Exercise 923Number theory › Divisibility and gcdOlympiadAdded 3 Nov 2026

Additive property of the p-adic valuation

#p-adic valuation#additivity#non-Archimedean

Statement

Let $p$ be prime and $v_p$ the $p$ -adic valuation. Show that for all $m,n\in\mathbb{Z}\setminus\{0\}$ :
$v_p(m+n)\ge\min(v_p(m),v_p(n)),$
with equality when $v_p(m)\neq v_p(n)$ .

Hint

Factor out $p^{\min(v_p(m),v_p(n))}$ .

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Solution

Setup. Let $k=\min(v_p(m),v_p(n))$ , so $m=p^km'$ , $n=p^kn'$ with at least one of $m',n'$ not divisible by $p$ . Then $m+n=p^k(m'+n')$ , so $v_p(m+n)=k+v_p(m'+n')\ge k$ .

Equality case. If $v_p(m)\neq v_p(n)$ , exactly one of $m',n'$ is divisible by $p$ , so $m'+n'\not\equiv0\pmod p$ , giving $v_p(m'+n')=0$ , i.e. $v_p(m+n)=k$ .

Conclusion. The inequality holds, with equality when the valuations differ.

Additive property of the p-adic valuation

Statement

Solution

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