Base case: if n = 1, then
1² - 1 = 0
which is even.
Induction hypothesis: assume the statement is true for n = k, namely that k ² - k is even. This means that k ² - k = 2m for some integer m.
Induction step: show that the assumption implies (k + 1)² - (k + 1) is also even. We have
(k + 1)² - (k + 1) = k ² + 2k + 1 - k - 1
… = (k ² - k) + 2k
… = 2m + 2k
… = 2 (m + k)
which is clearly even. QED
