# Show that $\displaystyle x^n+x^{\frac{1}{n}}+n = 0$ has no real zeros for $n\geqslant 2$

I'm unable to prove the following two statements:

1. $\displaystyle x^n+x^{\frac{1}{n}}+n = 0$ has no real zeros for all $n\geqslant 2.$

2. $\displaystyle x^n+x^{\frac{1}{n}}-n = 0$ has exactly one real zero for $n \geqslant 1$.

(where $n$ is an integer and $x$ is a positive real number).

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The second can't be true, since for odd $n$ the left-hand side is and odd function of $x$ which is not zero at $x=0$, so it has an even number of real zeros. Do you mean for $x \ge 0$? –  joriki Mar 5 '11 at 18:33
@joriki, Oops yes -- I should have said that! –  TCM Mar 5 '11 at 18:37

Let $y=x^{1/n}$. Then $x^{n} + x^{1/n} = y^{n^2} + y$.

In particular, for any constant $k$, you have $$\frac{d}{dy}\left(y^{n^2}+ y + k\right) = n^2y^{n^2-1} + 1.$$

For nonnegative $y$, this is always positive; so $x^n + x^{1/n}+a$ is strictly increasing on $[0,\infty)$. (Well, technically, $y^{n^2}+y+a$ is strictly increasing; but since $y$ is itself an increasing function of $x$ for $x\geq 0$ and $n\gt0$, then so is $x^n + x^{1/n}+a$). In particular, it has at most one zero on $[0,\infty)$.

If $a\gt 0$, then the function is positive at $0$, hence has no zeros on $[0,\infty)$. If $a\leq 0$, then the function is nonpositive at $0$, hence has exactly one zero on $[0,\infty)$.

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Here's a hint: observe that if $n < 0$ the function $f(x) = x^{n} + x^{1 \over n}$ is decreasing for $x \geq 0$, and if $n > 0$ then $f(x)$ is increasing for $x \geq 0$.

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Hint: For 1. the left hand side is always positive, and for 2. notice the function is monotonic on $(0,\infty)$. (You need to specify that $n\geq 1$ for the second one)

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