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

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 $$a_n \pi^n + a_{n-1}\pi^{n-1}+\cdots+a_0\pi^0 = e$$

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I believe the answer to this is not known. – Scaramouche Jul 31 '13 at 3:31
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 3:46
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 3:53
@WillJagy Did you receive my latest mail (a .pdf file)? – Pedro Tamaroff Jul 31 '13 at 3:57
According to Wikipedia, this is indeed an open problem en.wikipedia.org/wiki/… – Omnomnomnom Jul 31 '13 at 4:34

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".