#### Kaya Renger Coordinator

11/08/2017 at 14:54-
Denote A = n

^{2}(n^{4}- 1) = n^{2}(n^{2}- 1)(n^{2}^{ }+ 1) is the product of 3 consecutive integers,so \(A⋮3\).We have :\(\circledast A=n^2\left(n-1\right)\left(n+1\right)\left(n^2+1\right)\)

If n is even,then \(n^2⋮4\) and \(A⋮4\)

If n is odd,then n - 1 and n + 1 is even. So,\(A⋮4\)

Hence,\(A⋮4\)

\(\circledast A=n^2\left(n^2-1\right)\left(n^2-4\right)+5n^2\left(n^2-1\right)\)

\(=\left(n-2\right)\left(n-1\right)n^2\left(n+1\right)\left(n+2\right)+5n^2\left(n^2-1\right)\)

\(\left(n-2\right)\left(n-1\right)n^2\left(n+1\right)\left(n+2\right)\)include the product of 5 consecutive integers,so it's divisible by 5.Moreover, \(5n^2\left(n^2-1\right)⋮5\)

Hence,\(A⋮5\)

Since A is divisible by 3,4,5 and 3,4,5 are relatively prime numbers, \(A⋮3.4.5=60\)

**Kaya Renger**selected this answer.