13. A. Real numbers $a, b$ make the system of equations about $x, y$
$$
\left\{\begin{array}{l}
x y - x^{2} = 1, \\
x y^{2} + a x^{2} + b x + a = 0
\end{array}\right.
$$

have real solutions $(x, y)$.
(1) Prove: $|y| \geqslant 2$;
(2) Find the minimum value of $a^{2} + b^{2}$.
13. A. (1) From equation (1), we know $x \neq 0$, and $y=x+\frac{1}{x}$.
Therefore, $|y|=\left|x+\frac{1}{x}\right|=|x|+\frac{1}{|x|} \geqslant 2$.
(2) Substituting $x^{2}=x y-1$ into equation (2) gives $x\left(y^{2}+a y+b\right)=0$.
Thus, $y^{2}+a y+b=0$.
Since the system of equations has real solutions, the equation
$$
f(y)=y^{2}+a y+b=0
$$

has at least one real root in the range $|y| \geqslant 2$.
(i) When $|a| \leqslant 4$, we have
$$
f(-2) \leqslant 0 \text { or } f(2) \leqslant 0 \text {, }
$$

which means $2 a \geqslant b+4$ or $2 a \leqslant -(b+4)$.
If $b+4 \geqslant 0$, i.e., $b \geqslant -4$, then $|2 a| \geqslant b+4$.
Thus, $a^{2}+b^{2} \geqslant \frac{(b+4)^{2}}{4}+b^{2}$
$$
=\frac{5}{4}\left(b+\frac{4}{5}\right)^{2}+\frac{16}{5} \geqslant \frac{16}{5} \text {. }
$$

When $b=-\frac{4}{5}, a= \pm \frac{8}{5}$, the equality holds.
If $b+4<0$, i.e., $b<-4$, then $a^{2}+b^{2}>\frac{16}{5}$.
(ii) When $|a|>4$, $a^{2}+b^{2}>\frac{16}{5}$.

In summary, the minimum value of $a^{2}+b^{2}$ is $\frac{16}{5}$.