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Every integer can be expressed as the sum of five perfect cubes.
Let $n$ be an integer and $\dfrac{n^3-n}{6}=k$. Since $n^3-n$ is the product of three consecutive integers, it is a multiple of $6$, hence $k$ is also an integer.
\begin{align*}n^3-n&=6k\\&=(k+1)^3+(k-1)^3-k^3-k^3\\\therefore n&=n^3-(k+1)^3-(k-1)^3+k^3+k^3\\\end{align*} We can expressed $n$ as $$n=n^3+\bigg(-1-\dfrac{n^3-n}{6}\bigg)^3+\bigg(1-\dfrac{n^3-n}6\bigg)^3+\bigg(\dfrac{n^3-n}{6}\bigg)^3+\bigg(\dfrac{n^3-n}{6}\bigg)^3$$ Proof completed.
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Sat, 01 Sep 2018 10:49 GMT