**To Prove**

n^{2} + n is even ∀ 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) will also be even since multiplying even number with any integer gives even number.**n is odd**

If n is odd, then (n + 1) is even since adding odd number with odd number gives even number. This implies that n * (n + 1) will also be even since multiplying even number with odd number gives even number.

Therefore, we can say that n^{2} + n is even ∀ n ∈ Z.