Results 1 to 4 of 4

Math Help - [SOLVED] Prime Power Decomposition

  1. #1
    Member diddledabble's Avatar
    Joined
    Jul 2009
    Posts
    80

    Post [SOLVED] Prime Power Decomposition

    Prove that if (a,b)=1  ab=c^2 then there exists integers m and n such that a=m^2 b=n^2 I know you have to use prime power decomp to show this but I am stuck. Trying to prep for a final. Thanks
    Last edited by diddledabble; August 4th 2009 at 02:54 PM. Reason: Error in LaTeX
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517
    well you have pretty much already done it.

    Write out the prime factorization of c, then square it and you have a bunch of primes to even powers. Now since c^2=ab and factorization is unique, the factors have to be distributed among a and b. a and b cannot share any of the prime factors or else they would not be relatively prime since that prime would divide both of them. Thus each a and b is a product of even powered prime numbers, making them perfect squares.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member diddledabble's Avatar
    Joined
    Jul 2009
    Posts
    80
    I don't follow your logic with factoring c.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Gamma's Avatar
    Joined
    Dec 2008
    From
    Iowa City, IA
    Posts
    517
    The integers are a Unique Factorization Domain (in fact they are a Euclidean Domain). Every nonzero, and non-unit(things that are invertible, \pm1 in the integers) can be written as a product of irreducible elements (primes in this case since in PIDs and UFDs they are the same) unique up to multiplication by a unit (recall, that is \pm 1).

    This means every integer bigger than 1 has a factorization into primes which is unique up to multiplying it by like 6 negative ones, or 14 negative ones and 7 ones. Get it?

    So let c\in \mathbb{Z}. If c=\pm 1, you are done 1=(\pm 1)^2=1^21^2=1 and gcd(1,1)=1.
    0 is also clear.

    So now we know c is not a unit or zero, so write out its prime factorization.
    c=p_1^{e_1}p_2^{e_2}\cdots p_n^{e_n}.

    c^2=p_1^{2e_1}p_2^{2e_2}\cdots p_n^{2e_n}.

    Now if this is equal to ab two integers, they must have the same prime factorization. so suppose WLOG a is not a square, then one of those prime exponents must be odd, which means b must have an odd powered exponent. But that means that prime call it p_i divides both a and b, so 1=gcd(a,b)\geq p_i>1a contradiction, so both a nd b must be squares.
    Last edited by Gamma; August 5th 2009 at 11:15 AM. Reason: missed the = sign
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 3
    Last Post: February 17th 2011, 08:51 AM
  2. Prime decomposition in number rings
    Posted in the Number Theory Forum
    Replies: 0
    Last Post: March 11th 2010, 06:02 AM
  3. Prime Power Congruence
    Posted in the Number Theory Forum
    Replies: 4
    Last Post: April 24th 2009, 09:44 PM
  4. prime-power
    Posted in the Number Theory Forum
    Replies: 1
    Last Post: January 21st 2009, 09:48 AM
  5. Prime-power factorization
    Posted in the Algebra Forum
    Replies: 5
    Last Post: September 17th 2008, 05:04 PM

Search Tags


/mathhelpforum @mathhelpforum