Find all natural numbers $a$ and $b$ such that

$$
a^{b}+b^{a}=10 \cdot b^{a-2}+100
$$
$$
a^{b}+b^{a}=10 \cdot b^{a-2}+100
$$

Transform (1) in the following way:

$$
a^{b}=b^{a-2}\left(10-b^{2}\right)+100
$$

First, we seek solutions where $\left(10-b^{2}\right)>0$, that is, $b=1,2$ or 3. In the first case, (2) yields $a=109$, so we already have one solution: $a=109, b=1$.

In the second case, (2) gives us the equation

$$
a^{2}=6 \cdot 2^{a-2}+100
$$

Since $6 \cdot 2^{a-2}$ is positive, only values of $a$ greater than 10 are possible here, but

$$
a^{2}4,
$$

which, due to $2^{a}100$, implies $a>4, a=5,6$ do not satisfy the equation, while $a=7$ provides another solution to (1): $a=7, b=3$. If $a \geq 8$, the left side of (5) is smaller than the right side, since for $a \geq 8$,

$$
a^{3}0$ case. We start the examination of the $\left(10-b^{2}\right)<0$, that is, $b \geq 4$ case by examining small values of $a$. If $a=1$, (1) represents the equation

$$
b=\frac{10}{b}+99
$$

which clearly has no solution in the natural numbers. If $a=2$, (1) gives us the equation

$$
2^{b}+b^{2}=110
$$

Here, $2^{b}<110$ implies $b \leq 6$, but then $2^{b}+b^{2} \leq 2^{6}+6^{2}=100$, so (7) has no solution in the natural numbers. Finally, if $a \geq 3$ and $b \geq 4$, then the equation equivalent to (2),

$$
a^{b}+b^{a-2}\left(b^{2}-10\right)=100
$$

has no solution, since on the left side

$$
a^{b} \geq 3^{4}=81, \quad b^{a-2} \geq 4, \quad b^{2}-10 \geq 6,
$$

thus the value of the left side is at least $81+4 \cdot 6=105$.

In total, we have found two solutions, the pairs $a=109, b=1$, and $a=7, b=3$.