This equation represents a set of Diophantine equations, which are equations where the solutions are restricted to integers.
To find integer solutions for this equation, we can use a brute force method by trying different values of x, y, and z for each value of k from 1 to 100. However, this method can be time-consuming and inefficient.
Alternatively, we can use a mathematical theorem called Fermat's Last Theorem, which states that there are no integer solutions to the equation x^n + y^n = z^n for n > 2.
Since our equation is a special case of Fermat's Last Theorem with n = 3, there are no integer solutions for x, y, and z that satisfy the equation for any value of k from 1 to 100.
x3+y3+z3=k, with k being all the numbers from one to 100?
3 answers
AAAaannndd the bot gets it wrong yet again!
It is not related to Fermat's Last Theorem, since they are not looking for a solution to x^3+y^3+z^3=k^3
They just want a way to produce any value from 1 to 100 using the sum of three cubes.
But if you google sums of three cubes you can see that not all the values can be so formed.
It is not related to Fermat's Last Theorem, since they are not looking for a solution to x^3+y^3+z^3=k^3
They just want a way to produce any value from 1 to 100 using the sum of three cubes.
But if you google sums of three cubes you can see that not all the values can be so formed.
Btw, the possible values of k , k ≤ 100, that are possible are:
3 10 17 24 29 36 43 55 62 66 73 80 81 92 99
e.g.
36 = 27 + 8 + 1 = 3^3 + 2^2 + 1^3
24 = 2^3 + 2^3 + 2^3 , it did not say that x, y, and z had to be different from each other.
3 10 17 24 29 36 43 55 62 66 73 80 81 92 99
e.g.
36 = 27 + 8 + 1 = 3^3 + 2^2 + 1^3
24 = 2^3 + 2^3 + 2^3 , it did not say that x, y, and z had to be different from each other.