Book of Proof - Amazon S3

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9. For any integer n ≥ 0, it follows that 24 | (5 2n − 1).
Proof. The proof is by mathematical induction.
(1) For n = 0, the statement is 24 | (5 2·0 − 1). This is 24 | 0, which is true.
(2) Now assume the statement is true for some integer n = k ≥ 1, that is assume
24 | (5 2k − 1). This means 5 2k − 1 = 24a for some integer a, and from this we
get 5 2k = 24a + 1. Now observe that
5 2(k+1) − 1 =
5 2k+2 − 1 =
5 2 5 2k − 1 =
5 2 (24a + 1) − 1 =
25(24a + 1) − 1 =
25 · 24a + 25 − 1 = 24(25a + 1).
This shows 5 2(k+1) − 1 = 24(25a + 1), which means 24 | 5 2(k+1) − 1.
This completes the proof by mathematical induction.
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11. For any integer n ≥ 0, it follows that 3 | (n 3 + 5n + 6).
Proof. The proof is by mathematical induction.
(1) When n = 0, the statement is 3 | (0 3 + 5 · 0 + 6), or 3 | 6, which is true.
(2) Now assume the statement is true for some integer n = k ≥ 0, that is assume
3 | (k 3 + 5k + 6). This means k 3 + 5k + 6 = 3a for some integer a. We need to
show that 3 | ((k + 1) 3 + 5(k + 1) + 6). Observe that
(k + 1) 3 + 5(k + 1) + 6 = k 3 + 3k 2 + 3k + 1 + 5k + 5 + 6
= (k 3 + 5k + 6) + 3k 2 + 3k + 6
= 3a + 3k 2 + 3k + 6
= 3(a + k 2 + k + 2).
Thus we have deduced (k + 1) 3 − (k + 1) = 3(a + k 2 + k + 2). Since a + k 2 + k + 2 is
an integer, it follows that 3 | ((k + 1) 3 + 5(k + 1) + 6).
It follows by mathematical induction that 3 | (n 3 + 5n + 6) for every n ≥ 0.
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13. For any integer n ≥ 0, it follows that 6 | (n 3 − n).
Proof. The proof is by mathematical induction.
(1) When n = 0, the statement is 6 | (0 3 − 0), or 6 | 0, which is true.