# Any integer can be written as the sum of the cubes of 5 integers, not necessarily distinct

Question: Prove that any integer can be written as the sum of the cubes of five integers, not necessarily.

Solution:

We use the identity $6k = (k+1)^{3} + (k-1)^{3}- k^{3} - k^{3}$ for $k=\frac{n^{3}-n}{6}=\frac{n(n-1)(n+1)}{6}$, which is an integer for all n. We obtain

$n^{3}-n = (\frac{n^{3}-n}{6}+1)^{3} + (\frac{n^{3}-n}{6}-1)^{3} - (\frac{n^{3}-n}{6})^{3} - (\frac{n^{3}-n}{6})$.

Hence, n is equal to the sum

$(-n)^{3} + (\frac{n^{3}-n}{6})^{3} + (\frac{n^{3}-n}{6})^{3} + (\frac{n-n^{3}}{6}-1)^{3}+ (\frac{n-n^{3}}{6}+1)^{3}$.

More later,
Nalin Pithwa.

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