# Perfect square

1. Mar 1, 2006

### roger

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

2. Mar 1, 2006

### Zurtex

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 :tongue2:

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

Last edited: Mar 1, 2006
3. Mar 1, 2006

### Zurtex

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

4. Mar 1, 2006

### 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

Last edited: Mar 1, 2006
5. Mar 1, 2006

### Zurtex

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)

6. Mar 1, 2006

### roger

To be honest, I dont get what your doing.

7. Mar 1, 2006

### roger

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

8. Mar 1, 2006

### roger

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

9. Mar 1, 2006

### roger

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.

10. Mar 1, 2006

### bomba923

In other words, you are asking:

$$\text{For what values of }n \in \mathbb{N}\;\text{is }\sqrt {n! + 1} \in \mathbb{N}\;?$$

Last edited: Mar 1, 2006
11. Mar 1, 2006

### Zurtex

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