The familiar Pythagorean 3-4-5 triple gives one solution to (x-1)^n
+ x^n = (x+1)^n so what about other solutions for x an integer and
n= 2, 3, 4 or 5?

Find all 3 digit numbers such that by adding the first digit, the
square of the second and the cube of the third you get the original
number, for example 1 + 3^2 + 5^3 = 135.

Reciprocals

Stage: 5 Challenge Level:

Prove that for any positive numbers $x_1$, $x_2$,..., $x_n$