# Is there any sense with these squares?

#### kpkkpk

##### New member
1 = 1^2
1 = 1^2
9 = 3^2
1 = 1^2
81 = 9^2
729 = 27^2
225 = 15^2
324 = 18^2
X
82944 = 288^2
176400 = 420^2
215296 = 464^2
3444736 = 1856^2

So, I am trying to find short method to find factorials. In order to achieve this, I imagined factorials as squares, one edge of which corresponding square root of said factorial. However, as these square roots tend not to be natural numbers but have decimal extension, I chose the next bigger number mimicking factorial edge. That slightly bigger edge I then squared and from it I subtracted the real factorial value in a hope to find a series of natural numbers of some sense...And actually at first I was quite delighted while finding series presented above corresponding factorials 4!, 5!, 6!, 7!, 8!, 9!, 10!, 11!, 13!, 14!, 15!, 16!
As an example, please, look at number 225 (=15^2) in the list above. It associates with 10! (=3628800) in the following way:
10! squared (=10^(1/2)) = 1904,940944...
So, I chose 1905 as enlarged edge of square: 1905^2 = 3629025
From that I subtracted the real value of 10!: 3629025 - 3628800 = 225 = 15^2.

In an analogous way I found other squares presented in the list...but not that one corresponding 12! as 12!^(1/2) = 21886,10518... and 21887^2 - 12! = 39169 is not a square of any natural number. The same problem continued with bigger factorials (17!, 18!, 19!, perhaps more?) as well.

Can someone find any sense with this list or is this just natures cruel joke to lead us into desperation?

#### Opalg

##### MHB Oldtimer
Staff member
1 = 1^2
$\color{red}{n=4:\ \bigl(\lceil\sqrt {4!}\rceil\bigr)^2 - 4! = 25 - 24 =} 1 = 1^2$
$\color{red}{n=5:\ \bigl(\lceil\sqrt {5!}\rceil\bigr)^2 - 5! = 121 - 120 =} 1 = 1^2$
$\color{red}{n=6:\ 27^2 - 6! = 729 - 720 =} 9 = 3^2$
$\color{red}{n=7:\ 71^2 - 7! = 5041 - 5040 =} 1 = 1^2$
$\color{red}{n=8:\ 201^2 - 8! = 49491 - 40320 =} 81 = 9^2$
$\color{red}{n=9:\ }729 = 27^2$
$\color{red}{n=10:\ }225 = 15^2$
$\color{red}{n=11:\ }324 = 18^2$
X
$\color{red}{n=13:\ }82944 = 288^2$
$\color{red}{n=14:\ }176400 = 420^2$
$\color{red}{n=15:\ }215296 = 464^2$
$\color{red}{n=16:\ }3444736 = 1856^2$

So, I am trying to find short method to find factorials. In order to achieve this, I imagined factorials as squares, one edge of which corresponding square root of said factorial. However, as these square roots tend not to be natural numbers but have decimal extension, I chose the next bigger number mimicking factorial edge. That slightly bigger edge I then squared and from it I subtracted the real factorial value in a hope to find a series of natural numbers of some sense...And actually at first I was quite delighted while finding series presented above corresponding factorials 4!, 5!, 6!, 7!, 8!, 9!, 10!, 11!, 13!, 14!, 15!, 16!
As an example, please, look at number 225 (=15^2) in the list above. It associates with 10! (=3628800) in the following way:
10! squared (=10^(1/2)) = 1904,940944...
So, I chose 1905 as enlarged edge of square: 1905^2 = 3629025
From that I subtracted the real value of 10!: 3629025 - 3628800 = 225 = 15^2.

In an analogous way I found other squares presented in the list...but not that one corresponding 12! as 12!^(1/2) = 21886,10518... and 21887^2 - 12! = 39169 is not a square of any natural number. The same problem continued with bigger factorials (17!, 18!, 19!, perhaps more?) as well.

Can someone find any sense with this list or is this just natures cruel joke to lead us into desperation?
This is a neat investigation, based on the fact that for each integer $n$ from $4$ to $16$, apart from $12$, $\bigl(\lceil\sqrt {n!}\rceil\bigr)^2 - n!$ is a perfect square (where the ceiling symbols $\lceil\ \ \rceil$ indicate the next integer above the enclosed number). I have added some comments in red to the above quote, to emphasise the pattern.

So, why does this work for some integers but not for others?

We can write the equation $\bigl(\lceil\sqrt {n!}\rceil\bigr)^2 - n! = \text{square}$ as $n! = a^2-b^2$, where we want $a^2$ to be as close as possible to $n!$ (so that $a = \lceil\sqrt {n!}\rceil$) and therefore $b^2$ should be as small as possible. If we factorise the right side as $n! = (a+b)(a-b)$ then what we are trying to do is to express $n!$ as a product of two integers that are as close as possible to each other. This can be done as follows, starting with $n=4$: $$4! = 4\cdot 3\cdot 2\cdot 1 = (3\cdot 2)(4\cdot 1) = 6\cdot 4 = (5+1)(5-1) = 5^2 - 1^2.$$ In the same way, $$5! = (4\cdot 3)(5\cdot 2) = (11+1)(11-1) = 11^2 - 1^2,$$ $$6! = (5\cdot 3\cdot 2)(6\cdot 4\cdot 1) = (27+3)(27-3) = 27^2 - 3^2,$$ $$7! = (6\cdot 4\cdot 3)(7\cdot 5\cdot 2) = 71^2-1^2,$$ $$8! = (8\cdot 6\cdot 4\cdot 1)(7\cdot 5\cdot 3\cdot 2) = 201^2 - 9^2,$$ $$9! = (9\cdot 7\cdot 5\cdot 2)(8\cdot 6\cdot 4\cdot 3) = 603^2 - 27^2,$$ $$10! = (10\cdot 8\cdot 6\cdot 4\cdot 1)(9\cdot 7\cdot 5\cdot 3\cdot 2) = 1905^2 - 15^2,$$ $$11! = (11\cdot 8\cdot 6\cdot 4\cdot 3)(10\cdot 9\cdot 7\cdot 5\cdot 2) = 6318^2 - 18^2.$$ The way that I factorised $(n+2)!$ was to take the factorisation for $n!$, multiply the smaller of its two factors by $n+2$ and multiply the larger of its two factors by $n+1$. That way, I would hope to get a factorisation of $(n+2)!$ into two nearly equal factors. When we get to $n=12$, the result is $$12! = (12\cdot 9\cdot 7\cdot 5\cdot 3\cdot 2)(11\cdot 10\cdot 8\cdot 6\cdot 4\cdot 1) = 21900^2 - 780^2.$$ Here, for the first time, the larger of the two squares is not $\bigl(\lceil\sqrt {n!}\rceil\bigr)^2$. In fact $\lceil\sqrt {12!}\rceil = 21887 < 21900.$

So I think the best one can say is that (for $n\geqslant 4$) $n!$ can always be expressed as the difference of two squares, the larger of which is fairly close to $\bigl(\lceil\sqrt {n!}\rceil\bigr)^2$ but cannot necessarily be chosen to be equal to it.

#### kpkkpk

##### New member
Many thanks for you Opalg for detailed answer.
I also found this same series of squares from Online Ensyclopedia of Integer Sequences (OEIS) with code A038202.