# Can we find two positive integers $n$ and $m$ ($n,m>1$) such that $n^\pi = m$? [duplicate]

I came across this apparent random question in some math questions website. At first, I thought it was easy to show that there are no non-trivial integer solutions to this equation, but then I realized that the question is far beyond what I can answer.

Irrationality and transcendence of $\pi$ play no role I think, because if we change $\pi$ with $\log_2(9)$ there are non-trivial solutions. My question is, Do you know some tool to attack this kind of problem?

I'll immediately delete this question if you think it's outside the scope of MathOverflow, but I know there are really knowledgeable people here who could say something about this, and my "innocent" curiosity for this question made me post it here. Thank you for your attention.

PS: more accurate tags for the question are also welcomed.

• This follows from Schanuel's conjecture but it's probably hard to prove unconditionally. – Felipe Voloch Jun 14 '15 at 22:57
• @FelipeVoloch If is not too much to ask, could you outline the argument of why this should follow from Schanuel's conjeture? transcendence theory it's not my area. Thanks. – Héctor Jun 14 '15 at 23:03
• Apply Schanuel to $2\pi i, \log n, \log m$. The last two numbers are linearly independent over $\mathbb{Q}$ because of your hypothesis and the fact that $\pi$ is irrational. Then all three numbers are linearly independent over $\mathbb{Q}$ since $2\pi i$ is not real. Finally, the exponentials of all three numbers are rational, so Schanuel implies that the three numbers are algebraically independent over $\mathbb{Q}$ which is stronger than what you want. – Felipe Voloch Jun 14 '15 at 23:11
• @FelipeVoloch Thanks, very clear argument, i didn't see that. This satisfy me as an answer, but i'll leave this open in case someone has some great idea of how to show this without Schanuel. – Héctor Jun 14 '15 at 23:21

This follows from Schanuel's conjecture but it's probably hard to prove unconditionally.

Apply Schanuel to $2\pi i,\log n,\log m$. The last two numbers are linearly independent over $\mathbb Q$ because of your hypothesis and the fact that $\pi$ is irrational. Then all three numbers are linearly independent over $\mathbb Q$ since $2\pi i$ is not real. Finally, the exponentials of all three numbers are rational, so Schanuel implies that the three numbers are algebraically independent over $\mathbb Q$ which is stronger than what you want.

Unconditionally, this can essentially happen at most once. It is a consequence of the Five Exponentials Theorem, which itself follows from a combination of the classical Six Exponentials Theorem and Baker's Theorem on linear forms in logarithms.

Five Exponentials Theorem: (See Waldschmidt, Diophantine Approximation on Linear Alebgraic Groups, 2000, Section 11.3.3)

Suppose that $\lambda_0, \lambda_1, \lambda_2 \in \mathbb{C}^{\times}$ are chosen so that $e^{\lambda_i} \in \overline{\mathbb{Q}}$ for each $i = 0, 1, 2$. Suppose that $\lambda_1$ and $\lambda_2$ are linearly independent over $\mathbb{Q}$. Then for any non-zero algebraic number $\beta$, at least one of the two numbers $$e^{\beta \lambda_0 \lambda_1}, \quad e^{\beta \lambda_0\lambda_2}$$ is transcendental.

To apply to the present question, for positive integers $n_1$ and $n_2$ take $$\lambda_0 = \pi i, \quad \lambda_1 = \log n_1, \quad \lambda_2 = \log n_2, \quad \beta = -i,$$ and so $$e^{\beta \lambda_0 \lambda_1} = n_1^{\pi}, \quad e^{\beta \lambda_0 \lambda_2} = n_2^{\pi}.$$ By the Five Exponentials Theorem, if both $n_1^{\pi}$ and $n_2^{\pi}$ are algebraic, then $n_1$ and $n_2$ must be multiplicatively dependent. Thus up to multiplicative dependence there is at most one way for a solution to be found.