Is pi = log_a(b) for some integers a, b > 1? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T23:32:04Z http://mathoverflow.net/feeds/question/118523 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/118523/is-pi-log-ab-for-some-integers-a-b-1 Is pi = log_a(b) for some integers a, b > 1? Stefan Kohl 2013-01-10T11:37:22Z 2013-01-11T19:03:54Z <p>Are there integers $a, b > 1$ such that $\pi = \log_a(b)$?</p> <p>Or equivalently: are there integers $a,b > 1$ such that $a^\pi = b$?</p> <p>Note that the transcendence of $\pi$ makes this a problem - otherwise the Gelfond-Schneider theorem would tell the answer.</p> <p>When I asked this question about 10 years ago, I got an answer by Ignat Soroko (Minsk / Belarus) arguing that the problem is likely out of reach for present methods, cf. <a href="http://www.gap-system.org/DevelopersPages/StefanKohl/problems/a%5Epi=b.dvi" rel="nofollow">http://www.gap-system.org/DevelopersPages/StefanKohl/problems/a%5Epi=b.dvi</a>. Is this still the present state?</p> http://mathoverflow.net/questions/118523/is-pi-log-ab-for-some-integers-a-b-1/118654#118654 Answer by Timothy Chow for Is pi = log_a(b) for some integers a, b > 1? Timothy Chow 2013-01-11T17:49:49Z 2013-01-11T19:03:54Z <p>There is a reason why one can say a little bit more about this question in the case of $\pi$. Because $\pi$ is (essentially) the natural logarithm of a rational number, questions like this are easily derived from Schanuel's conjecture, which states:</p> <blockquote> <p>If $\alpha_1, \alpha_2, \ldots, \alpha_n$ are complex numbers that are linearly independent over $\mathbb Q$, then the transcendence degree of the field $\mathbb{Q}(\alpha_1, e^{\alpha_1}, \alpha_2, e^{\alpha_2}, \ldots, \alpha_n, e^{\alpha_n})$ over $\mathbb Q$ is at least $n$.</p> </blockquote> <p>Your proposed equation is $$\pi = \frac{\ln b}{\ln a}\qquad (*)$$</p> <p>and $(*)$ contradicts the $n=3$ case of Schanuel's conjecture as follows. Take $\alpha_1=\ln a$, $\alpha_2 = \ln b$, and $\alpha_3 = i\pi$. Then $(*)$ implies that $\alpha_1$ and $\alpha_2$ are linearly independent over $\mathbb Q$ (because $\pi$ is irrational), and $\alpha_3$ is trivially linearly independent of $\alpha_1$ and $\alpha_2$ because $\alpha_3$ is purely imaginary and $\alpha_1$ and $\alpha_2$ are real. But $e^{\alpha_1}$, $e^{\alpha_2}$, and $e^{\alpha_3}$ are all rational (even integral) so Schanuel's conjecture implies that $\alpha_1$, $\alpha_2$, and $\alpha_3$ are <i>algebraically</i> independent. This contradicts $(*)$.</p> <p>The $n=3$ case of Schanuel's conjecture is still open. There are various partial results known (see for example the final chapter of Baker's book <i>Transcendental Number Theory</i>), but I very much doubt that any of these partial results suffice to disprove $(*)$. Note that if we take $n=2$ and $\alpha_1 = \ln \beta_1$ and $\alpha_2 = \ln \beta_2$ for nonzero algebraic numbers $\beta_1$ and $\beta_2$ then we recover the Gelfond&ndash;Schneider theorem.</p> <p>It's a good exercise to derive various statements of this type (e.g., that $e+\pi$ or $\pi^e$ or whatever are transcendental) from Schanuel's conjecture. This way you will be able to figure out on your own whether the statement is likely to be known, or at least be able to approach an expert in transcendental number theory with a more targeted question.</p>