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Math Help - Some Prime Number Divisibility Questions

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    Super Member Aryth's Avatar
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    Some Prime Number Divisibility Questions

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

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

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

    Is it sufficient to say that:

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

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

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

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

    Then again, I have no clue what I'm doing.
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    Quote Originally Posted by Aryth View Post
    I was given this question in class and was never given an answer:

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

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

    Is it sufficient to say that:

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

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

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

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

    Then again, I have no clue what I'm doing.
    Without lose of generality say that a\leq b this means that \min(a,b)=a. You need to show that p^a || (m+n). Obviously, p^a|(n+m). Now if p^{a+1}|(m+n) we have two cases. The first case is that p^{a+1}|n, but if that happens then p^{a+1}|m and this is a contradiction. The second case is that p^{a+1} \not | n i.e. p^a || n \implies b=a. But in the second case it is possible that p^{a+1}|(m+n)! Here is a conterexample, p=3,m=15,n=21.
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