15. Express the polynomial $x^{4}+x^{3}+x^{2}+x+1$ as the difference of squares of two real-coefficient polynomials with different degrees and factorize it over the real numbers.
15. Let $x^{4}+x^{3}+x^{2}+x+1=[f(x)]^{2}-[g(x)]^{2}$, given that $f(x)$ is a quadratic polynomial and the degree of $g(x)$ is less than 2, we can set $f(x)=x^{2}+\frac{1}{2} x+a$.
So $[g(x)]^{2}=[f(x)]^{2}-x^{4}-x^{3}-x^{2}-x-1$
$$
=\left(2 a-\frac{3}{4}\right) x^{2}+(a-1) x+\left(a^{2}-1\right)
$$

Thus, $\Delta=(a-1)^{2}-4\left(2 a-\frac{3}{4}\right)\left(a^{2}-1\right)$
$$
=0 \Rightarrow\left\{\begin{array}{l}
2 a-\frac{3}{4}>0 \\
(a-1)\left(4 a^{2}+2 a-1\right)=0
\end{array}\right.
$$

We get $a=1$ so
$$
\begin{aligned}
x^{4}+x^{3}+x^{2}+x+1 & =\left(x^{2}+\frac{1}{2} x+1\right)^{2}-\frac{5}{4} x^{2} \\
& =\left(x^{2}+\frac{1}{2} x+1\right)^{2}-\left(\frac{\sqrt{5}}{2} x\right)^{2} \\
& =\left(x^{2}+\frac{1+\sqrt{5}}{2} x+1\right)\left(x^{2}+\frac{1-\sqrt{5}}{2} x+1\right)
\end{aligned}
$$

Taking $f(x)=x^{2}+\frac{1}{2} x+1, g(x)=\frac{\sqrt{5}}{2} x$, we have
$$
x^{4}+x^{3}+x^{2}+x+1=[f(x)]^{2}-[g(x)]^{2}
$$