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Let $m, n, p\in \mathbb R$, $n>0, p>0$. Prove that the following equation has exactly one positive solution:


Here is my attempt: Let $f(x)=x^5-mx^3-nx-p$, $f$ is continuous on $\mathbb R$ and $f(0)=-p<0, \lim_{x\to +\infty} f(x)=+\infty$. This implies, there exists $\xi>0$ such that $f(\xi)=0$ as a consequence of Bolzano-Cauchy theorem. Moreover, $$f'(x)=5x^4-3mx^2-n.$$ Since $(3m)^2+20n>0$ and $-5n<0$, we can easily see that $f'(x)=0$ has two roots $$x=\pm \frac{3m+\sqrt{9m^2+20n}}{10}.$$

I cannot continue to verify that $f(x)=0$ has only one positive solution?

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Actually "$x^2= \frac {3m \pm \sqrt {9m^2 +20n}}{10}$" in your post. – Vikrant Desai Jan 11 at 6:52

I think your result follows from the Descartes' rule of signs. The number of sign differences in your polynomial is 1, irrespective of whether $m$ is strictly positive, strictly negative or zero.

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This looks like a good argument to me. Perhaps the downvoter can explain where the problem is? – Martin R Jan 11 at 6:35

Here is how you can make your method work:

$f'(x) = 0$ has exactly one positive solution $$ x=\sqrt{ \frac{3m+\sqrt{9m^2+20n}}{10} } \, . $$

$f(0) = -p < 0$ and $f'(0) = -n < 0$, therefore $f$ has a local minimum in the interval $(0, r)$ where $r$ is the smallest positive root of $f$. So $f'(x) = 0$ for some $x \in (0, r)$.

If $f$ has another positive root $s > r$ then $f'(x) = 0$ for some $x \in (r, s)$, in contradiction to the fact that $f'(x) = 0$ has only one positive solution.

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Can you explain why "f(0)=−p<0 and f′(0)=−n<0, therefore ff has a local minimum in the interval (0,r)" – Richkent Jan 11 at 6:43
@Richkent: $f$ has a minimum on the interval $[0, r]$. Since $f$ is decreasing near zero, the minimum can not be at $x = 0$. It can also not be at $x = r$, therefore it is in the interior of the interval, and then the derivative vanishes at the minimum. – Martin R Jan 11 at 6:47
@Richkent Try to visualize this explanation. Draw a rough graph. – Vikrant Desai Jan 11 at 7:09

Let $f(x)=x^5-mx^3-nx-p$.

$f(0)<0, \lim_{x\to\infty}f(x)=\infty$, so the IVT gives that $f$ has a positive root.

Let's call the least such root $a$, noting thus that $a^5=ma^3+na+p$

Then, for $x\ge 0, f'(x)=5x^4-3mx^2-n$.

Note that $f'(a)=5a^4-(3ma^2+n)>5a^4-3(ma^2+n)=5a^4-3(a^4-\frac{p}{a})=2a^4+\frac{3p}{a}>0$

Furthermore, for $x\ge a$, $f'(x)=5x^4-3mx^2-n\ge x^2(5a^2-3m)-n\ge a^2(5a^2-3m)-n$.


  • $a^5=ma^3+na+p\implies a^2=m+\frac{n}{a^2}+\frac{p}{a^3}$

  • $\implies a^2(5a^2-3m)=a^2(2m+\frac{5n}{a^2}+\frac{5p}{a^3})\ge 5n$

  • $\implies f'(x)\ge5n-n=4n>0$ for $x\ge a$

So the function has a root at $a$, and increases to the right of $a$, giving uniqueness.

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