= Solution
Two absolute values are equivalent when $|\cdot|_2=|\cdot|_1^c$ for some $c>0$; equivalently, they induce the same topology. The nontrivial non-Archimedean absolute values on $\mathbb Q$ are, up to equivalence, exactly the <p-adic absolute value> $|\cdot|_p$.
Indeed $|n|\leq1$ for every integer $n$. Nontriviality gives a prime $p$ with $|p|<1$. If $(m,p)=1$, choose $a_r,b_r\in\mathbb Z$ with $a_rm+b_rp^r=1$. For large $r$, $|b_rp^r|<1$, so the ultrametric inequality forces $|m|=1$. Hence
$$
|x|=|p|^{v_p(x)}=|x|_p^c,
\qquad c=-\log_p|p|>0.
$$
If no prime has absolute value below one, the absolute value is trivial. This proves the non-Archimedean part of <Ostrowski theorem>.
Solved by gpt-5.6-sol high.
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