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Number Theory Question 2

  1. Feb 25, 2012 #1
    Let p be a prime and let m and n be natural numbers. Prove that p | mn implies pn | mn.

    Attempt:

    Since mn can be written out as a product of primes i.e: p1p2...pn in which p is a divisor.

    Raising mn means that there would exist pn primes for each factor of m: mn = m1m2...mn = (p1...pn)1(p1...pn)2....(p1...pn)n = p1a1p2a2...pnan

    which means pn | mn.

    Is there anything I'm missing to clean it up?

    Cheers for the help.
     
    Last edited: Feb 25, 2012
  2. jcsd
  3. Feb 25, 2012 #2
    It seems ok but could be written better.

    Like you said m is a product of primes so lets write m as [itex]m=p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_t^{\alpha_t}[/itex]. Then [itex]m^n = \left(p_1^{\alpha_1}p_2^{\alpha_2}\ldots p_t^{\alpha_t}\right)^n = p_1^{n\alpha_1}p_2^{n\alpha_2}\ldots p_t^{n\alpha_t}[/itex]. Since [itex]p|m^n[/itex], [itex]p|p_i[/itex] for [itex]1\leq i\leq t[/itex].
    Let [itex]p_j[/itex] be that p. Then [itex]p|p_j[/itex]. Since [itex]p_j[/itex] is prime, [itex]p=p_j[/itex].

    [itex]p|m^n=p_1^{n\alpha_1}p_2^{n\alpha_2}\ldots p_j^{n\alpha_j} p_t^{n\alpha_t}[/itex].

    [itex]n\alpha_j\geq n[/itex] so [itex]p^n|p_j^{n\alpha_j}\Rightarrow p^n|m^n[/itex]
     
  4. Feb 26, 2012 #3

    Deveno

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    Science Advisor

    i would prove p|m first. then pn|mn is rather obvious.

    in intuitive terms, we don't get "any new prime factors" when we take a power.

    again p|mn → p|m is a simple application of the fact:

    p|ab → p|a or p|b, which some people take as a definition of prime (it generalizes well to more general settings).

    (a "rigorous" proof of p|mn→p|m would use induction on n, but i feel that's over-kill).
     
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