Writing cubic sums a better way

I was asked the following problem:


What is the smallest positive integer n that can be written as a sum of three positive cubes in at least 12 ways?

Can this be coded better way such that it is easy to compute the smallest positive integer that is sum of m nth powers? I'm not a professional programmer, so I managed only to write one solution that works.

l = list()
for i in range(1,500):
    for j in range(i,500):
        for k in range(j,500):
            l.append(i**3+j**3+k**3)
l.sort()
for i in range(0,len(l)-12):
    if l[i] == l[i+12]:
        number = l[i]
for i in range(1,500):
    for j in range(i,500):
        for k in range(j,500):
            if i**3+j**3+k**3 == number:
                print(str(number)+"="+str(i)+"^3+"+str(j)+"^3+"+str(k)+"^3")

Instead of three nested for loops, you can use itertools to enumerate combinations. Also, I would personally prefer to keep a count in a dictionary, as opposed to storing the same numbers over and over.

import itertools

cube_counts = {}

for i, j, k in itertools.combinations(range(1, 500), 3):
    cube = i**3 + j**3 + k**3
    cube_counts[cube] = cube_counts.get(cube, 0) + 1

print(min([cube for cube, count in cube_counts.items() if count >= 12]))

StackExchange Code Review Q#69538, answer score: 5