= Solution
An additive <valuation> on a <field> is a map $v:K^\times\to\Gamma$, where $\Gamma$ is a totally ordered abelian group, with $v(xy)=v(x)+v(y)$ and $v(x+y)\geq\min(v(x),v(y))$ whenever $x+y\ne0$. Set $v(0)=\infty$. We consider nontrivial <valuations>; if the trivial <valuation> is allowed, it must be listed separately from the asserted classification. Two <valuations> are equivalent when their <valuation rings> agree, or equivalently their ordered value-group images identify in a way compatible with the maps. Real-valued equivalent nontrivial <valuations> differ by positive scaling.
For a <valuation> on $\mathbb Q$, every integer has nonnegative value, because it is a sum of copies of $1$ or its negative. If all <prime numbers> had value zero, <unique factorization> would make the <valuation> trivial. Thus some prime $p$ has $v(p)>0$. There cannot be two such primes: for distinct $p,q$, <Bézout's identity> gives $ap+bq=1$ with integers $a,b$, and the <valuation> inequality would give $v(1)>0$. All other prime values therefore vanish. Factoring the numerator and denominator of a rational number gives
$$
\boxed{v(x)=v(p)\,v_p(x)\quad(x\in\mathbb Q^\times).}
$$
The image is the cyclic ordered group generated by $v(p)>0$, proving equivalence to the <P-adic valuation>. This is the <classification of nontrivial valuations on the rational numbers>; an Archimedean absolute value is not an additive <valuation> satisfying this ultrametric inequality.
The simple-root form of <Hensel's lemma> says that for a <complete discrete valuation ring> $\mathcal O$ with maximal ideal $\mathfrak p$, a polynomial $f\in\mathcal O[X]$ and $a\in\mathcal O$ with $f(a)\equiv0\pmod{\mathfrak p}$ and $f'(a)\not\equiv0\pmod{\mathfrak p}$ have a unique root in $a+\mathfrak p$.
For $X^3+1$ over $\mathbb Q_7$, all roots are integral: a negative <valuation> would make the leading term the unique term of lowest <valuation>. Modulo $7$, the roots are $3,5,6$, and $3X^2$ is nonzero at each. <Hensel's lemma> gives three distinct lifts, and a cubic has no further roots. \b[The number is $3$.]
For $X^2+2X+4$ over $\mathbb Q_2$, a root is again integral and reduces to zero modulo $2$. Write it as $2y$ with $y\in\mathbb Z_2$. The equation becomes $y^2+y+1=0$, impossible modulo $2$. \b[The number is $0$.]
For $3X^3+X+3$ over $\mathbb Q_3$, a putative root of <valuation> $k<0$ gives term <valuations> $1+3k,k,1$. The first is uniquely minimal, a contradiction. Thus all roots are integral. Reduction modulo $3$ is $X$, whose root zero is simple since the derivative $9X^2+1$ is a unit everywhere. <Hensel's lemma> gives exactly one lift. \b[The number is $1$.]
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