Competition problem — Number theory

Library / /8 of 42

Number theory Difficulty 5.3 AIME, harder Prove it Romania

Find all positive integers $a$ and $b$ such that $a^{4a} = b^b$ .

Solution

If $b \ge 4a$ , then $b > a$ and $b^b > a^{4a}$ . Therefore, in this case the equality is impossible.

If $b < 4a$ , we have $b^b = a^{4a} = a^{4a-b} a^b$ , so $b^b$ is divisible by $a^b$ . Therefore, $b$ is divisible by $a$ . It follows that $b = n a$ , with $n = 1$ (I), $n = 2$ (II), or $n = 3$ (III).

We get $(a^4)^a = ((n a)^n)^a$ , then $a^4 = n^n a^n$ , or $a^{4-n} = n^n$ . In case (I), we obtain $a = b = 1$ , in case (II) we obtain $a = 2, b = 4$ , while $a = 27, b = 81$ in case (III).

Want a route through all this instead of an archive? The track puts 2,000 problems in a working order, from AMC 10 level to the IMO shortlist.

Source: MathNet, licensed CC-BY-4.0. Statement and solution reproduced as published; topic and difficulty added by this site.