Integers_Recreation_One

'''
PROBLEM: [From Codewars: https://www.codewars.com/kata/55aa075506463dac6600010d/train/python]

Divisors of 42 are : 1, 2, 3, 6, 7, 14, 21, 42.
These divisors squared are: 1, 4, 9, 36, 49, 196, 441, 1764.
The sum of the squared divisors is 2500 which is 50 * 50, a square!

Given two integers m, n (1 <= m <= n) we want to find all integers
between m and n whose sum of squared divisors is itself a square.
42 is such a number.

The result will be an array of arrays or of tuples (in C an array of Pair)
or a string, each subarray having two elements, first the number
whose squared divisors is a square and then the sum of the squared divisors.

#Examples:

list_squared(1, 250) --> [[1, 1], [42, 2500], [246, 84100]]
list_squared(42, 250) --> [[42, 2500], [246, 84100]]

'''


def list_squared(m, n):
    output = []
    for num in range(m,n+1):
        summ = sum([i*i for i in range(1,num+1) if num%i==0])
        if (summ**.5)-int(summ**.5) == 0:
            output.append([num,summ])
        else:
            continue
    return output


print(list_squared(42,250))