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Remainder theory proof

  1. Sep 18, 2003 #1
    If p^2 is exactly divisible by p+q, then proof q^2 is exactly divisible by p+q.

    How do I proof this, and how do I apply the remainder theorem?

    I know if f(x) = x^2 + 2x + 1, since f(-1) = 0 there fore (x + 1) is a factor of f(x).

    So in this case
    p^2 = p x p or p^2 x 1...
  2. jcsd
  3. Sep 18, 2003 #2


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    Since you specifically mention x2+ 2x+ 1, haven't you looked at (p+q)<sup>2</sup>= p<sup>2</sup>+ 2pq+ q<sup>2</sup>?
  4. Sep 19, 2003 #3
    This is what I've done afterwards, but I'm not sure if it is right.

    Let p^2 = x^2 + 2xy + y^2.

    where y^2 = y.y or -y.-y or x^2 = x.x or -x.-x.

    When y = -x, or x = -y, p^2 = 0

    Therefore (x + y) is a factor of p^2, also (x + y) = (p + q)

    Therefore p^2 = p^2 + 2pq + p^2. p^2 = 0 when p = -q,

    but (p)^2 = (-q)^2
    p^2 = q^2
    therefore (p+q)mod q^2 = 0

    Are there any flaw in my argument or have I made any calcuation error?
  5. Sep 20, 2003 #4
    Sorry, it's q^2mod(p+q) = 0
  6. Sep 20, 2003 #5


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    not p=-q
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