Math 327, Solutions to Hour Exam 2 1. Prove that the function f : N × N ! N defined by f(m; n) = 5m10n is one-to-one. Solution: Begin by observing that we may rewrite the formula for f as f(m; n) = 5m+n2n. Suppose f(m1; n1) = f(m2; n2), then 5m1+n1 2n1 = 5m2+n2 2n2 = k for some common integer k. By the fundamental theorem of arithmetic, the exponents in the prime decomposition of k are unique and hence m1 + n1 = m2 + n2 and n1 = n2: Substituting the second equality into the first we have m1 +n1 = m2 +n1 and hence m1 = m2. Therefore, (m1; n1) = (m2; n2), which shows that the function is one-to-one. 2a. Find all integers x, 0 ≤ x < 10 satisfying the congruence relation 5x ≡ 2 (mod 10). If no such x exists, explain why not. Solution: If the relation has an integer solution x, then there exists an integer k such that 5x − 2 = 10k, which can be rewritten as 2 = 5x − 10k = 5(x − 2k). But since x − 2k is an integer, this shows that 5 2, which is false. This contradiction shows that there are no solutions to the relation. 2b. Find all integers x, 0 ≤ x < 10 satisfying the congruence relation 7x ≡ 6 (mod 10). If no such x exists, explain why not. Solution: By Proposition 4.4.9 in the book, since 7 and 10 are relatively prime, there exists a unique solution to the relation. To find it, we use the Euclidean algorithm to see that 21 = 7(3) + 10(−2), which means that 7(3) ≡ 1(mod 10). Multiplying both sides of the relation by 3 we obtain x ≡ 18 (mod 10). Hence x = 18 − 10 = 8 is the desired solution. 2c. Find all integers x and y, 0 ≤ x; y < 4 that satisfies the pair of congruences 3x + 2y ≡ 3 (mod 4) x + 2y ≡ 1 (mod 4) : Solution: Subtracting the second line from the first, we obtain that 2x ≡ 2(mod 4). It can be seen that the solutions to this relation are x = 1; 3. When x = 1, we may substitute this into the bottom relation to obtain 2y ≡ 0 (mod 4), which has solution y = 0 and y = 2. Substituting the (x; y) = (1; 0) and (x; y) = (1; 2) into the top relation easily verifies that each pair is a solution. Turning to the case where x = 3, substituting this into the bottom relation leads to the relation 2y ≡ −2 (mod 4), which is equivalent to the relation 2y ≡ 2 (mod 4). This leads to the pairs (3; 1) and (3; 3) and it can be verified that they are indeed solutions by substitution into the top relation. In summary, the pairs of solutions are (1; 0), (1; 2), (3; 1) and (3; 3). 1 3. Find the smallest nonnegative integer x that satisfies the following system of congruences x ≡ 1 (mod 2) x ≡ 2 (mod 5) x ≡ 2 (mod 9) : Solution: We begin by solving the pair of relations y ≡ 1 (mod 2) y ≡ 2 (mod 5) which has a unique solution modulo 10 by the Chinese Remainder Theorem. Since 1 = 5(1) + 2(−2) we know that y must be congruent to 5(1)(1) + 2(−2)(2) = −3 ≡ 7 (mod 10). Thus if x is a solution to the system, it must also satisfy the pair of relations x ≡ 7 (mod 10) x ≡ 2 (mod 9) : This system has a solution which is unique modulo 90. To find it, we observe that 1 = 10(1) + 9(−1) and hence the solution must be congruent to (10)(2) + 9(−1)(7) = −43. The smallest nonnegative integer in the congruence class of −43 is x = −43 + 90 = 47. 4. Use mathematical induction to prove that 7n − 1 is divisible by 6 for any n ≥ 1. Solution: We first check the base step, that the statement is true for n = 1. Indeed, 71 − 1 = 6 which is clearly divisible by 6. For the induction step, we now assume that k ≥ 1 is an integer and that 6 7k − 1. In other words, there exists an integer q such that 7k − 1 = 6q. Therefore 7k+1 − 1 = 7 · 7k − 1 = 7(6q + 1) − 1 = 7(6q) + 7 − 1 = 7(6q) + 6 = 6(7q + 1): Since 7q + 1 is an integer, this shows that 6 7k+1 − 1. 5. Let x, a, and b be integers such that x ab. Suppose also that x and a are relatively prime. Use properties of the greatest common divisor to give a complete proof that x b. Solution: Since x and a are relatively prime, we know that there exist integers m; n such that 1 = ma+nx. Furthermore, since x ab there exists an integer q such that xq = ab. Multiplying both sides of this equa- tion by b, we obtain b = mab + nbx = mqx + nbx = (mq + nb)x. Since mq + nb is an integer, this shows x b as desired. Equivalently, we could have observed that x mab and x nbx, which implies that x mab + nbx, that is, x b. 2.