
#1
Apr908, 09:18 AM

P: 134

hello everyone,
I usually post my questions on one small czech mathematical forum, but here is an equation noone knows how to "solve". I`ve came to it by accident, when I made an mistake in one combinatorics equation. [tex]x! + (x3)! = 16x  24[/tex] 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 (x3)! 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. 



#2
Apr908, 11:28 AM

P: 11

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. 



#3
Apr908, 11:42 AM

P: 792

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.




#4
Apr1608, 06:20 AM

P: 259

Equation with factorials
I have also tried this in several different ways, I got:
x! = ( 8x (x2) (x1) (x1.5) )/( x (x1) ( x2) +1 ) which means no integer values...... Is this really all???? 



#5
Apr1608, 10:30 AM

P: 11

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. 


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