Kvant Math Problem 177

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Solve time: 7m44s
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Problem

Find all solutions of the equation $$\sqrt[\scriptstyle n~]{x^n-a^n}+\sqrt[\scriptstyle n~]{2a^n-x^n}=a,$$, where $a$ is a given real number, and $n$ is a natural number greater than one.

T. Temirov

Exploration

Consider the equation

$$\sqrt[n]{x^n - a^n} + \sqrt[n]{2a^n - x^n} = a,$$

with $a \in \mathbb{R}$ and $n > 1$ a natural number. The structure suggests symmetry around $x = a$, since $x^n - a^n$ and $2a^n - x^n$ sum to $a^n$. Begin by testing small $n$ to detect patterns. For $n = 2$, the equation becomes $\sqrt{x^2 - a^2} + \sqrt{2a^2 - x^2} = a$, which is only meaningful if $x^2 - a^2 \ge 0$ and $2a^2 - x^2 \ge 0$, implying $x \in [a, a\sqrt{2}]$. Compute $\sqrt{x^2 - a^2} + \sqrt{2a^2 - x^2}$ as a function of $x$ over this interval. At $x = a$, the sum equals $\sqrt{0} + \sqrt{a^2} = a$. At $x = a\sqrt{2}$, the sum equals $\sqrt{a^2} + \sqrt{0} = a$. The derivative suggests monotonicity; therefore, the equation might admit exactly two solutions for $n = 2$. For $n > 2$, test $x = a$. Then $\sqrt[n]{a^n - a^n} + \sqrt[n]{2a^n - a^n} = 0 + \sqrt[n]{a^n} = a$, which satisfies the equation. The boundary $x = 2^{1/n} a$ gives $\sqrt[n]{(2^{1/n}a)^n - a^n} + \sqrt[n]{2a^n - (2^{1/n}a)^n} = \sqrt[n]{(2 - 1)a^n} + \sqrt[n]{2a^n - 2a^n} = a + 0 = a$, which is valid. Negative solutions can be tested by substituting $x = -a$; then $\sqrt[n]{(-a)^n - a^n} + \sqrt[n]{2a^n - (-a)^n} = \sqrt[n]{(-1)^n a^n - a^n} + \sqrt[n]{2a^n - (-1)^n a^n}$. For even $n$, the first term is $\sqrt[n]{a^n - a^n} = 0$, the second is $\sqrt[n]{2a^n - a^n} = a$, yielding a solution $x = -a$. For odd $n$, the first term is $\sqrt[n]{-a^n - a^n} = \sqrt[n]{-2 a^n} = -2^{1/n} a$, the second term $\sqrt[n]{2a^n - (-a)^n} = \sqrt[n]{2a^n + a^n} = 3^{1/n} a$, so the sum is $(-2^{1/n} + 3^{1/n}) a$, which may equal $a$ only if $2^{1/n} + 1 = 3^{1/n}$, which is false. Therefore, negative solutions exist only for even $n$. This preliminary analysis indicates potential solutions $x = a$, $x = 2^{1/n}a$, and possibly $x = -a$ when $n$ is even.

The crucial point is to justify that no other $x$ can satisfy the equation. Because the function $f(x) = \sqrt[n]{x^n - a^n} + \sqrt[n]{2a^n - x^n}$ is increasing on $[a, 2^{1/n}a]$ and symmetric, any other solutions outside these candidate points must violate monotonicity or the domain restrictions. The main subtlety is the case $n$ odd versus even and ensuring no solutions exist in $(−\infty, a)$ or $(2^{1/n} a, \infty)$.

Problem Understanding

The problem asks to determine all real numbers $x$ satisfying a sum of $n$-th roots equation with parameters $a \in \mathbb{R}$ and $n \in \mathbb{N}$, $n > 1$. This is a Type A problem, as it requires identifying the complete solution set and proving no other solutions exist. The core difficulty is analyzing the sum of two $n$-th roots, which are defined only over restricted intervals, and ensuring that no unexpected solutions arise due to nonlinearity or parity of $n$. The preliminary exploration suggests the solutions are $x = a$, $x = 2^{1/n} a$, and $x = -a$ if $n$ is even, based on domain restrictions and symmetry.

Proof Architecture

Lemma 1 states that $x^n - a^n \ge 0$ and $2a^n - x^n \ge 0$ are necessary for real $n$-th roots to exist, implying $x \in [a, 2^{1/n} a]$ for $a \ge 0$; similarly, $x \in [2^{1/n} a, a]$ if $a \le 0$. Lemma 2 asserts that the function $f(x) = \sqrt[n]{x^n - a^n} + \sqrt[n]{2a^n - x^n}$ is strictly increasing on $[a, 2^{1/n} a]$ for $n$ even and $n$ odd, except possibly at $x = a$ or $x = 2^{1/n}a$, which guarantees at most two solutions in this interval. Lemma 3 checks the candidate points $x = a$, $x = 2^{1/n} a$, and $x = -a$ (if $n$ is even) and shows each satisfies the equation. The hardest direction is proving no other solutions exist, particularly for $n$ odd, as the derivative analysis requires care. The lemma most likely to fail under scrutiny is Lemma 2, specifically ensuring strict monotonicity over the interval for general $n$.

Solution

Let $f(x) = \sqrt[n]{x^n - a^n} + \sqrt[n]{2a^n - x^n}$. The $n$-th root is defined for real numbers only when the radicand is non-negative for even $n$, while for odd $n$ it is defined for all real numbers. Consider $n$ even. Then $x^n - a^n \ge 0$ and $2a^n - x^n \ge 0$ are necessary conditions, which yield $x \in [a, 2^{1/n} a]$ if $a > 0$ and $x \in [2^{1/n} a, a]$ if $a < 0$. For $x \in [a, 2^{1/n} a]$, define $u = x^n - a^n$ and $v = 2a^n - x^n$. Then $u, v \ge 0$ and $u + v = a^n$. The function $g(u) = \sqrt[n]{u} + \sqrt[n]{a^n - u}$ is strictly increasing in $u \in [0, a^n/2]$ and decreasing in $u \in [a^n/2, a^n]$, so the maximum occurs at $u = a^n/2$, giving $x^n = 3a^n/2$, which is not among boundary points. Testing boundary points, $u = 0$ yields $x = a$, giving $f(a) = 0 + \sqrt[n]{a^n} = a$, and $v = 0$ gives $x^n = 2a^n$, or $x = 2^{1/n} a$, giving $f(2^{1/n} a) = \sqrt[n]{a^n} + 0 = a$. The negative candidate $x = -a$ is valid for $n$ even: then $(-a)^n - a^n = 0$ for even $n$, and $2a^n - (-a)^n = 2a^n - a^n = a^n$, so $f(-a) = 0 + a = a$.

Consider $n$ odd. Then $x^n - a^n$ can be negative, but $\sqrt[n]{x^n - a^n}$ is defined. If $x > 2^{1/n} a$, then $2a^n - x^n < 0$, so $\sqrt[n]{2a^n - x^n} = -((x^n - 2a^n)^{1/n}) < 0$, so the sum is strictly less than $\sqrt[n]{x^n - a^n}$, which exceeds $a$ only at $x = 2^{1/n} a$. Similarly, if $x < a$, then $x^n - a^n < 0$, giving negative contribution. Therefore, candidates are only $x = a$ and $x = 2^{1/n} a$. Substituting $x = a$, we obtain