Find all pairs of rational numbers $(a,b)$ such that $0 < a < b$ and $a^a = b^b$ .

Solution 1

We claim that there is only one solution, which is the solution pair $(a,b)=(\frac{1}{4},\frac{1}{2})$ .

Let $b=ta$ , where $t$ is a rational number greater than $1$ . Then when we substitute that into the equation $a^a=b^b$ , we get $a^a={ta}^{ta}$ . We want to separate the variables from each other, so we can multiply by $\frac{1}{a}$ to get $a={ta}^t \implies a = t^t \cdot a^t$ . Moving $a$ to the other side, and using the exponent rule, we get $a^{1-t}=t^t$ . Multiplying this by $\frac{1}{1-t}$ , we get $a=t^{\frac{t}{1-t}}$ . Substituting that into $b=ta$ , we get $b=t \cdot t^{\frac{t}{1-t}}$ , so $b=t^{\frac{1}{1-t}}$ . Therefore our solution set is $(t^{\frac{t}{1-t}},t^{\frac{1}{1-t}}$ , but wait! The solution set contains irrational numbers, and the question only asks for rational numbers. Therefore, $\frac{t}{1-t}$ and $\frac{1}{1-t}$ both need to be integers. By inspection, we can see that the only solution is $t=2$ , giving us the solution set $(\frac{1}{4},\frac{1}{2})$ .