Tomongho A.K.

Find all increasing arithmetic progressions with a finite number of terms, the sum of which is 1, and each term has the form $1 / k$, where $k$ is a natural number.
Let $1 / m$ be the largest term of the progression. Then the progression has more than $m$ terms (otherwise the sum is less than 1). On the other hand, the common difference

$d \geq 1 / m - 1 / (m+1) = 1 / m(m+1)$, so there are no more than $m+1$ terms. Thus, the number of terms is $m+1$, and the arithmetic mean of all terms is $1 / (m+1)$. Since only one term of the progression is greater than the arithmetic mean, the progression has no more than three terms. Therefore, $m+1=3$, and the progression itself is $1 / 6, 1 / 3, 1 / 2$.

## Answer

## Polynomials (miscellaneous). $\quad]$ Problem 64668 Topics: [ Examples and counterexamples. Constructions ] [Identical transformations (trigonometry).]

Does there exist a polynomial $f(x)$ of degree 6 such that for any $x$ the equality $f(\sin x) + f(\cos x) = 1$ holds?

## Solution

Example 1. $f(x) = (4 x^{3} - 3 x)^{2}$. Indeed, $f(\sin x) + f(\cos x) = \sin^{2} 3x + \cos^{2} 3x = 1$.

Example 2. $f(x) = -2 x^{6} + 3 x^{4}$. Indeed, $f(\sin x) + f(\cos x) = -2 \sin^{6} x + 3 \sin^{4} x - 2 \cos^{6} x + 3 \cos^{4} x = $

$= -2 (\sin^{6} x + \cos^{6} x) + 3 (\sin^{4} x + \cos^{4} x) = -2 (\sin^{2} x + \cos^{2} x) (\sin^{4} x - \sin^{2} x \cos^{2} x + \cos^{4} x) + 3 ((\sin^{2} x + \cos^{2} x)^{2} - 2 \sin^{2} x \cos^{2} x) = $

$= -2 (1 - 3 \sin^{2} x \cos^{2} x) + 3 (1 - 2 \sin^{2} x \cos^{2} x) = 1$.

## Answer

There exists.