Diophantine Equation: $x^2 + 3y^2 = 11z^2$ I am having difficulty solving the following problem:


Prove rigorously that there is no integer solution for the Diophantine Equation $x^2 + 3y^2 = 11z^2$ except when $x = y=z = 0$.


Proving directly would seem far to difficult. I would assume proof by contradiction would be the most efficient way to solve this problem. To keep this simple, I shall assume that $(x,y,z)$ is a primitive integer solution. For any prime number $p$, there is a non-zero solution for the equation $x^2 + 3y^2 = 11z^2\pmod p$. Then simply find a good modulus $p$ to deduce a contradiction. 
This is where I have a difficult time, trying to deduce the contradiction. Any suggestions?
Thank you for your time, and thanks in advance for your feedback. 

Outline: Work modulo $3$. If $z$ is not divisible by $3$, then $11z^2\equiv 2\pmod{3}$. That is impossible, since $x^2$ cannot be congruent to $2$ modulo $3$. 
Thus $z$ is divisible by $3$, and therefore $x$ is, and therefore $y$ is. 
To use the idea in a formal proof, suppose that there is a non-trivial solution. Then there is a solution with $z\gt 0$ and as small as possible. Use the above reasoning to produce a  solution with a smaller positive $z$.
