Problem
NT-B1-M01-P013 Five Consecutive Factors
#13
★★★☆☆ Level 3 of 5
Prove that \(120\mid n(n^2-1)(n^2-4)\) for every integer \(n\).
Factor the expression into five consecutive factors.
We have \(n(n^2-1)(n^2-4)=n(n-1)(n+1)(n-2)(n+2)\), the product of five consecutive integers \(n-2,n-1,n,n+1,n+2\). Among them one number is divisible by \(5\), one is divisible by \(3\), and there are enough factors of \(2\) to make \(8\): among five consecutive integers, either one is divisible by \(4\) and another is even, or one is divisible by \(8\). Hence the product is divisible by \(8\cdot3\cdot5=120\).
Stronger than the usual divisibility by \(6\): the power of \(2\) must be controlled.