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**Aryth** I was given this question in class and was never given an answer:

Definition: If $\displaystyle p^n|q$ and $\displaystyle p^{n+1}\not | q$ then it is said that $\displaystyle p^n$ exactly divides q, denoted as $\displaystyle p^n || q$

Question: Show that if $\displaystyle p^a || m$ and $\displaystyle p^b || n$, then $\displaystyle p^{min(a,b)} || (m + n)$

Is it sufficient to say that:

$\displaystyle (m,n) = p^{min(a,b)}$

$\displaystyle (m, n+m) = p^{min(a,b)}$ since they have the same gcd.

Since $\displaystyle p^{min(a,b)}$ is a common divisor, then

$\displaystyle p^{min(a,b)} || (m+n)$

Then again, I have no clue what I'm doing.