Having just finished work on a major project milestone, I took it easy for a few days, allowing myself to spend time thinking about other things. That’s when I encountered an absolutely neat problem on Quora.

Someone asked a seemingly innocuous number theory question: are there two positive integers such that one is exactly the π-th power of the other?

Now wait a minute, you ask… We know that π is a transcendental number. How can an integer raised to a transcendental power be another integer?

But then you think about \(\alpha=\log_2 3\) and realize that although \(\alpha\) is a transcendental number, \(2^\alpha=3\). So why can’t we have \(n^\pi=m\), then?

As it turns out, we (probably) cannot, but the reason is subtle and it relies on a very important, but unproven conjecture from transcendental number theory.

But first, let us rewrite the equation by taking its logarithm:

$$\pi\log n = \log m.$$

We can also divide both sides by \(\log n\), which leads to

$$\pi = \frac{\log m}{\log n}=\log_n m,$$

but it turns out to be not very helpful. However, squaring the equation will help, as we shall shortly see:

$$\pi^2\log^2 n=\log^2 m.$$

Can this equation ever by true for positive integers \(n\) and \(m\), other than the trivial solution \(n=m=1\), that is?

To see why it cannot be the case, let us consider the following triplet of numbers:

$$(i\pi,\log n,\log m),$$

and their exponents,

$$(e^{i\pi}=-1, e^{\log n}=n, e^{\log m}=m).$$

The three numbers \((i\pi,\log n,\log m)\) are linearly independent over \({\mathbb Q}\) (that is, the rational numbers). What this means is that there are no rational numbers \(A, B, C, D\) such that \(Ai\pi+B\log n+C\log m + D=0\). This is easy to see as the ratio of \(\log n\) and \(\log m\) is supposed to be transcendental but both numbers are real, whereas \(i\pi\) is imaginary.

On the other hand, their exponents are all rational numbers (\(-1, n, m\)). And this is where the unproven conjecture, Schanuel’s conjecture, comes into the picture. Schanuel’s conjecture says that given \(n\) complex numbers \((\alpha_1,\alpha_2,…,\alpha_n)\) that are linearly independent over the rationals, out of the \(2n\) numbers \((\alpha_1,…,\alpha_n,e^{\alpha_1},…,e^{\alpha_n})\), at least \(n\) will be transcendental numbers that are algebraically independent over \({\mathbb Q}\). That is, there is no algebraic expression involving roots and powers of the \(\alpha_i\), \(e^{\alpha_i}\), and rational numbers that will yield 0.

The equation \(\pi^2\log^2 n=\log^2 m\), which we can rewrite as

$$(i\pi)^2\log^2 n + \log^2 m=0,$$

is just such an equation, and it can never be true.

I wish I could say that I came up with this solution but I didn’t. I was **this** close: I was trying to apply Schanuel’s conjecture, and I was of course using the fact that \(\pi=-i\log -1\). But I did not fully appreciate the implications and meaning of Schanuel’s conjecture, so I was applying it improperly. Fortunately, another Quora user saved the day.

Still I haven’t had this much fun with pure math (and I haven’t learned this much pure math all at once) in years.