15. Find all natural numbers $a$ and $b$ that satisfy the following equation: $\left[\frac{a^{2}}{b}\right]+\left[\frac{b^{2}}{a}\right]=\left[\frac{a^{2}+b^{2}}{a b}\right]+a b$.
15. Solution: By symmetry, we may assume $a \leqslant b$.

From $x-10$
In (1), if $b \geqslant a^{2}+2$, then $b^{3}-\left(a^{2}+1\right) b^{2}-2 a b+a^{3}-a^{2}$
$=b\left[b\left(b-a^{2}-1\right)-2 a\right]+a^{3}-a^{2} \geqslant b(b-2 a)+a^{3}-a^{2}$
$=(b-a)^{2}+a^{3}-2 a^{2} \geqslant\left(a^{2}-a+2\right)^{2}+a^{3}-2 a^{2}$
$=\left(a^{2}-a+2\right)^{2}-a^{2}+a^{2}(a-1)=\left(a^{2}+2\right)\left(a^{2}-2 a+2\right)+a^{2}(a-1)>0$, which contradicts (1).
Therefore, $b \leqslant a^{2}+1$.
In (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 $f(a)=a^{2}\left(-a^{2}+2 a-1\right)0$.
Thus, the equation $f(x)=0$ has one root in each of the intervals $(-a, 0)$, $(0, a)$, and $\left(a^{2}, a^{2}+1\right)$.
Hence, when $a \leqslant b \leqslant a^{2}$, $f(b)0 \Leftrightarrow b(b-1)^{2}>0$, so $b \geqslant 2=a^{2}+1$.
In summary, regardless of the value of $a$, we have $b \geqslant a^{2}+1$.
By $b \leqslant a^{2}+1$ and $b \geqslant a^{2}+1$, we know $b=a^{2}+1$.
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 is $a^{3}+2 a=a+a\left(a^{2}+1\right)$, which is an identity.
When $a=1$, the above equation is $\left[\frac{1}{2}\right]+\left[\frac{1+2+1}{1}\right]=\left[\frac{2}{1}+\frac{1}{2}\right]+(1+1)$, which is clearly true.
Therefore, when $a \leqslant b$, the solution to the original equation is $b=a^{2}+1, a \in \mathbf{N}$.
Thus, the solutions to the original equation are $b=a^{2}+1, a \in \mathbf{N}$, or $a=b^{2}+1, b \in \mathbf{N}$.