All-Soviet-Union Mathematical Olympiad

Solution:

Let $x_n = (1 + \sqrt{2} + \sqrt{3})^n$. Consider the conjugates of $1 + \sqrt{2} + \sqrt{3}$ under the field automorphisms fixing $\mathbb{Q}$:

Let $\alpha_1 = 1 + \sqrt{2} + \sqrt{3}$
Let $\alpha_2 = 1 + \sqrt{2} - \sqrt{3}$
Let $\alpha_3 = 1 - \sqrt{2} + \sqrt{3}$
Let $\alpha_4 = 1 - \sqrt{2} - \sqrt{3}$

Then, $x_n = \alpha_1^n$, and similarly define $y_n = \alpha_2^n$, $z_n = \alpha_3^n$, $w_n = \alpha_4^n$.

We can write:
$$x_n = a_n + b_n \sqrt{2} + c_n \sqrt{3} + d_n \sqrt{6}$$
Similarly, $y_n = a_n + b_n \sqrt{2} - c_n \sqrt{3} - d_n \sqrt{6}$
$z_n = a_n - b_n \sqrt{2} + c_n \sqrt{3} - d_n \sqrt{6}$
$w_n = a_n - b_n \sqrt{2} - c_n \sqrt{3} + d_n \sqrt{6}$

Now, solve for $a_n, b_n, c_n, d_n$:

Add all four equations:
$$x_n + y_n + z_n + w_n = 4a_n$$
So,
$$a_n = \frac{1}{4}(x_n + y_n + z_n + w_n)$$

Similarly,
$$x_n + y_n - z_n - w_n = 4b_n \sqrt{2} \implies b_n = \frac{1}{4\sqrt{2}}(x_n + y_n - z_n - w_n)$$
$$x_n - y_n + z_n - w_n = 4c_n \sqrt{3} \implies c_n = \frac{1}{4\sqrt{3}}(x_n - y_n + z_n - w_n)$$
$$x_n - y_n - z_n + w_n = 4d_n \sqrt{6} \implies d_n = \frac{1}{4\sqrt{6}}(x_n - y_n - z_n + w_n)$$

Now, as $n \to \infty$, $\alpha_1 > 1$, but $|\alpha_2| < 1$, $|\alpha_3| < 1$, $|\alpha_4| < 1$ (since $1 + \sqrt{2} - \sqrt{3} \approx 1 + 1.414 - 1.732 \approx 0.682$, $1 - \sqrt{2} + \sqrt{3} \approx 1 - 1.414 + 1.732 \approx 1.318$, $1 - \sqrt{2} - \sqrt{3} \approx 1 - 1.414 - 1.732 \approx -2.146$).

Thus, as $n \to \infty$, $x_n$ dominates, and the other terms tend to zero compared to $x_n$.

Therefore,
$$a_n \sim \frac{1}{4} x_n$$
$$b_n \sim \frac{1}{4\sqrt{2}} x_n$$
$$c_n \sim \frac{1}{4\sqrt{3}} x_n$$
$$d_n \sim \frac{1}{4\sqrt{6}} x_n$$

So,
$$\lim_{n \to \infty} \frac{b_n}{a_n} = \frac{1}{\sqrt{2}}$$
$$\lim_{n \to \infty} \frac{c_n}{a_n} = \frac{1}{\sqrt{3}}$$
$$\lim_{n \to \infty} \frac{d_n}{a_n} = \frac{1}{\sqrt{6}}$$