Answer :

Case I: Let n be an even positive integer.

Then, n = 2q

we have

n^{2} – n = (2q)^{2} – 2q

= 4q^{2} – 2q

= 2q (2q – 1 )

⇒ n^{2} – n = 2r, where r = q (2q – 1)

⇒ n^{2} – n is divisible by 2 .

Case: II

Let n be an odd positive integer.

Then, n = 2q + 1

⇒ n^{2} –n

= (2q + 1)^{2} – (2q + 1)

Taking (2q+1) common from both the terms

= (2q +1) (2q+1 –1)

= 2q (2q + 1)

n^{2} – n = 2r, where r = q (2q + 1)

n^{2} – n is divisible by 2.

∴ n ^{2} – n is divisible by 2 for every integer n, Hence proved

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