$14 \cdot 75$ Find all natural numbers $a$ and
$$
b:\left[\frac{a^{2}}{b}\right]+\left[\frac{b^{2}}{a}\right]=\left[\frac{a^{2}+b^{2}}{a b}\right]+a b .
$$
(37th International Mathematical Olympiad Preliminary Question, 1996)
［Solution］By symmetry, without loss of generality, assume $a \leqslant b$.
From $x-\mathrm{1}0 .\end{array}\right.$
First, consider (1),
If $b \geqslant a^{2}+2$, then
$$
\begin{aligned}
& b^{3}-\left(a^{2}+1\right) b^{2}-2 a b+a^{3}-a^{2} \\
= & b\left[b\left(b-a^{2}-1\right)-2 a\right]+a^{3}-a^{2} \geqslant b(b-2 a)+a^{3}-a^{2} \\
= & (b-a)^{2}+a^{3}-2 a^{2} \geqslant\left(a^{2}-a+2\right)^{2}+a^{3}-2 a^{2} \\
= & \left(a^{2}-a+2\right)^{2}-a^{2}+a^{2}(a-1) \\
= & \left(a^{2}+2\right)\left(a^{2}-2 a+2\right)+a^{2}(a-1)>0 .
\end{aligned}
$$

This contradicts (1). Therefore, $b \geqslant a^{2}+2$ is not true, so $b \leqslant a^{2}+1$.
Next, consider (2),
Let $f(x)=x^{3}-\left(a^{2}+1\right) x^{2}+a x+a^{3}-a^{2}$.
When $a \geqslant 2$, we have $f(-a)=-a^{3}-\left(a^{2}+1\right) a^{2}-a^{2}+a^{3}-a^{2}0$,
and $f(a)=a^{2}\left(-a^{2}+2 a-1\right)0$.
Therefore, the equation $f(x)=0$ has three roots in the intervals $(-a, 0),(0, a),\left(a^{2}, a^{2}+\right.$ 1) each.
Hence, when $a \leqslant b \leqslant a^{2}$, we have $f(b)0$,
i.e., $b(b-1)^{2}>0$, so $b \geqslant 2=a^{2}+1$.
In summary, if (2) holds, then $b \geqslant a^{2}+1$.
Therefore, we can only have $b=a^{2}+1$.
Substituting $b=a^{2}+1$ into the original equation, we get
$$
\left[\frac{a^{2}}{a^{2}+1}\right]+\left[\frac{a^{4}+2 a^{2}+1}{a}\right]=\left[\frac{a^{2}+1}{a}+\frac{1}{a^{2}+1}\right]+a\left(a^{2}+1\right) .
$$

When $a \geqslant 2$, the above equation simplifies to $a^{3}+2 a=a+a\left(a^{2}+1\right)$, which is an identity.
When $a=1$, the above equation simplifies to
$$
\left[\frac{1}{2}\right]+\left[\frac{1+2+1}{1}\right]=\left[\frac{2}{1}+\frac{1}{2}\right]+(1+1)
$$

which is clearly true.
Therefore, when $a \leqslant b$, all solutions to the original equation are
$$
b=a^{2}+1, a \in N \text {. }
$$

Similarly, when $b \leqslant a$, all solutions to the original equation are
$$
a=b^{2}+1, b \in N \text {. }
$$

Therefore, all solutions to the original equation are
$$
b=a^{2}+1, a \in N \text { or } a=b^{2}+1, b \in N \text {. }
$$