|\le \sqrt{|bc|}$, $|y|\le \sqrt{|ca|}$, and $|z|\le \sqrt{|ab|}$, so that one can search by computer (there are more effective algorithms, though, than just trying here). 
Reference for finding explicitly solutions: J. E. Cremona, D. Rusin: Efficient solution of rational conics. Math. Comp. 72 (2003), no. 243, 1417 1441.
