Are $\pi$ and $e$ algebraically independent?

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Update Edit : Title of this question formerly was "Is there a polynomial relation between $e$ and $\pi$?"

Is there a polynomial relation (with algebraic numbers as coefficients) between $e$ or $\pi$ ? For example does there exists algebraic numbers $a_1,a_2,..,a_n$ s.t. $$a_n e^n + a_{n-1}e^{n-1}+\cdots+a_0e^0 = \pi$$ or $$a_n \pi^n + a_{n-1}\pi^{n-1}+\cdots+a_0\pi^0 = e$$

I'm looking up things, I don't see anything decisive, but it's fair to say that the collection of similar problems that can be resolved is vanishingly small. — Will Jagy, Jul 31 '13 at 03:46
It's not known for coefficients from $\Bbb{Q}$: isn't this equivalent to coefficients from $\bar{\Bbb{Q}}$? — Kevin Carlson, Jul 31 '13 at 03:53
According to Wikipedia, this is indeed an open problem http://en.wikipedia.org/wiki/Algebraic_independence#Algebraic_independence_of_known_constants — Ben Grossmann, Jul 31 '13 at 04:34

score 32 · Accepted Answer · edited Apr 13 '17 at 12:58

32

According to Wikipedia, this is an open problem (as of $17$ years ago, anyway). A common phrase to describe the question (which will help with searches) is "are $\pi$ and $e$ algebraically independent".

an important related problem is the validity of Schanuel's conjecture.

A related thread from over at MO

edited Apr 13 '17 at 12:58

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answered Jul 31 '13 at 04:38

Ben Grossmann

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Feel free to update this answer with more useful/reliable sources. – Ben Grossmann Jul 31 '13 at 04:53

Are $\pi$ and $e$ algebraically independent?

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