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Perfect square

  1. Mar 1, 2006 #1
    hi

    I am wondering whether there is any systematic way to determine for what values of x this function is a perfect square : f(x) = square root ( x!+1)

    thanks for any advice.

    Roger
     
  2. jcsd
  3. Mar 1, 2006 #2

    Zurtex

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    Hmm, well, thinking about it for a moment it's the same as saying:

    [tex]x! + 1 = y^4 \quad \text{for some} \quad x \, , y \, \in \mathbb{N}[/tex]

    Clearly:

    [tex]y \neq i \quad \text{for} \quad i = 1, \ldots, x[/tex]

    Therefore we must have [tex]x! > x^4[/tex], this is satisfied for all [tex]x > 6[/tex]

    We can reduce the problem yet further to:

    [tex]x! = y^4 - 1[/tex]

    Or:

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

    Given that x > 6, then there must me at least a factor of 24, therefore y must be odd. So letting y = 2m + 1, giving us:

    [tex]x! = 8m(1 + m)(1 + 2m + 2m^2)[/tex]

    Also if x > 6 then 5 must be a factor of this. Take this polynomial modulo 5 and see what you get :tongue2:


    Just stuck that together, but it seems to be right :biggrin:


    EDIT: Well none of my LaTeX seems to have come out, don't know why, if a mod could help I'd be greatly appreciative
     
    Last edited: Mar 1, 2006
  4. Mar 1, 2006 #3

    Zurtex

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    Well assuming my LaTeX never comes through properly, here is the post again:

    Hmm, well, thinking about it for a moment it's the same as saying:

    x! + 1 = y4

    For some x,y natural numbers

    Clearly:

    y =\= i
    For: i = 1, ..., x

    Therefore we must have x! > x4, this is satisfied for all x > 6

    We can reduce the problem yet further to:

    x! = y4 - 1

    Or:

    x! = (y - 1)(y + 1)(y2 + 1)

    Given that x > 6, then there must me at least a factor of 24, therefore y must be odd. So letting y = 2m + 1, giving us:

    x! = 8m(1 + m)(1 + 2m + 2m2)

    Also if x > 6 then 5 must be a factor of this. Take this polynomial modulo 5 and see what you get :tongue2:


    Just stuck that together, but it seems to be right :biggrin:
     
  5. Mar 1, 2006 #4
    I have difficulty reading what's been written, but the first of all it's not
    x!=(y^4)-1 its x!=(y^2)-1
     
    Last edited: Mar 1, 2006
  6. Mar 1, 2006 #5

    Zurtex

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    Well you said:

    f(x) = square root ( x!+1) is a perfect square, which is the same as:

    square root ( x!+1) = y^2

    or:

    x! + 1 = y^4

    Please try and be more clear next time. But the same principle still applies, try and prove that this never is the case (which I am guessing is so)
     
  7. Mar 1, 2006 #6
    To be honest, I dont get what your doing.
     
  8. Mar 1, 2006 #7
    It was my mistake, it should have been square not (^4)
     
  9. Mar 1, 2006 #8
    I've re read what you wrote but , I cannot understand it. Please explain to me what your doing ?
     
  10. Mar 1, 2006 #9
    Ok, I will write it out again to avoid confusion.

    x!+1=y^2 obviously x=4 gives a perfect square, and so is x=5.

    All I'm trying to find is an easy way to find which values of x give me a perfect number, without keeping substituting the values in, that's all.
     
  11. Mar 1, 2006 #10
    In other words, you are asking:

    [tex]\text{For what values of }n \in \mathbb{N}\;\text{is }\sqrt {n! + 1} \in \mathbb{N}\;?[/tex]
     
    Last edited: Mar 1, 2006
  12. Mar 1, 2006 #11

    Zurtex

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    Yeah, I've just had a hack at my method and I don't seem to be coming up with anything, it seems mine was only appropriate for powers of 4

    I did a quick little computation and got:

    7! + 1 = 712

    However I think you will be stumped to find anymore, look near the bottom of this page:

    http://mathworld.wolfram.com/FactorialSums.html
     
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