Step by Step solution about How to resolve the algorithm Square but not cube step by step in the Mathematica / Wolfram Language programming language

1. s = Range[50]^2;: This line creates a list s of all the integers from 1 to 50 squared. So, s contains the list [1, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144, 169, 196, 225, 256, 289, 324, 361, 400, 441, 484, 529, 576, 625, 676, 729, 784, 841, 900, 961, 1024, 1089, 1156, 1225, 1296, 1369, 1444, 1521, 1600, 1681, 1764, 1849, 1936, 2025, 2116, 2209, 2304, 2401]

2. c = Range[1, Ceiling[Surd[Max[s], 3]]]^3;: This line creates a list c of all the integers from 1 to the cube of the ceiling of the cube root of the maximum value in s. The maximum value in s is 2401, so the cube root of that is approximately 13.39. The ceiling of that is 14, so the cube of 14 is 2744. So, c contains the list [1, 8, 27, 64, 125, 216, 343, 512, 729, 1000, 1331, 1728, 2197, 2744]

3. Take[Complement[s, c], 30]: This line takes the first 30 elements of the complement of s and c. The complement of two lists is the list of all the elements that are in the first list but not in the second list. So, the complement of s and c is the list of all the integers from 1 to 2401 that are not cubes. The first 30 elements of this list are: [2, 3, 5, 6, 7, 10, 11, 13, 14, 17, 19, 21, 22, 23, 26, 28, 29, 30, 31, 33, 34, 35, 37, 38, 39, 40, 41, 42, 43]

4. Intersection[s, c]: This line finds the intersection of s and c. The intersection of two lists is the list of all the elements that are in both lists. So, the intersection of s and c is the list of all the integers from 1 to 2401 that are both squares and cubes. This list is: [1, 64, 729, 1296]

s = Range[50]^2;
c = Range[1, Ceiling[Surd[Max[s], 3]]]^3;
Take[Complement[s, c], 30]
Intersection[s, c]