Prove that factorial of a number can never be a perfect square.

Question

inkyvoyd · Answer

n(n-1)*(n-2)*...(2)=n^2 assume that it can.

zarkon · Answer

1!

anonymous · Answer

Except that,

zarkon · Answer

0!

inkyvoyd · Answer

@zarkon, -.-

inkyvoyd · Answer

(n-1)(n-2)*...(2)=n

unklerhaukus · Answer

that is disproving it @zarkon

anonymous · Answer

@inkyvoyd 
Not necessarily n^2
Any k^2

anonymous · Answer

oops too late

inkyvoyd · Answer

ahhg.

anonymous · Answer

NOT 1, guys. Excewpt that

anonymous · Answer

Somewhere in the expansion of n! there is the largest prime factor.
(If n is prime, it is n itself, otherwise you have to go looking for it.)

But if it is the largest prime factor, it only occurs once, and therefore n! cannot be a square,
since in a square every prime factor occurs an even number of times.

anonymous · Answer

@inkyvoyd  be careful here, it is not saying that $n^2=n!$ but rather that $n^2=m!$

inkyvoyd · Answer

Yes, I just realized that, and I think sim got the answe r-.-

inkyvoyd · Answer

but wait, he didn't,.

anonymous · Answer

@simamura 
CAn you prove your conjecture?

anonymous · Answer

How about using Bertrand's Postulate?

inkyvoyd · Answer

i think we use mathematical induction for this one.

anonymous · Answer

Ohhh.
Nice FFM. That states that precisely, right?

anonymous · Answer

yeah but better make base case $n=2$ because as zarkon noted it is false for $n=1$

anonymous · Answer

True.
Its proved by induction using log right?

anonymous · Answer

this is not trivial there is stuff to be done

anonymous · Answer

There is a always a prime between $n$ and $2n$. This means there is  prime $p $  that divides $n!$ but $p^2$ does not then $ n!,n>1$ would have even number of divisors always, but a perfect square must have odd number of divisors hence ... QED.

anonymous · Answer

Trivial? Pretty complicated proof to here.

inkyvoyd · Answer

FFM, only, you used a theorem itself that is really hard to prove.

inkyvoyd · Answer

I'd like something more elegant, in the sense that we don't prove something "simpler" with something more difficult.

anonymous · Answer

Agreed.

anonymous · Answer

Theorems are meant to be used ;)

inkyvoyd · Answer

We are trying to prove that the square root of all factorials is not rational, correct? (besides 1)

anonymous · Answer

Correct.

inkyvoyd · Answer

This has something to do with the gamma function, I bet.

blockcolder · Answer

I wonder if we can use the fact that the product of consecutive integers is always one less than a perfect square.

blockcolder · Answer

4 consecutive integers, I meant.

inkyvoyd · Answer

That's it block, I think.

inkyvoyd · Answer

But, your statement is incorrect :S

blockcolder · Answer

It's correct and I can prove it.

anonymous · Answer

http://math.stackexchange.com/questions/31973/

inkyvoyd · Answer

(n-1)(n+1)(n)(n+2)=(n^3-n)(n+2)=n^4+2n^3-n^2-2n
n^4+2n^3-n^2-2n+1 does not appear to be a perfect square in any way.

inkyvoyd · Answer

Or does it? Gimme a sec.

anonymous · Answer

http://math.stackexchange.com/questions/12544/

blockcolder · Answer

I wrote "one less than a perfect square". :P

inkyvoyd · Answer

So, n^4+2n^3-n^2-2n+1 must be a perfect square if n is a natural number

anonymous · Answer

(n^2 +n + 1)^2

inkyvoyd · Answer

lool I fail

inkyvoyd · Answer

So, we can prove that 4 consecutive numbers are not a perfect square.

blockcolder · Answer

$$n(n+1)(n+2)(n+3)=(n^2+3n)(n^2+3n+2)=[(n^2+3n+1)-1][(n^2+3n+1)+1]\
=(n^2+3n+1)^2-1$$

anonymous · Answer

Wait, not quit...
(n^2+n -1)^2 = (n^2 +n -1)(n^2 +n -1) =
n^4 + n^3 -n^2 +n^3 +n^2 -n -n^2 -n +1 =
n^4 +2n^3-n^2-2n +1
There we go. It's (n^2 + n -1)^2
not (n^2 +n +1)^2 like I said originally.

blockcolder · Answer

My last equation got cut off, let me rewrite:
$$n(n+1)(n+2)(n+3)=(n^2+3n)(n^2+3n+2)\
=[(n^2+3n+1)-1][(n^2+3n+1)+1]\
=(n^2+3n+1)^2-1$$

anonymous · Answer

It seems like you can break this down into cases and use induction on each case... perhaps?

anonymous · Answer

Cases based on number of consecutive numbers being multiplied, that is.

inkyvoyd · Answer

(n)(n+1)(n+2)(n+3)=u^2-1
(n)(n+1)(n+2)(n+3)(n+3+1)
=u^2-1+(u^2-1)(n+3)

u^2-1+(u^2-1)(n+3)+(u^2-1)(n+4)

inkyvoyd · Answer

I might have it.

inkyvoyd · Answer

n(n+1)=n^2+n

a perfect square would be in the form n^2+2n+1
in other words, n^2+n=n^2+2n+1
n=-1, which is not an integer, so n(n+1) is never a perfect square.

inkyvoyd · Answer

Then, n(n+1)(n+2)
=n(n+1)(n)+n(n+1)(2), which is never a perfect square (someone give me the proof)

inkyvoyd · Answer

since that is never a perfect square,
n(n+1)(n+2)(n)+n(n+1)(n+2)(3) which is never a perfect square? I've not reduced the problem down any. -.-

inkyvoyd · Answer

maybe instead of n, i'll just have 1.