Easy to see that $3$ not divide the numerator of $x_n$ for all $n\in \mathbb{Z}^+$. So, it's enough to proof the following proposition: Quote: For any positive integers $p,q$ that $3\nmid q$ and $3(\frac{p}{q})^3+\frac{p}{q}=\frac{r^2}{s^2}$ for some positive integers $r,s$. Then, $\frac{p}{q}$ must be a square of a rational number. Note that we can WLOG $\gcd (p,q)=\gcd (r,s)=1$. From the equation, we get $s^2(3p^3+pq^2)=r^2q^3$. For any, if exists, prime number $d$ that $d\mid \gcd (q^3,3p^3+pq^2)$. We get $d\mid q\implies d\mid 3p^3\implies d\mid \gcd (q,3p)=1$, so no such $d$ exists. In other words, $\gcd (q^3,3p^3+pq^2) =1$. This gives $q^3\mid s^2$. Also, since $\gcd (s^2,r^2)=1$, we get $s^2\mid q^3$. Hence, $q^3=s^2$, so there exists $\ell \in \mathbb{Z}^+$ that $q=k^2$ and $s=k^3$. Note that $\gcd (p,k)=\gcd (p,q)=1$. Plugging in the equation gives $p(3p^2+k^4)=r^2$. Easy to see that $\gcd (p,3p^2+k^4)=1$, so $p$ must be a perfect square. Hence, both $p$ and $q$ are squares of positive integers, done.