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SOLUTIONS TO HOMEWORK 2 Due Wednesday, September 21 in class. 1(a) Use the Euclidean algorithm to find gcd(9797, 1649). Ans: We use division with remainder to find the gcd, as on p. 25. 9797 = 5 ∗ 1649 + 1552 1649 = 1 ∗ 1552 + 97 1552 = 16 ∗ 97 + 0 Hence gcd(9797, 1649) = 97. 1(b). Find integers m and n so that gcd(9797, 1649) = m · 9797 + n · 1649. Ans: One method is given in the example on p. 27: (9797, 1649) = (a, b) (1649, 1552) = (b, a − 5b) (1552, 97) = (a − 5b, b − (a − 5b)) From this we can read off 97 = −1 ∗ 9797 + 6 ∗ 1649. So, in the notation above, one possible solution is (m, n) = (−1, 6). 1(c). Write the continued fraction for 9797/1649 (see page 26 of the text). Ans: Here we follow pp. 25-26, using our three lines of computation from part (a): 5 ∗ 1649 + 1552 1552 1 9797 = =5+ =5+ 1649 1649 1649 1649/1552 1 1 1 =5+ =5+ 1 1649/1552 (1 ∗ 1552 + 97)/1552 1 + 16 √ 2. Let R be the collection of complex numbers m + n −3, with m and n integers. I’ll write assumptions like this as √ R = {m + n −3 | m, n ∈ Z}. 5+ 2(a). Explain why R is closed under addition and multiplication. √ √ Ans: Let r1 = m1 + n1 −3 and r2 = m2 + n2 −3 denote two arbitrary elements of R. We need to check that r1 + r2 ∈ R and r1 r2 ∈ R to show that R is closed under addition and multiplication, respectively. First, √ r1 + r2 = (m1 + m2 ) + (n1 + n2 ) −3, and since m1 + m2 ∈ Z and n1 + n2 ∈ Z, we have that r1 + r2 ∈ R. Hence, R is closed under addition. 1 2 SOLUTIONS TO HOMEWORK 2 Now we compute r1 r2 : √ √ (m1 + n1 −3)(m2 + n2 −3) = √ √ √ = (m1 m2 + n1 n2 −3 −3) + (m1 n2 + m2 n1 ) −3 √ = (m1 m2 − 3n1 n2 ) + (m1 n2 + m2 n1 ) −3. Because m1 m2 − 3n1 n2 ∈ Z and (m1 n2 + m2 n1 ) ∈ Z, we have that r1 r2 ∈ R, and so R is closed under multiplication. 2(b). Define a “norm” on R by √ km + n −3k = m2 + 3n2 . Prove that krk is a non-negative integer (for all r ∈ R), and that kr · sk = krk · ksk (r, s ∈ R). Ans: Because m, n ∈ Z by the definition of R, m2 and n2 are non-negative integers, and hence m2 + 3n2 is a non-negative integer. We’ll now use our calculation of r1 r2 above to prove kr1 · r2 k = kr1 k · kr2 k (r1 , r2 ∈ R). On one hand, kr1 · r2 k = (m1 m2 − 3n1 n2 )2 + 3 ∗ (m1 n2 + m2 n1 )2 = (m21 m22 + 9n21 n22 − 6m1 m2 n1 n2 ) + 3(m21 n22 + m22 n21 + 2m1 m2 n1 n2 ) = m21 m22 + 9n21 n22 + 3m21 n22 + 3m22 n21 On the other hand, kr1 k · kr2 k = (m21 + 3n21 )(m22 + 3n22 ) = m21 m22 + 9n21 n22 + 3m21 n22 + 3m22 n21 . This proves the desired result. Note: If you know some results from complex variables, there is a much cleaner solution. (The proofs of the two necessary facts are similar to the above proof: you write things out in terms of real and complex coordinates and expand.) Let z denote the complex conjugate of z. Here are the facts: for every x, y ∈ C, (1) xy = x ∗ y (2) xy = yx. Now, note that kxk = x ∗ x. Then using the two facts above, kxyk = xyxy = xxyy = kxkkyk. 2(c). Show that the only elements of R having a multiplicative inverse are ±1. Ans: First, ±1 both have multiplicative inverses: each is its own inverse. Now, √ consider any number r = m + n −3 with a multiplicative inverse r−1 . By part (b), we know k1k = kr ∗ r−1 k = krk ∗ kr−1 k. Since k1k = 1, we have 1 = krk ∗ kr−1 k. Also by part (b), we know that krk and kr−1 k are non-negative integers. But if the product of two non-negative integers is 1, then both of those non-negative integers are 1. The equation krk = 1 is equivalent to m2 + 3n2 = 1. Because m and n are SOLUTIONS TO HOMEWORK 2 3 integers, it’s clear that m = ±1, n = 0. So, r = ±1. Since our only restraint on r was that it have norm 1, we’re done. 2(d). Call an element r√of R prime if√it has exactly four divisors (namely ±1 and ±r). Prove that 2, 1 + −3, and 1 − −3 are all prime in R. Ans: The idea is similar to the idea in the previous problem. We write a factorization r = r1 r2 . We then compute norms of both sides, and get that r1 or r2 is ±1, and hence the only divisors are the listed ones. Here we go. Assume 2 = r1 r2 . Then k2k = kr1 k ∗ kr2 k. And we compute k2k = 4. Again, since norms are non-negative integers, we have kr1 k, kr2 k have possible values 1, 2, and 4, and their product must be 4. But m2 + 3n2 = 2 has no integer solutions. Hence one of kr1 k, kr2 k is 1, and the other is 4. From part (c), the only elements with norm 1 are ±1. If it’s 1, then the other element is 2. If it’s −1, the other element is −2. √ The √ exact same argument works for the other two cases, because k1 + −3k = k1 + −3k = k2k = 4. Hence, these three elements are all prime. 2(e). Prove that any element of R other than 0 and ±1 is a product of primes in R: so prime factorization is possible in R. Ans: For any non-zero element r ∈ R, clearly all of 1, −1, r, −r are divisors of it. If r 6= 0, 1, −1, then the four given elements are distinct. So, an element r 6= 0, 1, −1 which is not prime has more than 4 divisors. We want to show that every r 6= 0, 1, −1 has a prime factorization. We’ll prove this by induction on the norm of r. As we saw above, given r 6= 0, 1, −1, norms 0, 1, 2 √ are impossible. If krk = 3, then from the formula for the norm it’s clear that r = ± 3. A factorization r = r1 r2 leads to a factorization kr1 k ∗ kr2 k = krk = 3, and as in the previous problem, since norms are non-negative integers, one of r1 and r2 has norm 1 (this is true in general for any element r with prime norm [prime as an integer, that is]). Hence one of them is ±1, and hence the other is ±r. So, the base case for our induction is complete: if krk = 3, then r is a product of primes; in fact, r is prime in this case. For our inductive hypothesis, we now assume that for any r 6= 0, 1, −1 with krk < N (where N ≥ 4), r is a product of primes. Now consider an r with norm N . If r = r1 r2 only has solutions with either r1 or r2 equalling ±1, then r is prime and we’re done. Otherwise both kr1 k and kr2 k ≥ 2. Since kr1 k ∗ kr2 k = N , we have 2 ≤ kr1 k, kr2 k < N . We can then apply the inductive hypothesis to write both r1 and r2 as a product of primes, and thus also r = r1 r2 as a product of primes. 2(f ). What remark would you make about the equations √ √ 2 · 2 = 4 = (1 + −3)(1 − −3)? Ans: They show√that the factorization into primes is not unique, since we saw √ above that 2, 1 + −3 and 1 − −3 were all primes. 3. Suppose that a > b > 1 are relatively prime natural numbers. According to the Euclidean algorithm, it is possible to find integers x and y so that ax + by = 1. Prove that we can actually arrange 0 < x < b, −a < y < 0. 4 SOLUTIONS TO HOMEWORK 2 (Hint: page 34 of the text might help.) Note the correction to the originally posted homework: we now have a > b > 1 rather than a > b ≥ 1. Ans: Because a(−bq) + b(aq) = 0 for any integer q, the given equation ax + by = 1 implies a(x − bq) + b(y + aq) = 1. Applying division with remainder to x and b, we find q and r so that x = bq + r with 0 ≤ r < b. Then ar + b(y + aq) = 1. This r satisfies the required inequality, because if r = 0, then we see that b divides 1, which is impossible since we require b > 1. Set s = y + aq. We claim that −a < s < 0. Assume first s ≤ −a. Then 1 = ar + bs ≤ ar − ba = a(r − b) < 0, where the last inequality holds because r < b and a is positive. But we see 1 < 0, which is a contradiction. Now assume s ≥ 0. Then 1 = ar + bs ≥ ar ≥ a > 1, which yields another contradiction. 4. Problems 2.5.6 and 2.5.7 on page 33 of the text. Ans: 2.5.6: Note that q is odd, and thus is a prime different from 2. By the uniqueness of prime factorization, we know any divisor of 2p−1 q (which we see there written in terms of its prime factorization) has the form 2a q b where 0 ≤ a ≤ p − 1 and 0 ≤ b ≤ 1 are integers. We are only interested in the proper divisors, and so we remove 2p−1 q 1 from consideration. We are left precisely with the given list. Ans: 2.5.7 The equality is more evident when we write the left-hand side as (2 − 1)(1 + 2 + 22 + · · · + 2p−2 ). Expanding, we see that all the terms cancel except for −1 and 2p−1 . Hence 1 + 2 + 22 + · · · + 2p−2 = 2p−1 − 1. Now we add up all the proper divisors: 1 + 2 + · · · + 2p−1 + q(1 + 2 + · · · + 2p−2 ) = (1 + 2 + · · · + 2p−2 ) + 2p−1 + q(1 + 2 + · · · + 2p−2 ) = (2p−1 − 1) + 2p−1 + q(2p−1 − 1) = 2p − 1 + (2p − 1)(2p−1 − 1) = 2p − 1 − (2p − 1) + q2p−1 = 2p−1 q.