## 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)

Roger

 PhysOrg.com science news on PhysOrg.com >> Front-row seats to climate change>> Attacking MRSA with metals from antibacterial clays>> New formula invented for microscope viewing, substitutes for federally controlled drug
 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
 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