n = a^3 + b^3 + c^3 + d^3 + e^3

Every integer number can be represented as a sum of 5 cubes


(n + 1)^3 = n^3 + 3n^2 + 3n + 1
(n - 1)^3 = n^3 - 3n^2 + 3n - 1

(n + 1)^3 + (n - 1)^3 + (-n)^3 + (-n)^3 + 0^3

n^3 + 3n^2 + 3n + 1 + n^3 - 3n^2 + 3n - 1

So,

(n + 1)^3 + (n - 1)^3 + (-n)^3 + (-n)^3 + 0^3 = 6n

6n + 1 = 6n + 1^3 = (n + 1)^3 + (n - 1)^3 + (-n)^3 + (-n)^3 + 1^3

6n + 2 = 6(n - 1) + 8 = n^3 + (n - 2)^3 + (-n + 1)^3 + (-n + 1)^3 + 2^3

6n + 3 = 6(n + 5) - 27 = (n + 6)^3 + (n + 4)^3 + (-n - 5)^3 + (-n - 5)^3 + (-3)^3

6n + 4 = 6(n + 2) - 8 = (n + 3)^3 + (n + 1)^3 + (-n - 2)^3 + (-n - 2)^3 + (-2)^3

6n + 5 = 6(n + 1) - 1 = (n + 2)^3 + n^3 + (-n - 1)^3 + (-n - 1)^3 + (-1)^3