Problem 1441

1. Given the sequence $(x_n)$ defined by $x_{n+1} = 3x_n^3 + x_n$ for all $n \in \mathbb{N}$ and $x_1 = \frac{a}{b}$ where $a, b$ are positive integers such that $3 \nmid b$. We need to prove that if $x_m$ is a square of a rational number for some positive integer $m$, then $x_1$ is also a square of a rational number.

2. First, observe that $3$ does not divide the numerator of $x_n$ for all $n \in \mathbb{Z}^+$. This is because the sequence is defined recursively and the denominator $b$ is not divisible by $3$.

3. We need to prove the following proposition:
For any positive integers $p, q$ such that $3 \nmid q$ and $3\left(\frac{p}{q}\right)^3 + \frac{p}{q} = \frac{r^2}{s^2}$ for some positive integers $r, s$, then $\frac{p}{q}$ must be a square of a rational number.

4. Assume without loss of generality that $\gcd(p, q) = \gcd(r, s) = 1$. From the given equation, we have:
$$s^2(3p^3 + pq^2) = r^2q^3$$

5. Consider any prime number $d$ that divides $\gcd(q^3, 3p^3 + pq^2)$. We have:
$$d \mid q \implies d \mid 3p^3 \implies d \mid \gcd(q, 3p) = 1$$
Therefore, no such $d$ exists, implying that $\gcd(q^3, 3p^3 + pq^2) = 1$.

6. Since $\gcd(q^3, 3p^3 + pq^2) = 1$, it follows that $q^3 \mid s^2$. Also, since $\gcd(s^2, r^2) = 1$, we get $s^2 \mid q^3$.

7. Hence, $q^3 = s^2$, so there exists $\ell \in \mathbb{Z}^+$ such that $q = k^2$ and $s = k^3$. Note that $\gcd(p, k) = \gcd(p, q) = 1$.

8. Substituting these into the equation gives:
$$p(3p^2 + k^4) = r^2$$
It is easy to see that $\gcd(p, 3p^2 + k^4) = 1$, so $p$ must be a perfect square.

9. Therefore, both $p$ and $q$ are squares of positive integers, which means $\frac{p}{q}$ is a square of a rational number.

$\blacksquare$