Ordered Pairs!

We see that the equation is symmetrical in a way that the values of $a, b$ are interchangeable.

Without a loss in generality, we can let $a\geq b$

$\therefore a^b+b^a\geq b^b+b^b=2b^b$

$2b^b\leq 100$

$b^b\leq 50$

$b=1,2,3$

When $b=1$ , $a^1+1=100 \Rightarrow a=99$

When $b=2$ ,

$a^2+2^a=100$

$a\geq b=2$

Also, $a<7$ as $2^7=128>100$

After some trials, we can verify that $a=6$ .

When $b=3$ ,

$a^3+3^a=100$

$a\geq b=3$

$a<5$ as $5^3=125>100$

After some trials, we can verify that there is no solution.

Summing up, we have

$a=99,b=1$

$a=6,b=2$

$a=2,b=6$

$a=1,b=99$

The additional two solutions are due to interchangeability of values.

6^2+2^6 = 100 : Possible ordered pairs (2,6), (6,2)

1^99+99^1 = 100 : Possible ordered pairs (1,99), (99,1)

Since 2^6 + 6^2 = 100 we have two ordered solution pairs.

But it took me a while longer to see that we must also include 99^1 + 1^99 for two more ordered pairs.