Find $n(R\cap Q)$ if $R=\{(x,y)$:$x^2$+$3y^2$=$28,x$,$y\in\mathbb{Z}\}$ and $Q$=$\{(x,y)$:$x$>$y,x$,$y\in\mathbb{Z}\}$

Question

Th relations $R=\{(x,y):x^2+3y^2=28,x,y\in\mathbb{Z}\}$ and $Q=\{(x,y):x>y,x,y\in\mathbb{Z}\}$, then the number of elements in $R\cap Q$ is ?

how do I find the integer solutions of the equation $x^2+3y^2=28$ ?

Note: If it was a first order Diophantine equation I know how to solve, but what is the easiest way to solve second order Diophantine equations of such types ?

Answer 1

As $x^2+3y^2=28$ you can easily see that $x^2=28-3y^2\le28$, so $|x|\le \sqrt{28}=2\sqrt{7}$. Similarly you can get the estimate for $y$. Then you just have to enumerate all the possibilities.