Exercise 2.1.26 (The product of three consecutive integers is divisible by $504$ if the middle one is a cube)

Proof. If we write $n^3$ the central cube, the product of the three consecutive integers is

The factorisation of $P$ is

Since $504=8\times 7\times 9=2^3\cdot 3^2\cdot 7,$ we must prove $7\mid P,2^3\mid P,3^2\mid P$.

By (little) Fermat’s theorem,

• If $n$ is even, $2^3\mid n^3$, thus $2^3\mid P$.
• If $n$ is odd, $2\mid n-1,2\mid n+1$ and $2\mid n^2+n+1$, thus $2^3\mid P$.

• If $n\equiv 0(\mathit{mod}3)$, $3^3\mid n^3$, thus $3^3\mid P=n^3(n^6-n)$.
• If $n\equiv 1(\mathit{mod}3)$, $3\mid x-1$ and $3\mid x^2+x+1$, thus $3^2\mid P=n^3(n-1)(n^2+n+1)(n+1)(n^2-n+1)$.

• If $n\equiv -1(\mathit{mod}3)$, $3\mid x+1$ and $3\mid x^2-x+1$, thus $3^2\mid P$.

  Since $7\mid P,2^3\mid P,3^2\mid P$, where $7,2^3,3^2$ are relatively prime by pairs, we obtain

  $$504=2^3\cdot 3^2\cdot 7\mid (n^3-1)n^3(n^3+1).$$

  The product of three consecutive integers is divisible by $504$ if the middle one is a cube.