Is there a fifth-degree polynomial such that on the interval $(-1,1)$, its maximum is +1, its minimum is -1, both are taken at two different points, and its value is 0 at the endpoints of the interval?
A fifth-degree polynomial can have at most 4 different critical points - provided that its derivative, which is a fourth-degree polynomial, has all four of its roots real and distinct. Among the 4 critical points, only two maxima and two minima can alternate, because between any adjacent pair of them, the value of the derivative is either constantly positive or constantly negative. Since in the case of the sought $p(x)$, the maxima are positive and the minima are negative, there is also a root of $p(x)$ between each maximum and minimum, totaling 3.

![](https://cdn.mathpix.com/cropped/2024_05_02_a64cd8971427da8b8454g-1.jpg?height=605&width=672&top_left_y=386&top_left_x=716)

It will suffice to construct a $p(x)$ such that one of the intermediate roots is $x=0$, and the other two are $\pm c$, where $0 < c < 1$. The derivative of $p(x)$ is then

$$
p^{\prime}(x)=5 A\left(x^{4}-\frac{3\left(1+c^{2}\right)}{5} x^{2}+\frac{c^{2}}{5}\right)
$$

which can be rewritten as

$$
p^{\prime}(x)=5 A\left(x^{4}-\frac{3\left(1+c^{2}\right)}{5} x^{2}+\frac{c^{2}}{5}\right)
$$

The discriminant of this quadratic equation in $x^2$ is

$$
D=9-2 c^{2}+9 c^{4}=9\left(1-\frac{c^{2}}{9}\right)^{2}+\frac{80}{9} c^{4}>0
$$

($x=0$ is not a root of (2), since $c \neq 0$.) Let the positive roots of this equation be denoted by $x_{1}$ and $x_{2}$, and the other two by $-x_{1}$ and $-x_{2}$. Then, by the relationship between the roots and coefficients of (2) (considered as a quadratic equation in $x^2$),

$$
x_{1}^{2}+x_{2}^{2}=\frac{3\left(1+c^{2}\right)}{5} ; \quad x_{1}^{2} x_{2}^{2}=\frac{c^{2}}{5}
$$

and $p^{\prime}(x)=5 A\left(x^{2}-x_{1}^{2}\right)\left(x^{2}-x_{2}^{2}\right)$. From this, we can see that $p^{\prime}(x)$ changes sign at all four roots, so $p(x)$ indeed has critical points at these points.

We want, for example,

$$
p\left(x_{1}\right)=+1 ; \quad p\left(x_{2}\right)=-1
$$

Since $p(x)$ is an odd function, this implies

$$
p\left(-x_{1}\right)=-1 ; \quad p\left(-x_{2}\right)=+1
$$

It suffices to choose $c$ such that

$$
p\left(x_{1}\right)+p\left(x_{2}\right)=0
$$

is satisfied, because the constant $A$ can be chosen so that $p\left(x_{1}\right)=1$, and then (5) implies $p\left(x_{2}\right)=-1$. Since $x_{1}$ and $x_{2}$ are roots of (2),

$$
5 x_{i}^{5}=3\left(1+c^{2}\right) x_{i}^{3}-c^{2} x_{i} \quad(i=1,2)
$$

Thus, from (3), $p\left(x_{1}\right)+p\left(x_{2}\right)$ can be expressed without knowing the values of $x_{1}$ and $x_{2}$:

$$
\begin{aligned}
& \frac{5}{A}\left[p\left(x_{1}\right)+p\left(x_{2}\right)\right]=\left(5 x_{1}^{5}+5 x_{2}^{5}\right)-5\left(1+c^{2}\right)\left(x_{1}^{3}+x_{2}^{3}\right)+5 c^{2}\left(x_{1}+x_{2}\right)= \\
= & 4 c^{2}\left(x_{1}+x_{2}\right)-2\left(1+c^{2}\right)\left(x_{1}^{3}+x_{2}^{3}\right)=2\left(x_{1}+x_{2}\right)\left[2 c^{2}-\left(1+c^{2}\right)\left(x_{1}^{2}-x_{1} x_{2}+x_{2}^{2}\right)\right]= \\
= & 2\left(x_{1}+x_{2}\right)\left[2 c^{2}-\left(1+c^{2}\right)\left(\frac{3+3 c^{2}}{5}-\frac{c}{\sqrt{5}}\right)\right] .
\end{aligned}
$$

(We used the fact that from (3), $x_{1} x_{2}=\frac{c}{\sqrt{5}}$ follows because $x_{1}, x_{2}, c$ are positive.)

This can only be 0 if $c$ is a root of the equation

$$
2 c^{2}-\left(1+c^{2}\right)\left(\frac{3+3 c^{2}}{5}-\frac{c}{\sqrt{5}}\right)=0
$$

Dividing this by $c^{2}$ and substituting $y$ for $c+\frac{1}{c}$, we get the quadratic equation

$$
2-y\left(\frac{3 y}{5}-\frac{1}{\sqrt{5}}\right)=0
$$

whose roots are

$$
y_{1}=\sqrt{5}, \quad y_{2}=-\frac{2 \sqrt{5}}{3}
$$

but since $c+\frac{1}{c} \geq 2$, only $y_{1}$ is usable. From the quadratic equation

$$
c+\frac{1}{c}=\sqrt{5}
$$

we get

$$
c_{1}=\frac{\sqrt{5}-1}{2} ; \quad c_{2}=\frac{\sqrt{5}+1}{2}
$$

and only $c_{1}$ meets our goal, because $c_{2}>1$. $c_{1}$ is indeed a root of (6), so with it, (5) holds for $p(x)$.

We only need to show that with $c_{1}$, $x_{1}$ and $x_{2}$ also belong to the interval $(0,1)$. Indeed,

$$
x_{1}^{2}=\frac{10-4 \sqrt{5}}{10}<1, \quad \text { and } \quad x_{2}^{2}=\frac{5+\sqrt{5}}{10}<1
$$

Finally, determining $A$ and being satisfied with approximate but comprehensible values of the constants, (1) is replaced by the following polynomial:

$$
p(x)=12.4\left(x^{5}-1.38 x^{3}+0.38 x\right)
$$

Remark. It can be shown that apart from $p(x)$ and $-p(x)$, there are no other solutions to the problem.