<s>[INST] Equation in rational numbers? Is it true that this equation $6=\frac{x^2}{y^2+1}$ has no solutions in rational numbers?
If so, why?
It is quite evident that it has no solutions in integers (because $y^2+1$ never divides $3$).
 [/INST] Equivalently, we are looking for a solution of $6(a^2+b^2)=c^2$ in integers not all $0$.  
Suppose to the contrary that there is such a solution. Let $3^k$ be the highest power of $3$ that divides both $a$ and $b$. Let $a=3^ks$ and $b=3^k t$. 
Then $3$ does not divide $s^2+t^2$, so the highest power of $3$ that divides $6(a^2+b^2)$ is $3^{2k+1}$. It follows that $6(a^2+b^2)$ cannot be a perfect square. 
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