4・13 Let the roots of the equation $x^{2}-p x+q=0$ be $r$ and $s$, and neither of them is zero. Try to find the equation whose roots are $r^{2}+\frac{1}{s^{2}}$ and $s^{2}+\frac{1}{r^{2}}$ (no need to solve the original equation).
[Solution] The required equation should be
$$
\left[x-\left(r^{2}+\frac{1}{s^{2}}\right)\right]\left[x-\left(s^{2}+\frac{1}{r^{2}}\right)\right]=0
$$

which simplifies to $x^{2}-\left(r^{2}+s^{2}+\frac{1}{r^{2}}+\frac{1}{s^{2}}\right) x+\left(r^{2} s^{2}+2+\frac{1}{r^{2} s^{2}}\right)=0$.
However, $r$ and $s$ are the roots of $x^{2}-p x+q=0$, and by the relationship between roots and coefficients, we know: $r+s=p, r s=q$, so
$$
\begin{array}{l} 
r^{2}+s^{2}+\frac{1}{r^{2}}+\frac{1}{s^{2}}=\frac{\left(r^{2} s^{2}+1\right)\left(r^{2}+s^{2}\right)}{r^{2} s^{2}} \\
=\frac{\left(q^{2}+1\right)\left(p^{2}-2 q\right)}{q^{2}}, \\
r^{2} s^{2}+2+\frac{1}{r^{2} s^{2}}=q^{2}+2+\frac{1}{q^{2}}=\left(\frac{1}{q}+q\right)^{2},
\end{array}
$$

Substituting these results into (1), we get
$$
x^{2}-\frac{\left(q^{2}+1\right)\left(p^{2}-2 q\right)}{q^{2}} x+\left(q+\frac{1}{q}\right)^{2}=0 .
$$