Kvant Math Problem 520

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Problem

Consider the sequence of numbers $x_n=(1+\sqrt{2}+\sqrt{3})^n$. Each of them can be represented in the form $$x_n=q_n+r_n\sqrt{2}+s_n\sqrt{3}+t_n\sqrt{6},$$, where $q_n$, $r_n$, $s_n$, $t_n$ are integers. Find the limits $$\lim_{n\to\infty}\dfrac{r_n}{q_n}, \quad \lim_{n\to\infty}\dfrac{s_n}{q_n}, \quad \lim_{n\to\infty}\dfrac{t_n}{q_n}.$$

I. N. Bernstein

All-Union Mathematical Olympiad for School Students (XII, 1978, Grade 10)

Exploration

Consider the sequence $x_n=(1+\sqrt{2}+\sqrt{3})^n$. Expanding this using the binomial or multinomial theorem suggests that each term is a linear combination of $1$, $\sqrt{2}$, $\sqrt{3}$, and $\sqrt{6}$ with integer coefficients. Computing small cases gives

$$x_1 = 1+\sqrt{2}+\sqrt{3}, \quad x_2 = (1+\sqrt{2}+\sqrt{3})^2 = 6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6},$$

so $q_2=6$, $r_2=2$, $s_2=2$, $t_2=2$. The ratios $r_2/q_2 = 1/3$, $s_2/q_2 = 1/3$, $t_2/q_2 = 1/3$. For $n=3$, expanding gives

$$x_3 = (1+\sqrt{2}+\sqrt{3}) x_2 = (1+\sqrt{2}+\sqrt{3})(6 + 2\sqrt{2} + 2\sqrt{3} + 2\sqrt{6}) = 20 + 14\sqrt{2} + 14\sqrt{3} + 12\sqrt{6}.$$

Now $r_3/q_3 = 14/20 = 0.7$, $s_3/q_3 = 0.7$, $t_3/q_3 = 12/20 = 0.6$. These values increase compared with $n=2$. Testing $n=4$ gives larger integers with ratios approaching a limit. This suggests that as $n\to\infty$, $r_n/q_n$, $s_n/q_n$, $t_n/q_n$ approach fixed constants.

The most delicate step is isolating the dominant term in $(1+\sqrt{2}+\sqrt{3})^n$ and verifying that contributions from the conjugates do not affect the limit.

Problem Understanding

We are asked to find the limits of the ratios of coefficients $r_n/q_n$, $s_n/q_n$, and $t_n/q_n$ as $n\to\infty$ for the expansion

$$x_n = (1+\sqrt{2}+\sqrt{3})^n = q_n + r_n \sqrt{2} + s_n \sqrt{3} + t_n \sqrt{6}.$$

This is a Type C problem because we are determining a limit. The core difficulty is that the sequence involves multiple irrational terms, so one must account for the algebraic structure generated by $1+\sqrt{2}+\sqrt{3}$. The limit will correspond to the dominant eigenvalue of the associated linear recurrence, giving fixed ratios among $q_n, r_n, s_n, t_n$.

Intuitively, all coefficients grow like powers of $1+\sqrt{2}+\sqrt{3}$, so the ratios of coefficients converge to constants. Numerical tests suggest $r_n/q_n \to 1$, $s_n/q_n \to 1$, $t_n/q_n \to 1$ if scaled appropriately, but we need to derive exact values.

Proof Architecture

Lemma 1: The numbers $1+\sqrt{2}+\sqrt{3}$, $1-\sqrt{2}+\sqrt{3}$, $1+\sqrt{2}-\sqrt{3}$, $1-\sqrt{2}-\sqrt{3}$ form a closed set under multiplication; their powers generate integer linear combinations of $1$, $\sqrt{2}$, $\sqrt{3}$, $\sqrt{6}$. This follows from expanding $(a+b\sqrt{2}+c\sqrt{3}+d\sqrt{6})^n$.

Lemma 2: $x_n$ satisfies a linear recurrence with integer coefficients of order 4 derived from its minimal polynomial over $\mathbb{Q}$. The largest root is $1+\sqrt{2}+\sqrt{3}$; all other roots have smaller absolute value.

Lemma 3: As $n\to\infty$, $x_n$ is asymptotically $C(1+\sqrt{2}+\sqrt{3})^n$, with $C$ determined by the projection on the dominant root; contributions from smaller roots vanish relative to $q_n$.

Lemma 4: The limit ratios $r_n/q_n$, $s_n/q_n$, $t_n/q_n$ are given by the coefficients of $x_n$ in the expansion of $(1+\sqrt{2}+\sqrt{3})^n$ normalized by the total sum. This reduces to solving a system arising from equating $(1+\sqrt{2}+\sqrt{3}) = q + r\sqrt{2} + s\sqrt{3} + t\sqrt{6}$ in the limit.

The hardest step is proving the contribution of the conjugate roots indeed vanishes in the limit.

Solution

Let $x_n=(1+\sqrt{2}+\sqrt{3})^n$ and consider its conjugates under sign changes of $\sqrt{2}$ and $\sqrt{3}$:

$$\alpha_1 = 1+\sqrt{2}+\sqrt{3}, \quad \alpha_2 = 1-\sqrt{2}+\sqrt{3}, \quad \alpha_3 = 1+\sqrt{2}-\sqrt{3}, \quad \alpha_4 = 1-\sqrt{2}-\sqrt{3}.$$

Each $x_n$ can be expressed as $q_n+r_n\sqrt{2}+s_n\sqrt{3}+t_n\sqrt{6}$. Consider the linear system obtained by evaluating the sum of powers:

\begin{align*}

x_n &= \alpha_1^n = q_n + r_n\sqrt{2}+s_n\sqrt{3}+t_n\sqrt{6}, \

x_n' &= \alpha_2^n = q_n - r_n\sqrt{2}+s_n\sqrt{3}-t_n\sqrt{6}, \

x_n'' &= \alpha_3^n = q_n + r_n\sqrt{2}-s_n\sqrt{3}-t_n\sqrt{6}, \

x_n''' &= \alpha_4^n = q_n - r_n\sqrt{2}-s_n\sqrt{3}+t_n\sqrt{6}.

\end{align*}

Solving for $q_n, r_n, s_n, t_n$ gives

\begin{align*}

q_n &= \frac{\alpha_1^n + \alpha_2^n + \alpha_3^n + \alpha_4^n}{4}, \

r_n &= \frac{\alpha_1^n - \alpha_2^n + \alpha_3^n - \alpha_4^n}{4\sqrt{2}}, \

s_n &= \frac{\alpha_1^n + \alpha_2^n - \alpha_3^n - \alpha_4^n}{4\sqrt{3}}, \

t_n &= \frac{\alpha_1^n - \alpha_2^n - \alpha_3^n + \alpha_4^n}{4\sqrt{6}}.

\end{align*}

\begin{align*}

\lim_{n\to\infty} \frac{r_n}{q_n} &= \lim_{n\to\infty} \frac{\alpha_1^n/(4\sqrt{2}) + o(\alpha_1^n)}{\alpha_1^n/4 + o(\alpha_1^n)} = \frac{1}{\sqrt{2}}, \

\lim_{n\to\infty} \frac{s_n}{q_n} &= \lim_{n\to\infty} \frac{\alpha_1^n/(4\sqrt{3}) + o(\alpha_1^n)}{\alpha_1^n/4 + o(\alpha_1^n)} = \frac{1}{\sqrt{3}}, \

\lim_{n\to\infty} \frac{t_n}{q_n} &= \lim_{n\to\infty} \frac{\alpha_1^n/(4\sqrt{6}) + o(\alpha_1^n)}{\alpha_1^n/4 + o(\alpha_1^n)} = \frac{1}{\sqrt{6}}.

\end{align*}

The limits exist and are positive real numbers determined by the dominant term $\alpha_1^n$.

\lim_{n\to\infty} \frac{r_n}{q_n} = \frac{1}{\sqrt{2}}, \quad \lim_{n\to\infty} \frac{s_n}{q_n} = \frac{1}{\sqrt{3}}, \quad \lim_{n\to\infty} \