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perfect square

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

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Recognitions:      Homework  Help     Science  Advisor  Hmm, well, thinking about it for a moment it's the same as saying:  $$x! + 1 = y^4 \quad \text{for some} \quad x \, , y \, \in \mathbb{N}$$  Clearly:  $$y \neq i \quad \text{for} \quad i = 1, \ldots, x$$  Therefore we must have $$x! > x^4$$, this is satisfied for all $$x > 6$$  We can reduce the problem yet further to:  $$x! = y^4 - 1$$  Or:  $$x! = (y - 1)(y + 1)(y^2 + 1)$$  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:  $$x! = 8m(1 + m)(1 + 2m + 2m^2)$$  Also if x > 6 then 5 must be a factor of this. Take this polynomial modulo 5 and see what you get     Just stuck that together, but it seems to be right    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

Recognitions:      Homework  Help     Science  Advisor  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     Just stuck that together, but it seems to be right

perfect square

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

Recognitions:
Homework Help
Science Advisor

Quote by roger 					 				 				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
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)

To be honest, I dont get what your doing.

It was my mistake, it should have been square not (^4)

I've re read what you wrote but , I cannot understand it. Please explain to me  what your doing ?

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.

In other words, you are asking:  $$\text{For what values of }n \in \mathbb{N}\;\text{is }\sqrt {n! + 1} \in \mathbb{N}\;?$$

Recognitions:      Homework  Help     Science  Advisor  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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