tmua.fyi Paper 2 Challenge 2 · Question 7

Which statement best justifies the claim that $n^3-n$ is divisible by $6$ for every integer $n$ ?

n^3

and

n

always have the same parity.

6

divides

n^3

for every integer

n

n^3-n=n(n^2-1)

, and

n^2-1

is always even.

DThe result holds for

n=1,2,3

and therefore for all

n

n^3\equiv n\pmod6

because

6

has no repeated prime factor.

n^3-n=(n-1)n(n+1)

, and a product of three integers is always divisible by

6

n^3-n

is even, and every even number is divisible by

6

n^3-n=(n-1)n(n+1)

is a product of three consecutive integers, so one of them is a multiple of

3

and at least one is even.

examday