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It is well-known that the logarithms of prime numbers are linearly independent over $\mathbb Q$. It is also known that the question whether the logarithms are algebraically independent over $\mathbb Q$ is an open problem.

What is known about the next to linear by complexity case? Are the logarithms of primes quadratically independent over $\mathbb Q$, i.e. $$\sum_{ij\le N}a_{ij} \log p_i \log p_j = 0, \quad a_{ij} \in \mathbb Q, \quad \implies a_{ij} = -a_{ji} $$?

user
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    It's clear what you mean by the title, but you should fix the equation since e.g. $\log(2)\log(3) - \log(3)\log(2) = 0$ is not interesting. – Ricardo Buring Mar 22 '19 at 13:51
  • @RicardoBuring Thank you for catching this. In fact the motivation for my question was exactly the implication $a_{ij}=-a_{ji}$. The question is edited to clarify this point. – user Mar 22 '19 at 15:07

1 Answers1

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It seems highly probable that this is an open question, for the following indirect reason.

For non-negative integers $m_p, n_p$ ($p$ prime), almost all zero, if $a = \prod_pp^{m_p}$ and $b = \prod_pp^{n_p}$, then \begin{multline*} \log_2a = \log_3b \iff \frac{\sum_p m_p\log{p}}{\log2} = \frac{\sum_p n_p\log{p}}{\log3} \\ \iff -n_2(\log2)^2 + (m_2 - n_3)\log2\log3 + m_3(\log3)^2 \\ - \sum_{p\geqslant5}n_p\log2\log{p} + \sum_{p\geqslant5} m_p\log3\log{p} = 0, \end{multline*} and if the logarithms of the primes were known to be quadratically independent over $\mathbb{Q}$, this would imply $a = 2^n$, $b = 3^n$ for some non-negative integer $n$; but as this would settle the notorious open problem If $2^x $and $3^x$ are integers, must $x$ be as well?, someone would surely have noticed by now!

Calum Gilhooley
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    And you are right. Such a connection would be noticed: M. Waldschmidt, in "Colloquium De Georgi 2013 and 2014", p. 134. – user Mar 24 '19 at 19:30
  • Thank you for the reference. Google wouldn't let me look, but there is a PDF copy here. From page 6: "Even the nonexistence of quadratic relations among logarithms of algebraic numbers is not proved. For instance, Schanuel's Conjecture implies that a relation like $\log\alpha_1\log\alpha_2=\log\alpha_3$ among nonzero logarithms of algebraic numbers is not possible. A special case would be the transcendence of the number $e^{\pi^2}$ - it is not yet proved that this number is irrational." – Calum Gilhooley Mar 24 '19 at 20:16
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    Of course the non-existance of quadratic relations among logarithms of integer numbers can in principle be a much easier problem, similarly as in the linear case (cf. Baker's theorem for the general algebraic case). However there is another remark: "A very special case [of the nonexistence of nontrivial homogeneous quadraticrelations between logarithms of algebraic numbers], which is open yet, is to prove that if $t$ is a real number such that $2^t$ and $3^t$ are integers, then $t$ is an integer". – user Mar 24 '19 at 20:30
  • Then vote this :-) ahahah.... – Sebastiano Feb 09 '21 at 23:03