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Do there exist integers $a,b,c$ such that at least one of them is non-zero and $ae+b\pi = c$?

Please note that, I am only asking if e and $\pi$ can satisfy a non-trivial diophantine equation $ax + by = c$. I am not interested in any arbitrary polynomial relation between them. I think this is false. Any help is welcome!

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Shubhamoy Nandan
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1 Answers1

10

If no such integers $a, b, c$ exist, then in particular there are no $a, b, c \in \mathbb{Z}$ with $a = b$ such that $a e + b \pi = c$ so

$$e + \pi \not= \frac{c}{a}$$

for all $a, c \in \mathbb{Z}$. Thus, $e + \pi$ is irrational. According to #22 in the list of open problems linked from this answer, it is not known if $e + \pi$ is irrational, so this would resolve an open problem.

On the other hand, if there exist $a, b, c \in \mathbb{Z}$ (not all zero) such that $a e + b \pi = c$, then since $e$ and $\pi$ are irrational, both $a, b \not= 0$. Then

$$\frac{e}{\pi} = \frac{c}{a \pi} - \frac{b}{a}$$

Since the reciprocal of a transcendental number is transcendental, $\frac{1}{\pi}$ is transcendental. If $c \not= 0$, this implies that $\frac{e}{\pi}$ is transcendental. If $c = 0$, then $\frac{e}{\pi}$ is rational and, therefore, is not transcendental. Determining if $\frac{e}{\pi}$ is transcendental an open problem (by the reference above) which we resolve one way or the other depending on whether or not $c = 0$.

There is also another argument which shows that if there exist $a, b, c \in \mathbb{Z}$ (not all zero) such that $a e + b \pi = c$, then $e \pi$ is transcendental. Consider the quadratic polynomial

$$(x-ae)(x-b\pi) = x^2 - (ae+b\pi)x + (ae)(b\pi)$$

Since its roots $ae$ and $b\pi$ are transcendental, the coefficents $ae+b\pi$ and $(ae)(b\pi)$ cannot both be algebraic. Thus, if $ae + b\pi$ is an integer (and hence algebraic), then $e\pi$ is transcendental. Determining if $e\pi$ is transcendental is also open, so this resolves yet another open problem.

Thus, either the existence or non-existence of the described diophantine relation would resolve an open problem.

Thanks to @hardmath for making comments and edits that greatly improved this answer.

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