If $$n$$ is positive integer, show that $$n(n^2-1)(n^2-5n+26)$$ is divisible by 120

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Uniteasy Admin · Accepted Answer

$$n(n^2-1)(n^2-5n+26)$$
$$=(n-1)n(n+1)[(n-2)(n-3)+20]$$
$$=(n-2)(n-2)(n-1)n(n+1)+20(n-1)n(n+1)$$
The first term is 5 consecutive integers, which must have 2 even numbers and one of them must be multiples of 4. There also must be a number that is divisible by 3. There also must be a number that is divisible by 5. 
Therefore it must be divisible by $$2	imes 4	imes 3	imes 5=120$$
The second term is the product of 20 and 3 consecutive numbers, in which there must be 1 even number and a multiple of 3. 
Therefore it must be divisible byy $$20	imes2	imes3=120$$

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