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Need help with a proof on divisors.

  1. Mar 5, 2006 #1
    Ok, the proof to be done is pretty simple:
    Prove that a number is a square only when the number of positive divisors is odd.

    I pretty sure I know the answer, I'm just not sure how to go about writing it out...

    If c is the number then you can write:
    c=ab.
    a and b are divisors of c. If a doesn't equal b then you have two different divisors. If a=b, then c=ab can be rewritten as c=a^2, and you only have one divisor. Because of this any number that can be written as a square of another number has an odd number of divisors -- all the pairs of factors that equal the number plus the 1 divisor that is squared to make the number.

    I guess I just need to know how to write this out better, and how to make sure I haven't assumed to much as given/proven at the outset.
    Thanks,
    -GM-
     
  2. jcsd
  3. Mar 5, 2006 #2

    shmoe

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    Science Advisor
    Homework Helper

    That's pretty much it, you've got the essential idea of pairing divisors c=ab with a,b distinct, which leaves out the oddball [tex]\sqrt{c}[/tex] out. You could try organizing the divisors c=ab with the assumption that a<=b, but this isn't necessarily any better.

    For an alternate way, you could look at the prime factorization of c and count the divisors that way. (this isn't a 'better' way, but gives a different point of view to consider if you run across other divisor problems)
     
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