22. Find all natural numbers $a$ and $b$ that satisfy the following equation:
$$
\left\lceil\frac{a^{2}}{b}\right\rceil+\left\lceil\frac{b^{2}}{a}\right\rceil=\left\lceil\frac{a^{2}+b^{2}}{a b}\right\rceil+a b .
$$
Solution: By symmetry, without loss of generality, assume $a \leqslant b$.
Given $x-10 .\end{array}\right.$
First, consider (1). If $b \geqslant a^{2}+2$, then
$$
\begin{array}{l}
b^{3}-\left(a^{2}+1\right) b^{2}-2 a b+a^{3}-a^{2} \\
=b\left[b\left(b-a^{2}-1\right)-2 a\right]+a^{2}-a^{2} \\
\geqslant b(b-2)-a^{2}+a^{2} \\
=(b-a)^{2}+a^{3}-2 a^{2} \\
\geqslant\left(a^{2}-a+2\right)^{2}+a^{3}-2 a^{2} \\
=\left(a^{2}-a+2\right)^{2}-a^{2}+a^{2}(a-1) \\
=\left(a^{2}+2\right)\left(a^{2}-2 a+2\right)+a^{2}(a-1) \\
>0 .
\end{array}
$$

This contradicts (1).
Thus, $b \leqslant a^{2}+1$.
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 $f(-x)0$,
$$
\begin{array}{l}
f(a)=a^{2}\left(-a^{2}+2 a-1\right)0 .
\end{array}
$$

Therefore, $f(x)=0$ has one root in each of the intervals $(-x, 0)$, $(0, a)$, and $\left(a^{2}, a^{2}+1\right)$, and $f(x)=0$ has at most three roots. Hence, when $a \leqslant b \leqslant a^{2}$, $f(b)0$,
$$
i.e., b(b-1)^{2}>0.
$$

Thus, $b \geqslant 2=a^{2}+1$.
In summary, if (2) holds, then $b \geqslant a^{2}+1$.
Combining the above, the only solution is $b=a^{2}+1$. Substituting into the original equation, we get
$$
\begin{array}{l}
\left\lceil\frac{a^{2}}{a^{2}+1}\right\rceil+\left\lceil\frac{a^{4}+2 a^{2}+1}{a}\right\rceil \\
\left.\quad=\left\lceil\frac{a^{2}+1}{a}+\frac{1}{a^{2}+1}\right\rceil+a\left(a^{2}+1\right) .
\end{array}
$$

Verification shows that $a=1$ is a solution to the above equation.
When $a \geqslant 2$, the above equation simplifies to
$$
a^{3}+2 a=a+a\left(a^{2}+1\right),
$$

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

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

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