1. Given is the sequence $x_{1}=1, x_{2}=4$ and $x_{n+2}=4 x_{n+1}-x_{n}$ for $n \geqslant 1$. Find all natural numbers $m$ such that the number $3 x_{n}^{2}+m$ is a perfect square for every natural number $n$.
1. The characteristic polynomial of the given recurrence sequence is $x^{2}-4 x+1$, whose roots are $x_{1,2}=2 \pm \sqrt{3}$. Therefore, the sequence $\left(x_{n}\right)$ is of the form $x_{n}=A(2+\sqrt{3})^{n}+B(2-$ $\sqrt{3})^{n}$ for some constants $A$ and $B$; it is easy to find that $A=-B=\frac{1}{2 \sqrt{3}}$, i.e.,

$$
x_{n}=\frac{1}{2 \sqrt{3}}\left((2+\sqrt{3})^{n}-(2-\sqrt{3})^{n}\right)
$$

Now, $3 x_{n}^{2}=\frac{1}{4}\left((2+\sqrt{3})^{2 n}-2+(2-\sqrt{3})^{2 n}\right)$, so

$$
3 x_{n}^{2}+1=y_{n}^{2} \quad \text { for } \quad y_{n}=\frac{1}{2}\left((2+\sqrt{3})^{n}+(2-\sqrt{3})^{n}\right) \text {. }
$$

Note that the number $y_{n}$ is an integer. Thus, $m=1$ satisfies the condition of the problem.

Suppose there exists $m \neq 1$ that also satisfies the condition of the problem. Then $3 x_{n}^{2}+m=y_{n}^{2}+(m-1)=z_{n}^{2}\left(z_{n} \geqslant 0\right)$ is a perfect square for all $n$, so $m-1=\left(z_{n}-y_{n}\right)\left(z_{n}+y_{n}\right)$ has infinitely many divisors, which is impossible. It follows that $m=1$ is the only solution.