When can we get $v_p(x+y) > \operatorname{min}\{v_p(x),v_p(y)\}$ in $p$-adic valuation?

Question

Suppose $x,y\in\mathbb{Q}$. We can write $x=p^n(c/d)$, $y=p^m(a/b)$ with $p$ not dividing $a,b,c$ or $d$. When can we get $v_p(x+y) > \operatorname{min}\{v_p(x),v_p(y)\}$ with the inequality strict? We have, if we assume wlog that $n>m$, that $v(x+y)= v(p^m(p^{n-m}+1)=v(y)$, so I'm not sure when we'd get a strict inequality here. What am I missing?

Answer 1

You cannot assume wlog that $n > m$ because $n$ might be equal to $m$. (Indeed this phenomenon does not occur when $n \neq m$ and you have given a correct proof of that.)

For a counterexample, take $p = 7$ and then $14 + 35 = 49$.

Answer 2

$v_p(x+y) > \min v_p(x),v_p(y)$ iff $x=p^n a/b,y=p^n c/d, p\nmid abcd, p |ad+bc$.

Proof: $x+y= p^n \frac{ad+bc}{bd}$, $v_p(x+y)=n+v_p(ad+bc)$.