5-116 Express the polynomial $x^{4}+x^{3}+x^{2}+x+1$ as the difference of squares of two real-coefficient polynomials of different degrees.
[Solution] If the polynomial $W(x)=x^{4}+x^{3}+x^{2}+x+1$ is expressed as $U^{2}(x) - V^{2}(x)$, where $U(x)$ and $V(x)$ are two polynomials of different degrees, then the polynomial $U(x)$ must be quadratic, and the polynomial $V(x)$ must be linear or constant. Therefore, the polynomial $V^{2}(x)$ does not contain terms of degree higher than 2, which means that the highest degree and the next highest degree terms of the polynomials $U^{2}(x)$ and $W(x)$ must be the same, so $U^{2}(x)$ has the form $x^{4}+x^{3}+\cdots$. Thus, $U(x)$ can be expressed as $U(x)=x^{2}+\frac{1}{2} x+a$. Therefore,
$$
\begin{aligned}
V^{2}(x)=U^{2}(x)-W(x)= & \left(x^{2}+\frac{1}{2} x+a\right)^{2}-\left(x^{4}+x^{3}+x^{2}+x\right. \\
& +1) \\
= & \left(2 a-\frac{3}{4}\right) x^{2}+(a-1) x+a^{2}-1 .
\end{aligned}
$$

Since $2 a-\frac{3}{4}=0$ and $a-1=0$ cannot hold simultaneously, it follows from the expression of $V^{2}(x)$ that $V(x)$ cannot be a constant. Therefore, $V(x)$ must be a linear polynomial. This indicates that $V^{2}(x)$ is a quadratic trinomial with a positive leading coefficient and a discriminant of zero, i.e.,
$$
2 a-\frac{3}{4}>0, (a-1)^{2}-4\left(2 a-\frac{3}{4}\right)\left(a^{2}-1\right)=0 \text { . }
$$

The second condition can be simplified to
$$
(a-1)\left(4 a^{2}+2 a-1\right)=0 \text {, }
$$

Solving this gives $a=1, a=\frac{1}{4}(-1 \pm \sqrt{5})$. Among these, only $a=1$ satisfies the condition
$$
2 a-\frac{3}{4}>0 \text { . }
$$

Thus, the original polynomial can be expressed as
$$
x^{4}+x^{3}+x^{2}+x+1=\left(x^{2}+\frac{1}{2} x+1\right)^{2}-\left(\frac{1}{2} \sqrt{5} x\right)^{2} \text {, }
$$

and the representation is unique.