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No idea where to start.

  1. Apr 7, 2005 #1
    Prove that there are no positive integers x & y such that

    [tex]\frac{1}{x^2} + \frac{1}{xy} + \frac{1}{y^2} = 1 [/tex].

    I really have no idea where to start. I tried rearranging the equation, but I could not see anything useful in any of the rearrangements.
  2. jcsd
  3. Apr 7, 2005 #2
    [tex]\frac{1}{x^2} + \frac{1}{xy} + \frac{1}{y^2} = 1[/tex]

    [tex] \Longleftrightarrow x^2 + xy + y^2 = x^2y^2[/tex]

    [tex]\Longleftrightarrow x^2 + 2xy + y^2 = (x+y)^2 = x^2y^2 + xy =xy(xy+1) [/tex]

    ie. it implies

    [tex](x+y)^2 = xy(xy+1)[/tex]

    see a contradiction? :smile:
  4. Apr 7, 2005 #3
    *smacks head*

    Of course, consecutive numbers cannot be squares.
  5. Apr 7, 2005 #4
    Right, products of consecutive positive integers can't be squares (I assume that's what you meant~). See an easy proof of that?
  6. Apr 7, 2005 #5
    Oh yeah, I phrased it wrongly. :redface:

    An easy proof? No, I can't find an EASY one, but this is the best I can do:

    a-1, a and a+1 are consecutive integers.

    [tex](a - 1)^2 = a^2 - 2a + 1[/tex]
    [tex](a + 1)^2 = a^2 + 2a + 1[/tex]

    a^2 - a and this is bigger than a^2 - 2a + 1 but smaller than a^2 + 2a + 1, since there is only square number between these two boundaries, a^2, this means that a * (a - 1) cannot be a square number.

    The same reasoning, when applied to a * (a + 1), and using a + 2 reveals that it can't be a square number too.
    Last edited: Apr 7, 2005
  7. Apr 7, 2005 #6
    I can't see most of that post right now, and I'm too tired to think at the moment :smile:

    I'll check back in the morning~
  8. Apr 7, 2005 #7
    [tex](x+y)^2 = xy(xy+1)[/tex]
    can be a square number if x=y=0.
    [tex]a=xy(xy+1)[/tex], i.e.[tex]a=z(z+1)[/tex]
    So, if a is in this form, it can be a square number.
    However, referring to the case, x cannot be 0 and y cannot be 0 neither.
  9. Apr 7, 2005 #8
    Yep, if one of them is zero it's okay. As you mentioned, in this case both of them are positive (ie. nonzero), though.

    Your proof is fine, recon, though you can make it simpler:

    If [itex]a>0[/itex] then

    [tex]a^2 \leq a^2 + a = a(a+1) \leq a^2 + 2a + 1 = (a+1)^2[/tex]

    so if [itex]a(a+1) = b^2[/itex] then [itex]a<b<a+1[/itex], but there's no integer satisfying this if [itex]a[/itex] is an integer.
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