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

  • 15
    $\begingroup$ I believe the answer to this is not known. $\endgroup$ Jul 31, 2013 at 3:31
  • 4
    $\begingroup$ 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. $\endgroup$
    – Will Jagy
    Jul 31, 2013 at 3:46
  • 1
    $\begingroup$ It's not known for coefficients from $\Bbb{Q}$: isn't this equivalent to coefficients from $\bar{\Bbb{Q}}$? $\endgroup$ Jul 31, 2013 at 3:53
  • 5
    $\begingroup$ According to Wikipedia, this is indeed an open problem en.wikipedia.org/wiki/… $\endgroup$ Jul 31, 2013 at 4:34

1 Answer 1


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

  • 1
    $\begingroup$ Feel free to update this answer with more useful/reliable sources. $\endgroup$ Jul 31, 2013 at 4:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.