Math Is Fun Forum

farah345

Guest

n!

Can

be the square of any natural number?
yes or no...?
also why..??

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: n!

Well, 0! = 1! = 1 = 1².
Higher than that though, I'm pretty sure there aren't any square factorials.

There definitely aren't any if this is true:
"For any n>1, there is a prime number strictly between n and 2n."

I can't remember if that's a theorem or just a probably-true-conjecture.

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Why did the vector cross the road?
It wanted to be normal.

farah345

Guest

Re: n!

There definitely aren't any if this is true:
"For any n>1, there is a prime number strictly between n and 2n."

ohk fine..lets say that this is true so how will u proof that  there aren't any square factorials from this..??

mathsyperson

Moderator

Registered: 2005-06-22

Posts: 4,900

Re: n!

If that's the case, then the prime factorisation of any n! will have one prime that there's exactly one of.
For a number to be square there needs to be an even amount of each prime factor, so therefore there can't be any square factorials.

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Why did the vector cross the road?
It wanted to be normal.

bobbym

bumpkin

From: Bumpkinland

Registered: 2009-04-12

Posts: 109,606

Re: n!

Hi farah345;

The only proof I have ever seen is very complicated and involves bertrands postulate which says there is always a prime between m and 2m. It is true that no factorial greater than 1 is ever a square.

Last edited by bobbym (2009-08-22 13:25:41)

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In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.