**To Prove**

n^{2} – n is divisible by 2 ∀ n ∈ Z

**Proof**

n^{2} – n = n * (n – 1)

For the value of n, we have 2 cases

**n is even**

If n is even, then n * (n – 1) is even since multiplying even number and odd number gives even number. Therefore, n^{2}– n is divisible by 2 if n is even.**n is odd**

If n is odd, then (n – 1) is even. This implies that n * (n – 1) is also even since multiplying even number and odd number gives even number. Therefore, n^{2}– n is divisible by 2 if n is odd.

Therefore, we can say that n^{2} – n is divisible by 2 ∀ n ∈ Z

**Alternative Proof**

Let n be an even number. If n is even, then we can write n as

n = 2k, ∀ k ∈ Z

n^{2} – n = (2k)^{2} – (2k)

= 4k^{2} – 2k

= 2 * (2k^{2} – k)

Therefore, n^{2} – n is divisible by 2 if n is even.

Let n be an odd number. If n is odd, then we can write n as

n = 2k + 1, ∀ k ∈ Z

n^{2} – n = (2k + 1)^{2} – (2k + 1)

= 4k^{2} + 1 + 4k – 2k – 1

= 4k^{2} + 2k

= 2 * (2k^{2} + k)

Therefore, n^{2} – n is divisible by 2 if n is odd.

Using the above 2 proofs, we can say that n^{2} – n is divisible by 2 ∀ n ∈ Z.