Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

For the small values of n I have been able to check, it seems that for $n>3$, there exist whole numbers $x,y$ s.t. $n! = x^2 - y^2$. For example ..

$4! = 5^2 - 1^2$

$5! = 11^2 - 1^2$

$6! = 27^2 - 3^2$

$7! = 71^2 - 1^2$

$8! = 201^2 - 9^2$

$9! = 603^2 - 27^2$

$10! = 1905^2 - 15^2$

$11! = 6318^2 - 18^2$

$12! = 21888^2 - 288^2$

In most of the cases above, the $x$ value is just the next integer larger than $\sqrt{n!}$, though at $n=12$ and $n=17$ it's the one following that. With the tools at hand I've only been able to check this as far as $n=17$.

I expect there's probably already a name for this, but not knowing that name, googling was coming up dry.

share|improve this question
    
Seen this? –  J. M. Dec 17 '11 at 17:41
    

1 Answer 1

up vote 34 down vote accepted

If $n >3$, then $n!$ is divisible by $4$. So $n!=4k=(2)(2k)$ for some integer $k$. Note now that $$4k=(k+1)^2-(k-1)^2.$$ If $n$ is large, there are many representations of $n!$ as a difference of two squares. For let $2a$ and $2b$ be any two even numbers whose product is $n!$. Then $$n!=4ab=(a+b)^2-(a-b)^2.$$

Comment: Let $a$ be an odd integer. Then $a+1$ and $a-1$ are even, and therefore $(a+1)/2$ and $(a-1)/2$ are integers. We have $$a=\left(\frac{a+1}{2}\right)^2-\left(\frac{a-1}{2}\right)^2,$$ so $a$ is a difference of two squares.

If $a$ is divisible by $4$, the argument we gave above shows that $a$ is a difference of two squares.

If $a$ is even but not divisible by $4$, then $a$ is not a difference of two squares. For a difference of two even squares is divisible by $4$, and a difference of two odd squares is divisible by $8$.

share|improve this answer
    
so .. the take-away is, all multiples of 4 can be written as the difference of squares, and for all n>3, n! is a multiple of 4. thanks! –  JustJeff Dec 18 '11 at 14:31
    
@JustJeff: Yes. And there is more, since for any integer $N$, one can find an expression for the number of representations of $N$ as a difference of squares in terms of the structure of the prime power factorization of $N$. –  André Nicolas Dec 18 '11 at 15:04
    
How does "Let a be an odd integer." fit to "If a is divisible by 4, ... If a is even"? –  draks ... Jun 26 '12 at 22:38
    
They are several different cases, discussed independently. –  André Nicolas Jun 26 '12 at 22:47

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.