Represent any number with two factorials

Question

I was wondering if it is possible to represent any positive integer with x! - y! ? If not, is there any proof?

Answer 1

For $x, y > 1$, $x! - y!$ is always an even number, so the only odd numbers representable in this form are numbers that are smaller than $n!$ by $1$ for some $n$. Taking $k = 3$, as $k+1 = 4 \ne n!$ for any $n$, we see that $3$ is not representable in the form $x! - y!$.

Answer 2

If $x \in \mathbb{N}, x>1 \implies 2|x!$ It means, that some odd numbers can not be represented as $x!-y!$

e.g. number 7. You have only 1 possibility:

i) 8 - 1 (but $\not \exists x\in \mathbb{N} : 8 = x!$)

Answer 3

Hint: For all $y>x$ $$y!-x!\geq x!$$ Now are there $x,y\in\mathbb N$ such that $y!-x!=3?$ If there are, $x$ must be equal to 1 or 2.

Answer 4

Hint: Say that $x = y + 1$. Then $$ x! - y! = (y+1)! - y! = y!(y+1 - 1) = y!\cdot y $$ Now try to express $3$ as $x! - y!$.

Answer 5

You didn't say that $x$ and $y$ have to be integers!

So, for any real $z$, since the factorial can be written in terms of the gamma function ($x! = \Gamma(x+1)$), and $\Gamma(x)$ is continuous for $x > 0$, any real number can be written as the difference of two factorials.

It's always a requirements problem.