1999 OIM Problems/Problem 1
Problem
Find all positive integers that are less than 1000 and satisfy the following condition: the cube of the sum of their digits is equal to the square of that integer.
~translated into English by Tomas Diaz. ~orders@tomasdiaz.com
Solution
Insight: Every number that satisfies this must be a cube itself
Proof/reasoning: let the sum of digits be 
and the original number be 
. Then 
. If weren’t a cube, neither would 
, but it is. Therefore, is a cube.
Now we list out all cubes that are smaller than 
and 
.
So the only integers that satisfy this condition are 
and 
~Archieguan
