x!+y!+z!=a^b?

Question

5 unknown

Answer 1

This has infinitely many solutions. You have put no restrictions at all on the variables, although factorial is typically assumed to have a non-negative integer as its argument. Here are a few solutions.

1! + 1! + 1! = 3^1
1! + 1! + 1! = [sqrt(3)]^2
2! + 3! + 4! = 2^5
3! + 5! + 7! = 3.39196^7 {where 3.39196 is just a decimal approximation of 5166^(1/7)}

You can just choose values for x, y, z, and either a or b. Here are solutions for a and b, assuming the other is chosen.

a = (x! + y! + z!)^(1/b)
b = log_a(x! + y! + z!)

If you'd rather leave x, y, or z to be solved, it's quite a bit harder. The factorial function can be extended to have negative or non-integer arguments, in the form of the gamma function, but solving for it is extremely difficult.