Can the equation x! + (x-3)! = 16x - 24 be solved for x?

[Alesak]

hello everyone,
I usually post my questions on one small czech mathematical forum, but here is an equation no one knows how to "solve". I`ve came to it by accident, when I made an mistake in one combinatorics equation.


$$x! + (x-3)! = 16x - 24$$

its fairly simple to solve in one way(x has to be greater than 2, and from some point left side is greater than right side, because (x-3)! is always greater then 0 and we don`t have to care about -24 on the right side, so we can check for which x is x! > 16x. this leaves us only very few possibilities for x to check).

this is nice, but I`d like to know if its possible to get it in form x = something. I can`t think of any way how to do it.

also, this equation doesn`t have any solution, its rather theoretical question.

 

[Roberto Bramb]

If x can be a real number (using Gamma function) you can find 4 solutions betweeen 1 and 4.5 (by plotting method or std numeric method).
x1=1.1837
x2=1.3134
x3=2.1222
x4=4.4099
There are also 5 negative solution... None is integer.

 

[qspeechc]

It is simple to check that there is no integer solution. Factorials increase far more rapidly than a linear function, more rapidly than exponentials even. We know x>2. For x=4, the LHS is smaller than the RHS, for x=5 it is larger, thus there is no integer x where they are equal.

 

[mubashirmansoor]

I have also tried this in several different ways, I got:

x! = ( 8x (x-2) (x-1) (x-1.5) )/( x (x-1) ( x-2) +1 )

which means no integer values...

Is this really all?

 

[Roberto Bramb]

Plotting the function (remebering x!=Gamma(x+1))
G(x)=Gamma(x + 1) + Gamma(x - 2) - 16x + 24
you can see that for x>0 there are only four zeros.
For x<0 there are infinite solutions near 0,-1,-2,-3...none exacly integer.
There Gamma function is singular (has a pole) and change sign.