Show that for each integer k ≥ 0, if P(k) is true, then P(k + 1) is true: Let k be any integer with k ≥ 0, and suppose that P(k) is true. Select P(k) from the choices below. 5 is divisible by 7^k - 2^k, 5 is a multiple of 7^k - 2^k, 7^k - 2^k < 5, 7^k - 2^k is divisible by 5. [This is P(k), the inductive hypothesis.] We must show that P(k + 1) is true. Select P(k + 1) from the choices below. 5 is divisible by 7^{k + 1} - 2^{k + 1}, 7^{k + 1} - 2^{k + 1} is divisible by 5, 7^{k + 1} - 2^{k + 1} < 5, 5 is a multiple of 7^{k + 1} - 2^{k + 1}. By the inductive hypothesis and the definition of divisibility, there exists an integer r such that 7^k - 2^k = 5r. Then 7^{k + 1} - 2^{k + 1} = 7 * 7^k - 2 * 2^k = (5 + 2) * 7^k - 2 * 2^k. Continue simplifying the right-hand side of the equation, apply the induction hypothesis, and express the result in terms of k and r: 7^{k + 1} - 2^{k + 1} = 5 * (5r) and this quantity is an integer because k and r are integers. Since 7^{k + 1} - 2^{k + 1} is divisible by 5, then P(k + 1) is true, which completes the inductive step. [Thus both the basis and the inductive steps have been proved, and so the proof by mathematical induction is complete.]