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Math Help - Fundamental Theorem of Arithmetic Help

  1. #1
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    Fundamental Theorem of Arithmetic Help

    If anyone could help, I will appreciate it!

    This are the problem:

    1. Show that all of the powers in the prime-power factorization of an integer n are even if and only if n is a perfect square.

    2. Show that every positive integer can be written as the product of possibly a square and a square-free integer. A "square-free integer" is an integer that is not divisible by any perfect squares other than 1.

    Thank you!
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  2. #2
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    Quote Originally Posted by jmc1979 View Post
    If anyone could help, I will appreciate it!

    This are the problem:

    1. Show that all of the powers in the prime-power factorization of an integer n are even if and only if n is a perfect square.
    If N is a perfect square the there exists an n such that n^2=N.

    Let n=p1^a1 p2^a2...pk^ak

    be the prime decomposition of n, then:

    N=n^2= p1^(2*a1) p2^(2*a2)...pk^(2*ak),

    so the prime decomposition of N contains only even powers.

    The other half of the proof goes through in a similar way, suppose
    that the prime decomposition of N contains only even powers of primes,
    then:

    N = p1^(2*a1) p2^(2*a2)...pk^(2*ak),

    so if n = p1^a1 p2^a2...pk^ak, N=n^2, so N is a square.

    RonL
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  3. #3
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    Quote Originally Posted by jmc1979 View Post
    2. Show that every positive integer can be written as the product of possibly a square and a square-free integer. A "square-free integer" is an integer that is not divisible by any perfect squares other than
    Let the prime decomposition of an integer N be:

    N = p1^a1 p2^a2...pk^ak = p1^[2*floor(a1/2)+(a1 mod 2)] p2^[2*floor(a2/2)+(a2 mod 2)] ... pk^[2*floor(ak/2)+(ak mod 2)]

    .......= {p1^[2*floor(a1/2)] p2^[2*floor(a2/2)] ... pk^[2*floor(ak/2)]} {p1^[a1 mod 2] p2^[a2 mod 2] ... pk^[ak mod 2]}

    .......= A B

    where:

    A = p1^[2*floor(a1/2)] p2^[2*floor(a2/2)] ... pk^[2*floor(ak/2)]

    is a square, and:

    B = {p1^[a1 mod 2] p2^[a2 mod 2] ... pk^[ak mod 2]}

    is square free.

    RonL
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  4. #4
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    Thank you!

    Thank you for your help Capt. Black.
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  5. #5
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    Another one

    This another problem:

    Shot that if p^a||m then p^ka||m^k for any positive integer k.

    I did this problem but dont know if the answer is correct.

    Solution:
    Suppose that p^a||m then we know that m=p^a X, where X is a prime other than p. Then if (m)^k=(p^a X)^k = m^k=p^ka X^k. Since p does not divide X^k for any positive integer k, then we know that p^ka||m^k.

    Is this correct? Thank you to anyone!
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  6. #6
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    Quote Originally Posted by jmc1979 View Post
    This another problem:

    Shot that if p^a||m then p^ka||m^k for any positive integer k.
    Just use the fact that if,
    a|b and c|d then ac|bd.

    Thus, for any positive k, we have k relations:

    p^a | m
    p^a | m
    ....
    p^a | m

    Multiply them out,
    p^(ka)| m^k

    Ah! I believe you are trying to show the converse, that is,
    if p^k | m^k then, p | m.
    (Which is not true always, but for primes, yes).
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  7. #7
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    Thanks

    Thank you for the help!
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