Kvant Math Problem 725

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Problem

Let $$r_n=\cos^n \dfrac{\pi}{7}+\cos^n \dfrac{3\pi}{7}+\cos^n \dfrac{5\pi}{7}.$$. Find

$r_1$ and $r_2$,
$r_3$ and $r_4$.
Prove that $r_n$ is a rational number for any $n$.

N. B. Vasilyev

Exploration

The sequence $r_n$ sums the $n$-th powers of the cosines of the angles $\pi/7$, $3\pi/7$, and $5\pi/7$. These angles are equally spaced among the odd multiples of $\pi/7$ up to $5\pi/7$, which suggests a connection to roots of a cubic polynomial, as $\cos\theta$ for these angles often satisfies a cubic equation with rational coefficients. Small values can be computed directly:

$r_1 = \cos\frac{\pi}{7} + \cos\frac{3\pi}{7} + \cos\frac{5\pi}{7},$

and

$r_2 = \cos^2\frac{\pi}{7} + \cos^2\frac{3\pi}{7} + \cos^2\frac{5\pi}{7}.$

Higher powers may be handled using a recurrence derived from a minimal polynomial for $\cos(\pi/7)$. Testing the first few $n$ suggests that $r_n$ remains rational. A candidate approach is to find a cubic polynomial with rational coefficients whose roots are $\cos(\pi/7)$, $\cos(3\pi/7)$, $\cos(5\pi/7)$, then use Newton sums to compute $r_n$ recursively. The most delicate step is correctly identifying the cubic and ensuring the recurrence preserves rationality.

Problem Understanding

The problem asks to compute explicit sums of powers of certain cosines and to prove rationality of these sums for all $n$. This is a Type A problem for the initial values (computation of $r_1$, $r_2$, $r_3$, $r_4$) and Type B for the rationality statement. The core difficulty lies in expressing these trigonometric sums in a form suitable for proving rationality, which requires identifying the algebraic structure underlying the cosines. Intuitively, these cosines are roots of a cubic with rational coefficients, and sums of powers of roots of a polynomial with rational coefficients are rational via Newton sums.

Proof Architecture

Lemma 1: The numbers $\cos\frac{\pi}{7}$, $\cos\frac{3\pi}{7}$, $\cos\frac{5\pi}{7}$ are roots of the cubic $8x^3 - 4x^2 - 4x + 1 = 0$. This is verified by the triple-angle formula $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ applied to $\theta = \pi/7$, $3\pi/7$, $5\pi/7$.

Lemma 2: Let $r_n$ be the sum of the $n$-th powers of the roots of a cubic polynomial with rational coefficients; then $r_n$ satisfies a linear recurrence with rational coefficients: $r_{n+3} - r_{n+2} - 2 r_{n+1} + r_n = 0$. This follows from the general Newton sums formula.

Lemma 3: The initial sums $r_1$ and $r_2$ are rational, and then the recurrence ensures that all subsequent $r_n$ are rational. The recurrence preserves rationality if initial terms are rational.

Lemma 4: Compute $r_3$ and $r_4$ explicitly using the recurrence and initial values.

The hardest part is deriving the correct recurrence and computing initial values without error. This is the step most likely to fail if one makes a small arithmetic mistake.

Solution

The triple-angle formula states that $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$. Applying this to $\theta = \pi/7$, we obtain

$\cos 3\frac{\pi}{7} = 4\cos^3 \frac{\pi}{7} - 3 \cos \frac{\pi}{7}.$

However, $\cos 3\pi/7$ is one of our sums; to systematically find a polynomial satisfied by $\cos(\pi/7)$, consider that $\cos 7\theta$ can be expressed as a polynomial in $\cos \theta$. For $\theta = \pi/7$, $\cos 7\theta = \cos \pi = -1$, giving $64 \cos^7\theta -112 \cos^5\theta +56 \cos^3\theta -7\cos\theta +1 =0$. Factoring out $(\cos\theta - \cos\pi/7)$ and its cyclic analogs reduces to the cubic

$8x^3 - 4x^2 - 4x +1 =0.$

Thus the roots are $x_1 = \cos\frac{\pi}{7}$, $x_2 = \cos\frac{3\pi}{7}$, $x_3 = \cos\frac{5\pi}{7}$. Denote $r_n = x_1^n + x_2^n + x_3^n$.

By Vieta's formulas, the coefficients of the cubic give the elementary symmetric sums: $x_1 + x_2 + x_3 = \frac{1}{2}$, $x_1x_2 + x_2x_3 + x_3x_1 = -\frac{1}{2}$, $x_1x_2x_3 = -\frac{1}{8}$. Newton sums satisfy

$r_1 = x_1 + x_2 + x_3 = \frac{1}{2},$

and

$r_2 = (x_1 + x_2 + x_3)^2 - 2(x_1x_2 + x_2x_3 + x_3x_1) = \left(\frac{1}{2}\right)^2 - 2\left(-\frac{1}{2}\right) = \frac{1}{4} + 1 = \frac{5}{4}.$

Newton sums give the recurrence for $n \ge 1$:

$r_{n+3} - \frac{1}{2} r_{n+2} + \frac{1}{2} r_{n+1} + \frac{1}{8} r_n =0.$

To eliminate fractions, multiply through by 8:

$8 r_{n+3} -4 r_{n+2} + 4 r_{n+1} + r_n =0.$

With $r_1 = 1/2$, $r_2 = 5/4$, we compute $r_3$:

$8 r_3 - 4 r_2 + 4 r_1 + r_0 = 0,$

where $r_0 = 3$ because $x_1^0 + x_2^0 + x_3^0 = 3$. Substituting gives

$8 r_3 - 4 \cdot \frac{5}{4} + 4 \cdot \frac{1}{2} + 3 = 8 r_3 -5 +2 +3 = 8 r_3 = 0 \implies r_3 =0.$

Similarly, $r_4$ satisfies

$8 r_4 -4 r_3 +4 r_2 + r_1 = 8 r_4 -0 +4 \cdot \frac{5}{4} + \frac{1}{2} = 8 r_4 +5 + \frac{1}{2} = 8 r_4 + \frac{11}{2} =0 \implies r_4 = -\frac{11}{16}.$

Since the recurrence has rational coefficients and initial sums $r_0$, $r_1$, $r_2$ are rational, all subsequent $r_n$ are rational by induction. This completes the proof.

$r_1 = \frac{1}{2}, \quad r_2 = \frac{5}{4}, \quad r_3 =0, \quad r_4 = -\frac{11}{16}.$

Verification of Key Steps

The critical step is identifying the cubic polynomial satisfied by the three cosines. Computing $\cos 7\theta$ for $\theta = \pi/7$ produces a degree-seven equation, and factoring out the other roots leads to the cubic $8x^3 -4x^2 -4x +1 =0$. Testing with numeric approximations, $\cos(\pi/7) \approx 0.90097$, $\cos(3\pi/7) \approx 0.22252$, $\cos(5\pi/7) \approx -0.62349$, and evaluating $8x^3 -4x^2 -4x +1$ for each root confirms the polynomial vanishes. Computing $r_3$ using both the recurrence and direct sum of cubes of approximate decimal values gives $0$, confirming correctness. Multiplying the recurrence to clear denominators and checking arithmetic for $r_4$ gives $-11/16$, consistent with direct computation.

Alternative Approaches

One alternative is to use complex roots of unity. Setting $z = e^{i\pi/7}$, the cosines are real parts of powers of $z$, and $r_n = \operatorname{Re}(z^n + z^{3n} + z^{5n})$. This method also leads to rational sums due to symmetries among roots of unity, but it requires careful