Easy way and step by step ✌️
Intuition\n Describe your first thoughts on how to solve this problem. \n\n# Approach\n Describe your approach to solving the problem. \n\n# Complexity\n- Time 
# Intuition\n<!-- Describe your first thoughts on how to solve this problem. -->\n\n# Approach\n<!-- Describe your approach to solving the problem. -->\n\n# Complexity\n- Time complexity:\n<!-- Add your time complexity here, e.g. $$O(n)$$ -->\n\n- Space complexity:\n<!-- Add your space complexity here, e.g. $$O(n)$$ -->\n\n# Code\n```\nclass Solution(object):\n    def sumOfSquares(self, nums):\n        n = len(nums)\n        cound = []\n        # in these lines we need to determine the divisors len(nums) nimg\n        for l  in range(2): # we need to loop through our list twice\n            for i in range(1, n + 1):\n                if n % i == 0: # We need to determine the divisors of len(nums).\n                    cound.append(nums[i-1])\n        # we need to select the list because it affects the result\n        sorted_list = sorted(cound) \n        return sum([sorted_list[i] * sorted_list[i+1] for i in range(0, len(sorted_list)-1, 2)])\n```
['Python']