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 $3$).


$6 = \frac{x^2}{y^2+1} \iff 6y^2+6=x^2$. Given a solution $(x,y) \in \mathbb{Q}^2$, we can multiply out the denominators and get a solution to $6m^2+6n^2=l^2$, for $l,m,n \in \mathbb{Z}$ with $(l,m,n)=1$.

Reduce modulo 3 to see that $3\mid l$, and then dividing by the common factor of 3 on both sides and reducing modulo 3 again gives that $3 \mid m , n$ (look at possible sums of squares modulo $3$). This contradicts $(l,m,n)=1$.

You'll want to work through the details yourself...


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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