Can the equation ( 3x²+ 3x +1)^(1/3) have a solution within the natural numbers?

Answer

Bye Philip

This question has been open for a while; maybe because it’s not quite clear what you mean exactly. I try to answer a few possible interpretations. There is no equation so what do you want to solve? If you are looking for zero values ​​(solutions of the equation (3x²+3x+1)^1/3 = 0) then you will not find one, because (3x²+3x+1)^1/3 is strictly positive.

You may be looking for values ​​of x for which the expression (3x²+3x+1)^1/3 becomes a natural number? This is possible for any natural number n ≥ 1, that comes down to solving 3x²+3x+1 = n³ and that is a quadratic equation in x with discriminant 12n³-3 and that is strictly positive for all natural numbers except 0 In this way you always find two x-values, one positive and one negative, for which the cube root neatly becomes a natural number.

The solutions to x that you find in this way will generally not be natural (or integers, or rational) numbers. If you are looking for natural numbers x for which that cube root also becomes a natural number; then you have for example that for x = 0 the expression (3x²+3x+1)^1/3 becomes equal to 1. If x is also allowed to be integer, you find the same value for the cube root as x = -1.

Kind regards
Tom Dorissen