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Reflecting on this recent MSE question, I was led to the following conjecture :

Let $A=\left\lbrace x^y \mid x,y\in{\mathbb Q}_+ \right\rbrace$. If $\alpha,\beta,\gamma\in A$ are pairwise $\mathbb Q$-linearly independent (i.e. $(\alpha,\beta),(\alpha,\gamma)$ and $(\beta,\gamma)$ are $\mathbb Q$-linearly independent), then $(\alpha,\beta,\gamma)$ is ("globally") $\mathbb Q$-linearly independent.

Does anyone have an idea on how to prove or disprove this conjecture?

Some thoughts : There are positive rationals $a,b,c$ and an integer $r\geq 2$ such that $\alpha=a^{\frac{1}{r}},\beta=b^{\frac{1}{r}},\gamma=c^{\frac{1}{r}}$. Multiplying by a suitable common denominator, we may assume that $a,b,c$ are integers.

EDIT(10/12/2014) Perhaps one can use induction on $t$, the number of primes diving $abc$. Indeed, the base case $t=1$ follows from the irreducibility of $X^r-p$ for a prime $p$ (remember Eisenstein’s criterion).

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Ewan Delanoy
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I just found that this property follows from a proposition in the following paper by Iurie Boreico(see the "Higher powers" section on page 91) : http://www.thehcmr.org/issue2_1/mfp.pdf.

Note that in Boreico’s theorem all the coefficients in the linear relation are assumed to be nonnegative except the one before $1$. This is not a problem when we have three elements : if $q_1a_1^{\frac{1}{r}}+q_2a_1^{\frac{1}{r}}+q_3a_3^{\frac{1}{r}}=0$ with $q_1,q_2,q_3$ rational, then the $q_i$ are not all of the same sign. So, for example, $q_1$ and $q_2$ share the same sign while $q_3$ is of the opposite sign. Then

$$ \frac{-q_1}{q_3}\bigg(\frac{a_1}{a_3}\bigg)^{\frac{1}{r}}+ \frac{-q_2}{q_3}\bigg(\frac{a_1}{a_3}\bigg)^{\frac{1}{r}}=1, $$

and now we can apply Boreico’s theorem.

Ewan Delanoy
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