# Show that the sum of the cubes of any number of consecutive integers is divisible by the sum of these integers.

Let n + 1, n + 2, .....(n + m) be m consecutive integers

Let denote the sum of the cubes of these integers and denote the sum of these integers

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=

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= [1 + 2 + 3 +........ + n + (n + 1) + ..... + (n + m)] - [1 + 2 + ....... + n]

which is an integer because (n + m) (n + m + 1) and n(n + 1) are both even numbers being the product of two consecutive integers. Hence, the sum of cubes of any number of consecutive integers is divisible by the sum of these integers.