Easy Number Theory

You need no number theory here, except the fact that a divisor cannot be greater than the number itself. In other words,

If $x | y$ then we also have $x \leq y$ , or $x/y \leq 1$ .

Consider the sequence $a/b^2, a^6/b^7, a^{11}/b^{12}, ...$ In each step we multiply by $a^5/b^5$ . If $a>b$ , this is exponential growth, which is unbounded. From a certain point onward, the fractions will be greater than 1, showing that the numerator cannot be a divisor of the denominator. Therefore $a\leq b$ .

Similarly, the second sequence shows that $a\geq b$ .

It follows that $a = b$ .

This implies that for any positive integer $k$ , $a^{5k+1} | b^{5k+2}$ and $b^{5k+4} | a^{5k+5}$ .

Focusing on the first condition, given an arbitrary prime $p$ , we know that $(5k+1)v_{p}(a) \leq (5k+2)v_{p}(b)$

for all $k$ , so that $v_{p}(a) \leq \lim_{k \to \infty}{\frac{5k+2}{5k+1}}v_{p}(b) = v_{p}(b)$

Similarly, the $b^{5k+4} | a^{5k+5}$ condition implies that $v_{p}(b) \leq v_{p}(a)$ . Therefore, $a = b$ which implies that there is only one possible value for $b$ - which is $b = 20$ .