Example 5 Find all natural numbers $a, b$, such that $\left[\frac{a^{2}}{b}\right]+$
$$
\left[\frac{b^{2}}{a}\right]=\left[\frac{a^{2}+b^{2}}{a b}\right]+a b
$$
$$
\begin{array}{l}
\text { Solution: } \because a, b \text { are symmetric, } \therefore \text { without loss of generality, assume } a \leqslant b, \\
\because x-10 .
\end{array}\right. \\
\end{array}
$$

If $b > a^2 + 1$, then from $0 > b^3 - (a^2 + 1) b^2 -$
$$
\begin{array}{l}
2 a b + a^3 - a^2 = b \left[b (b - a^2 - 1) - 2 a \right] + a^3 - a^2 \\
\geqslant b (b - 2 a) + a^3 - a^2 = (b - a)^2 + a^3 - 2 a^2 \geqslant (a^2 - a + 2)^2 - a^2 + a^3 - a^2 = (a^2 + 2)(a^2 - 2 a + 2)
\end{array}
$$
$+ a^2 (a - 1) > 0$, which is a contradiction, so $b \leqslant a^2 + 1$.
Let $f(x) = x^3 - (a^2 + 1) x^2 + a x + a^3 - a^2$, by $f(b) > 0$, when $a \geqslant 2$, $f(-\infty) < 0, f(a) = a^2 (-a^2 + 2 a - 1) < 0$, hence the cubic equation $f(x) = 0$ has one root in each of $(-\infty, 0)$, $(0, a)$, and $(a^2, a^2 + 1)$. When $a \leqslant b \leqslant a^2$, $f(b) < 0$, which is a contradiction, so $b \geqslant a^2 + 1, b = a^2 + 1$.
Substituting into the original equation, we get: $\left[\frac{a^2}{a^2 + 1}\right] + \left[\frac{a^4 + 2 a^2 + 1}{a}\right] =$
$$
\left[\frac{a^2 + 1}{a} + \frac{a}{a^2 + 1}\right] + a (a^2 + 1).
$$

Clearly, $a = 1$ is a solution. When $a \geqslant 2$, the above expression is $a^3 + 2 a = a + a (a^2 + 1)$, which is an identity.

Therefore, when $a \leqslant b$, the solution to the original equation is $b = a^2 + 1$, $a \in \mathbf{N}^*$, and when $a \geqslant b$, the solution to the original equation is $a = b^2 + 1$, $b \in \mathbf{N}^*$. Thus, the solution to the original equation is $b = a^2 + 1$, $a \in \mathbf{N}^*$ or $a = b^2 + 1$, $b \in \mathbf{N}^*$.