PRMO 2019 question 19

Prerequisites:
Divisibility by 13.

11. Find the largest value of $a^b$ such that the positive integers $a, b > 1$ satisfy

$$a^b \cdot b^a + a^b + b^a = 5329$$

Solution:
Rewrite the given equation

$$a^b\,b^a \;+\; a^b \;+\; b^a \;=\; 5329$$

as

$$(a^b+1)(b^a+1)\;=\;5329 + 1 \;=\;5330.$$

Thus if we set $x=a^b$ and $y=b^a,$ then $(x+1)(y+1)=5330.$

Next one factors $5330$ (which is $2 \times 5 \times 13 \times 41$) and looks for factor‐pairs $(p,q)$ of $5330$ such that $x=p-1$ and $y=q-1$ can both be positive perfect powers $a^b$ and $b^a$. Among the possible divisors, the only workable pairs turn out to be

$$(p,q) \;=\; (65,\,82)\quad\text{and}\quad(82,\,65),$$

which give

$$x=64,\;y=81\quad\text{or}\quad x=81,\;y=64.$$

Translating back to $(a,b)$, these correspond to

$$(a,b)=(4,3)\quad\text{(giving }a^b=64\text{)} \quad\text{or}\quad (a,b)=(3,4)\quad\text{(giving }a^b=81\text{).}$$

Since the problem asks for the largest value of $a^b$, the answer is

$$\boxed{{\displaystyle 81}}\mathrm{.}$$

Notes:
A key point here is how to check if 533 is divisible by 13.
One way is to see that it's close to 520 and if we add 13 to it we get 533.
Another is to remember the 13 divisibility rule, multiply the last digit with 9 and subtract from the remaining integer. If the number thus obtained is divisible by 13 the original number is also divisible by 13.