Schanuel’s conjecture and the Mandelbrot Competition

April 19, 2012 by Qiaochu Yuan

A student I’m tutoring was working unsuccessfully on the following problem from the 2011 Mandelbrot Competition:

Let $a, b$ be positive integers such that $\log_a b = (\log 23)(\log_6 7) + \log_2 3 + \log_6 7$ . Find the minimum value of $ab$ .

After some tinkering, I concluded that the problem as stated has no solution. I am now almost certain it was printed incorrectly: $\log 23$ should be replaced by $\log_2 3$ , and then we can solve the problem as follows:

$\log_a b + 1 = (\log_2 3 + 1)(\log_6 7 + 1) = \log_2 6 \log_6 42 = \log_2 42$ .

It follows that $\log_a b = \log_2 21$ . Since $a, b$ are positive integers we must have $a \ge 2$ , and then it follows that the smallest solution occurs when $a = 2, b = 21$ . But what I’d like to discuss, briefly, is the argument showing that the misprinted problem has no solution.

Transcendental number theory

A general problem of the above type would appear at first glance to be quite difficult. However, conditional on Schanuel’s conjecture, these problems are in principle relatively straightforward.

Schanuel’s conjecture asserts that if $z_1, ... z_n$ are complex numbers which are linearly independent over $\mathbb{Q}$ , then

$\mathbb{Q}(z_1, ... z_n, e^{z_1}, ... e^{z_n})$

has transcendence degree at least $n$ over $\mathbb{Q}$ . It generalizes, among other things, both the Lindemann-Weierstrass theorem (the case where the $z_i$ are all algebraic) and the Gelfond-Schneider theorem (the case where $n = 2, z_1 = \log \alpha, z_2 = \beta \log \alpha$ for $\alpha \neq 0, 1$ algebraic and $\beta$ algebraic irrational), two basic tools for proving transcendentality of various numbers.

For our purposes we want to specialize Schanuel’s conjecture as follows. If $p_1, ... p_n$ are distinct primes, then by unique factorization $\ln p_1, ... \ln p_n$ are linearly independent over $\mathbb{Q}$ , so it follows from Schanuel’s conjecture that

$\mathbb{Q}(\ln p_1, ... \ln p_n)$

has transcendence degree exactly $n$ ; in other words, that the logarithms of the primes are algebraically independent. Even this special case of Schanuel’s conjecture is, to my knowledge, wide open. It follows in particular that the ring

$\mathbb{Q}[\ln 2, \ln 3, \ln 5, ...]$

is a polynomial ring on $\ln 2, \ln 3, \ln 5, ...$ .

The misprinted problem

After writing everything in terms of natural logarithms of primes and clearing denominators, the misprinted problem is to find positive integers $a, b$ such that the polynomial

$\displaystyle \ln b (\ln 2 + \ln 5) (\ln 2 + \ln 3) \ln 2$

in $\mathbb{Q}[\ln 2, \ln 3, ... ]$ is equal to the polynomial

$\displaystyle \ln a \left( \ln 23 \ln 7 \ln 2 + (\ln 2 + \ln 5)(\ln 2 + \ln 3) \ln 3 + \ln 7 (\ln 2 + \ln 5) \ln 2 \right)$ .

But the polynomial ring in countably many variables is a UFD and $\ln a, \ln b$ are both linear, hence irreducible, so the part of the RHS not containing $\ln a$ is necessarily divisible by either $\ln 2$ or $\ln 2 + \ln 5$ , and it is straightforward to verify that neither of these holds. Thus there are no such $a, b$ .

The correct problem

Especially since the term $\log_2 3$ already appears in the problem, the likeliest typo was that $\log 23$ actually meant $\log_2 3$ . In that case, after writing everything in terms of natural logarithms of primes and clearing denominators, the correct problem is to find positive integers $a, b$ such that

$\displaystyle \ln b (\ln 2 + \ln 3) \ln 2 = \ln a \left( \ln 7 \ln 3 + (\ln 2 + \ln 3) \ln 3 + \ln 7 \ln 2 \right)$ .

This time the RHS is divisible by $\ln 2 + \ln 3$ , and dividing out gives $\ln b \ln 2 = \ln a \ln 21$ as before.

Posted in math.AC | 5 Comments

5 Responses

on May 30, 2012 at 7:15 pm | Reply John Baez

Is ‘Gelfond’ of the Gelfond-Schneider theorem a different guy than the famous mathematician ‘Gelfand’? I guess so… but who is he? By the way, the name ‘Gelfand’ is etymologically related to ‘elephant’.

Also by the way, if you don’t know Zilber’s truly awesome work on Schanuel’s conjecture, you would like these slides, which explain it:

Click to access BerkeleyLC09.pdf

Zilber wrote down a list of axioms for addition, multiplication and exponentiation that the complex numbers might satisfy – including Schanuel’s conjecture.

Then, he proves that there’s exactly one structure obeying these axioms that has the same cardinality as the complex numbers. As a field, this structure is isomorphic to the complex numbers. But the exponentiation might be nonstandard.

So, either the complex numbers with its usual exponential obeys Schanuel’s theorem… or it does with some weird ‘pseudoexponentiation’ that has lots of the same properties as the standard one!

This is a nice example of how one can make a result seem very plausible without quite proving it. Bravo!
- on May 30, 2012 at 7:53 pm | Reply Qiaochu Yuan
  
  Wikipedia gives Gelfond’s first name as Alexander. That’s interesting!
on May 8, 2012 at 3:03 pm | Reply hyh

For the misprinted problem, the lower bound of an empty set is actually ∞ (infinity)… 🙂
on May 7, 2012 at 7:31 pm | Reply Konstantinos

Isn’t that a high school level competition?
- on May 7, 2012 at 7:35 pm | Reply Qiaochu Yuan
  
  Yes. The solution to the correct problem is completely elementary.

Comments RSS

	ZW on The double commutant theo…
	auto on More about the Schrödinger equ…
	Jaromir Kuben on Meditation on the Sylow theore…
	Qiaochu Yuan on Meditation on the Sylow theore…
	Qiaochu Yuan on Some remarks on the infinitude…

Annoying Precision

"A good stock of examples, as large as possible, is indispensable for a thorough understanding of any concept, and when I want to learn something new, I make it my first job to build one." – Paul Halmos