field theory - Are logarithms of prime numbers quadratically independent over $mathbb Q$?

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.
$$\sum_{ij\le N}a_{ij} \log p_i \log p_j = 0, \quad a_{ij} \in \mathbb Q, \quad \implies a_{ij} = -a_{ji} $$?

Answer

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!