**Question:**

**Prove by direct method that for any integer ‘n’, n3 – n is always even.**

**Solution:**

Given n3-n

Let us assume, n is even

Let n = 2k, where k is natural number

n3 – n = (2k)3– (2k)

n3 – n = 2k (4k2-1)

Let k (4k2 – 1) = m

n3 – n = 2m

Therefore, (n3-n) is even.

Now, let us assume n is odd

Let n = (2k + 1), where k is natural number

n3 – n = (2k + 1)3 – (2k + 1)

n3 – n = (2k + 1) [(2k + 1)2 – 1]

n3 – n = (2k + 1) [(4k2 + 4k + 1 – 1)]

n3 – n = (2k + 1) [(4k2 + 4k)]

n3 – n = 4k (2k + 1) (k + 1)

n2 – n = 2.2k (2k + 1) (k + 1)

Let λ = 2k (2k + 1) (k + 1)

n3 – n = 2λ

Therefore, n3-n is even.

Hence, n3-n is always even