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 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 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 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 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.

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