Solving $a^b = b^a$ for $a,b \in \Bbb N$ where $a,b$ are distinct

Question

While preparing for the Putnam math competition my teacher listed the following problem: Solving $a^b = b^a$ for $a,b \in \Bbb N$ where $a,b$ are distinct. I suspect the answer is that there is only one solution: $a=2$ and $b=4$. Now I start with the obvious that $b*ln(a) = a*ln(b)$ and clearly since a,b are distinct then either $a<b$ or $a>b$. Now suppose $b>a$ then $\frac{a}{b} = \frac{lna}{lnb}$ . Now here is where I am stuck. Any hints or ideas are much appreciated.

Edit: Upon more thought it seems that a and b most both be even or odd since otherwise $a^b$ would be odd while $b^a$ would be even and vice versa.

Look at $\frac{\ln x}{x}$. Where is it increasing and where is it decreasing? Then the answer for integers will fall out. I am not writing an answer because I am sure I have seen the problem (more than once) on MSE. — André Nicolas, Jul 14 '16 at 04:12
@AndréNicolas it is increasing from $(0,e)$ and decreasing from $(e, \infty)$ — IntegrateThis, Jul 14 '16 at 04:19
That's right, so we can get $x^y=y^x$ with $x\ne y$ only with $x$ before $e$ and $y$ after, or the other way. Thus with integers the only candidates for the smaller one are $x=1$ or $x=2$, and $x=2$ works. — André Nicolas, Jul 14 '16 at 04:21
This provides an answer, albeit not the most direct possible answer. — Ben Grossmann, Jul 14 '16 at 04:26
You can find an entire sequence of solutions for $k=1,2,3,\ldots$ if you assume that $a=kb$. — John Wayland Bales, Jul 14 '16 at 04:35

score 6 · Accepted Answer · answered Jul 14 '16 at 04:38

6

A trivial answer would be $a=b$.

For other solutions, we can assume $a>b$ without loss of generality.

Consider $\left(\frac{a}{b}\right)^b$. Because $a^b=b^a$, we have $\left(\frac{a}{b}\right)^b = b^{a-b}$. Note that $b^{a-b}$ has to be a natural number because $a-b \in \mathbb{N}$. Thus $\left(\frac{a}{b}\right)^b \in \mathbb{N}$.

This allows one to write $a=c b$ for some $c \in \mathbb{N}$. Hence $a^b=b^a$ becomes

$$(cb)^b=b^{cb}$$

which implies

$$b=c^{\frac{1}{c-1}}.$$

This yields a solution $b \in \mathbb{N}$ if and only if $c=2$. Furthermore, this results in $b=2$, $a=4$.

answered Jul 14 '16 at 04:38

K. Ghusinga

128

why can you write a = cb for some $c \in \Bbb N$ – IntegrateThis Jul 16 '16 at 20:55
1

Because $(\frac{a}{b})^b \in \mathbb N$, we have $\frac{a}{b} \in \mathbb N$ as well. Thus, $a$ can be written as $a = c b$. – K. Ghusinga Jul 16 '16 at 22:00
I should add to clarify that $b^{cb} = (b^{c-1}b)^{b} \implies c = b^{c-1} \implies b = c^{1/(c-1)}$ and that $c$ has to be larger than 1. – IntegrateThis Jul 01 '21 at 21:28

Solving $a^b = b^a$ for $a,b \in \Bbb N$ where $a,b$ are distinct

1 Answers1