n⁹ - 6n⁷ + 9n⁵ - 4n³ is divisible by 8640.

8640=2⁶*3³*5

By Fermat's little theorem, n⁵-n0(mod 5)

n⁹-6n⁷+9n⁵-4n³n⁹-n⁷-n⁵+n³(mod 5)n⁵-n³-n⁵+n³(mod 5)0(mod5)

n⁹-6n⁷+9n⁵-4n³ =n⁵(n²-3)²-4n³ =n³(n³-3n-2)(n³-3n+2)

If 3|n, 27|n³, it's ok.

n³-3n-2=(n+1)²(n-2)

If n=2(mod 3), 27|(n³-3n-2)

n³-3n+2=(n-1)²(n+2)

If n=1(mod 3), 27|(n³-3n+2)

 n⁹-6n⁷+9n⁵-4n³ =n³(n-1)²(n+1)²(n-2)(n+2)

If n=0(mod 2), 64|n³(n-2)(n+2), as 8|n(n-2)

 If n=1(mod 2), 64|(n-1)²(n+1)², as 8|(n-1)(n+1)

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