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Let $n\ge 2$ be an integer and $a$ be a real number such that $|a|\le (n-1)n^{-\frac n{n-1}}$. Can we find the real solutions of the algebraic equation $$ x^n -x =a. $$

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1 Answer 1

First, if $n$ is odd, then $f(x) = x^n-x$ is unbounded above and below, and so $f(x)=a$ has a solution for any $a$.

If $n$ is even, $f(x)$ is unbounded above and is convex, since $f''(x) = n(n-1)x^{n-2}\geq 0.$ Any local minimum of $f$ is therefore its global minimum.

Step 1: Find the global minimum of $f$ by solving $f'(x) =0.$

Step 2: What is the value of $f$ at this minimum?

Step 3: What can you conclude about the range of $f(x)?$

In neither case will you in general be able to solve for $x$ analytically.

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You can find closed-form solutions; they'll just be terribly complicated, and will involve special functions... –  J. M. Jun 1 '13 at 5:37
    
Yes! I want a closed form. Thank you! –  Chung. J Jun 1 '13 at 16:15

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