Solutions of the Diophantine equation $x^2(x^2+10)=3y^2(y^2+10)$ I am looking for the solutions of the Diophantine equation 
$$x^2(x^2+10)=3y^2(y^2+10).$$
Is there any solution of this equation except when $(x,y)=(0,0)$?
Or
Any computer programme such as MAGMA could solve this problem?
Thank you very much.
 A: If $x$ and $y$ are not both $0$, let $3^k$ be the highest power of $3$ that divides both $x$ and $y$. Let $x=3^ks$ and $y=3^k t$. Then $3$ cannot be a common divisor of $s$ and $t$.
Substitute and cancel. We get $s^2(x^2+10)=3t^2(y^2+10)$. Since $3$ cannot divide $x^2+10$, it must divide $s$. Say $s=3u$. Then $9u^2(x^2+10)=3t^2(y^2+10)$, and therefore $3u^2(x^2+10)=t^2(y^2+10)$. But then $3$ divides $t$, contradicting the fact that $3$ is not a common divisor of $s$ and $t$.   
A: Hint $\rm\ 3\nmid f(n)=n^2\!+10\:$ since $\rm\ mod\ 3\!:\ n\equiv 0,1,2\ \ but\ \ f(0)\equiv 1,\ f(1)\equiv 2\equiv f(2).\:$ Thus the $3$'s in the unique prime factorization of $\rm\:x^2(x^2\!+\!10)\,$ are those in $\rm\,x^2,\,$ an even number, but the $3$'s in $\rm\,3y^2(y^2\!+10)\,$ are those in $\rm\,3y^2,\,$ an odd number. Thus $\rm\,x^2(x^2\!+10)\ne3y^2(y^2\!+10)\,$  for $\rm\,x,y\in\Bbb Z.$
Remark $\, $ So, ignoring the factors $\rm\,x^2\!+\!10,\ y^2\!+\!10,\,$ which play no role, being never divisible by $3$, the proof is precisely the same as one of the standard proofs of the irrationality of $\sqrt{3}.$
