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Suppose that we are told that four numbers a, b, c, d lie between -5 and 5. Suppose also that the numbers are constrained so that
5< a+b < 10 \quad\mbox{ and }\quad -10< c+d < -5
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.