Tell HN: 2^4 == 4^2, extended to rationals

Last week there was a discussion about 2^4 == 4^2 (my search-fu is letting me down - I can't find it again, or I'd link it). It proved that that is the only pair of integers with that relationship.

But what about rationals?

First, remember that (a^b)^c) == a^(bc).

Let a = (3/2)^2 and b = (3/2)^3. Then a^b == b^a. (a and b are rational, but a^b is not.)

More generally, for any natural number n, let a = ((n+1)/n)^n and b = ((n+1)/n)^(n+1). Then a^b == b^a.

Note that the first element of this series has a = 2 and b = 4. Also note that as n increases, the series converges to both a and b equal to e.

But when I said "for any natural number n", that was actually an unnecessary restriction. The math works for any n - it doesn't have to be an integer. We could, for instance, use n = 1/2. Then a = ((3/2)/(1/2))^(1/2) and b = ((3/2)/(1/2))^(3/2). a and b are no longer rational, but a^b still equals b^a.

There are infinete number of rationals that satisfy the condition.

The integer uniquenes proof relied on the shape of the function between 1 and e non inclusive, and 2 is the only integer in the open interval.

I feel rationals is probably the easiest case to prove, still clever though! Interesting (though not surprising) how there's a countably infinite number of rationals that satisfy this.

If we exclude 0, the only solutions are 2^1 = 1^2 + 1 and 3^2 = 2^3 + 1

For general solution you can check https://math.stackexchange.com/questions/3072828/why-didnt-t..., which is related to Lambert W function.