Prove by direct method that for
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

n– n = (2k)3– (2k)

n– n = 2k (4k2-1)

Let k (4k– 1) = m

n– n = 2m

Therefore, (n3-n) is even.

Now, let us assume n is odd

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

n– n = (2k + 1)– (2k + 1)

n– n = (2k + 1) [(2k + 1)– 1]

n– n = (2k + 1) [(4k+ 4k + 1 – 1)]

n– n = (2k + 1) [(4k+ 4k)]

n– n = 4k (2k + 1) (k + 1)

n– n = 2.2k (2k + 1) (k + 1)

Let λ = 2k (2k + 1) (k + 1)

n– n = 2λ

Therefore, n3-n is even.

Hence, n3-n is always even

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