Result: ${v: \mathbb{Q}^* \rightarrow \mathbb Z}$ is a discrete valuation satisfying the following properties:

• ${v}$ is surjective.
• ${v(ab) = v(a) + v(b) \quad \text{for all} \quad a,b\in \mathbb Q^*}$
• ${v(a+b) \geq \{ v(a), v(b) \} \quad \text{provided} \quad a+b \neq 0}$

Then $v=v_p$ for some prime $p$, given by, $v_p\displaystyle\left(p^r \displaystyle\frac{a}{b} \right) = r$ for $(a,p)=(b,p)=1$.

Proof: It is a fact (cf. Dummit & Foote Ex 39 Sec. 7.4) that ${R = \{ x \in \mathbb{Q}^* : v(x) \geq 0 \}}$ is a local ring with a unique maximal ideal ${\mathfrak m}$ of elements of positive valuation. (Recollect that an element of ${R}$ is a unit iff its valuation is zero.) Now, ${v(1)=0 \Rightarrow 1 \in R \Rightarrow \mathbb Z \leq R}$.
Claim: ${\mathbb Z \cap \mathfrak m = (p)}$. Clearly, ${v}$ being surjective, ${\mathbb Z \cap \mathfrak m \neq (0)}$ because otherwise, each nonzero integer would have valuation zero and so would every nonzero rational. If ${\mathbb Z \cap \mathfrak m = (n)}$ then we factorize ${n=ab}$ with ${a,b}$ integers. Then ${ab\in \mathfrak m \Rightarrow a\in \mathfrak m }$ or ${b \in \mathfrak m}$. Thus ${\mathbb Z \cap \mathfrak m}$ must be a prime ideal. So the claim is justified.
Now given ${\displaystyle\frac{a}{b}}$, write ${\displaystyle\frac{a}{b}=p^r \frac{a'}{b'}}$ with ${(p,a')=(p,b')=1}$. I claim that ${a'}$ and ${b'}$ are units. Since ${(p,a')=1}$, we can write ${px+a'y=1}$. If ${a'}$ is not a unit then ${a'\in \mathfrak m}$ and thus ${1\in \mathfrak m}$, a contradiction. A similar argument suggests that ${b'}$ is also a unit. Hence,
$\displaystyle v\displaystyle\left(\frac{a}{b}\right)=rv(p)+v(a')-v(b')=rv(p)$

Now ${v}$ surjective implies, there exists ${\displaystyle\frac{a}{b}=p^r \frac{a'}{b'}}$ such that ${v\displaystyle\left(\displaystyle\frac{a}{b}\right)=rv(p)=1}$. This leaves two possibilities, namely, ${v(p)=\pm 1}$. But ${v(p)=-1}$ gives an easy contradiction: ${-1=v(p)\geq \min\{v(1),v(p-1)\} = 0}$. Thus the only possibility is that ${v(p)=1}$ and thus, ${v=v_p}$. $\blacksquare$

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