The square of this positive integer equals the sum of cubes of its digits.

Find this number and show that there are no other solutions.

1 is the only solution

2 - 10 are obviously not solutions.

cubes of 1 - 9 are respectively : 1,8,27,64,125,216,343, 512, 729.

For 11 - 19, the sum of digit cubes are 1 + (1,8,27,64,125,216,343, 512, 729) = (2,9,28,65,126,217,344, 513, 730). Other than 9 (but it is not square of 12), they are not even square numbers;

For 21 - 29, the sum of digit cubes are 8 + (1,8,27,64,125,216,343, 512, 729). Only 9 and 16 are square numbers, which are too small to be the square of 21, 22;

For 31 - 39, the sum of digit cubes are 27 + (1,8,27,64,125,216,343, 512, 729). None of them are square numbers.

40+ are not possible, because their squares are always much more than the sum of digit cubes of each  

Posted by chun on 2017-11-02 09:49:16