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Hey, I need a bit of help with a proof, please!

  1. Jan 4, 2006 #1
    Here's the item needing to be proofed (this is not homework, but I'm very interested in it).

    Show that for each odd prime number y, there is exactly one positive integer x such that x(x+y) is a perfect square.


    Thanks for any help/leads, etc..

    -Steve
     
  2. jcsd
  3. Jan 4, 2006 #2

    shmoe

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    If x(x+y) is a perfect square, show that x and x+p are relatively prime. What can you then say about x?
     
  4. Jan 17, 2006 #3
    hmm....this looks stalled, so I'll write the whole proof :smile:
    Since [tex]x(x+y)[/tex] is a perfect square, [tex]x[/tex] must be a perfect square.

    Proof: [tex]x(x+y)[/tex] can be of [tex]a^2b^2[/tex] form or [tex]a^2 b^2 c^2[/tex] form. If [tex]x[/tex] isn't a perfect squre, [tex]x[/tex] can be of one of forms: [tex]x=a[/tex] or [tex]x=a^2 b[/tex], and then [tex]x+y = a b^2[/tex] or [tex]x+y=b[/tex] or [tex]x+y=b c^2[/tex] respectively, but in any case, [tex]x[/tex] and [tex]x+y[/tex] has a common measure. This contradicts the fact [tex]y[/tex] is a prime number.

    Now we can write [tex]x = a^2[/tex] and so [tex]x+y=d^2[/tex]. Then [tex]y =(x+y) - x = d^2 - a^2 = (d+a)(d-a)[/tex]. Now that [tex]y[/tex] is a prime number, [tex]d-a[/tex] must be 1 (notice [tex]y[/tex] cannnot be factorized)!! So [tex]d = a+1[/tex] and [tex]y = d+a = 2a+1[/tex]. So [tex]x = a^2 = ((y-1)/2)^2[/tex] and this is the only possible value of [tex]x[/tex].

    This is just my solution, so there may be better solutions.
     
    Last edited: Jan 18, 2006
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