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Appendix a Solutions and Hints to Exercises Appendix A Solutions and Hints to Exercises Chapter 1 1.1 1. Suppose that there are only finitely many prime p ≡ 3 (mod 4), and label them p1,...,pn. Then consider N = 4p1 ...pn − 1. This number is clearly odd, and has remainder 3 after dividing by 4. So its prime divisors are all odd, and they can’t all be 1 (mod 4), as the product of numbers which are 1 (mod 4) is again 1 (mod 4). So N has a prime divisor p ≡ 3(mod4).But p must be different from all the pi , since p|N but pi N for all i = 1,...,n. So there must be infinitely many primes p ≡ 3(mod4). 2. Suppose that there are only finitely many prime p ≡ 5 (mod 6), and label them p1,...,pn. Then consider N = 6p1 ...pn − 1. This number is clearly odd, and has remainder 5 after dividing by 6. So (6, N) = 1, and the prime divisors of N can’t all be 1 (mod 6), as the product of numbers which are 1 (mod 6) is again 1 (mod 6). So N has a prime divisor p ≡ 5(mod6).But p must be different from all the pi , since p|N but pi N for all i = 1,...,n. So there must be infinitely many primes p ≡ 5(mod6). 1.2 1. Suppose that p|x2 +1. If x is even, clearly p is odd. Then x2 ≡−1(modp) (and as p = 2, we have x2 ≡ 1(modp)), and so x4 ≡ 1(modp). Thus x has order 4 modulo p. So the order of the multiplicative group is a multiple of 4, which is the order of x. But the multiplicative group has order p − 1. Thus 4|p − 1, and so p ≡ 1(mod4). 2. Suppose that there are only finitely many prime p ≡ 1 (mod 4), and label them p1,...,pn. 2 Then consider N = (2p1 ...pn) + 1. This number is clearly odd. Let p be a prime divisor of N.By(1),p ≡ 1(mod4).But p must be different from all the pi , since p|N but pi N for all i = 1,...,n. So there must be infinitely many primes p ≡ 1(mod4). 1.3 Suppose that x ≡ 1 (mod 3). Then x2 + x + 1 ≡ 1(mod6).Letp|x2 + x + 1. Then (x2 + x + 1)(x − 1) = x3 − 1 ≡ 0(modp). Thus x3 ≡ 1(modp). As F. Jarvis, Algebraic Number Theory, Springer Undergraduate 257 Mathematics Series, DOI: 10.1007/978-3-319-07545-7, © Springer International Publishing Switzerland 2014 258 Appendix A: Solutions and Hints to Exercises above, 3|p − 1, and so p ≡ 1(mod3). Now suppose that there are only finitely many prime p ≡ 1 (mod 6), and label them p1,...,pn. 2 Then consider N = (6p1 ...pn) + (6p1 ...pn) + 1. Let p be a prime divisor of N. By the above, p ≡ 1 (mod 3). Also, N is odd, so p must be odd too, and so p ≡ 1(mod6).Butp must be different from all the pi , since p|N but pi N for all i = 1,...,n. So there must be infinitely many primes p ≡ 1(mod6). 1.4 1. If rn+1 < rn/2, then certainly rn+2 < rn+1.Ifrn+1 = rn/2, in the next step, rn+2 = 0. Finally, if rn+1 > rn/2, in the next step, rn+2 = rn −rn+1 < rn/2. So in all cases, rn+2 < rn/2. 2. For every two steps of the algorithm, the remainder halves (at least). 3. If b < 2n, this means that the remainder after 2n steps is less than 1, so must be 0, and the algorithm has terminated. Thus numbers which are at most 2n require at most 2n applications of the division algorithm. 1.5 1. Allowing negative remainders, we have 630 = 5 × 132 − 30 132 = 4 × 30 + 12 30 = 2 × 12 + 6 12 = 2 × 6 + 0 (so only one step is saved in this case). 2. Similar to the last exercise. 1.6 The existence of the division algorithm just follows from long division of poly- nomials, and the Euclidean algorithm is again simply a repetition of the division algorithm. For the example, the first step is x5+4x4+10x3+15x2+14x+6 = x(x4+4x3+9x2+12x+9)+x3+3x2+5x+6 and then repeating in the same way as the Euclidean algorithm gives x4 + 4x3 + 9x2 + 12x + 9 = (x + 1)(x3 + 3x2 + 5x + 6) + x2 + x + 3 and finally x3 + 3x2 + 5x + 6 = (x + 2)(x2 + x + 3) + 0, so the highest common factor is x2 + x + 3. 1.7 We’ll use the modified algorithm of Exercise 1.5 as an example. 999 = 1 × 700 + 299 700 = 2 × 299 + 102 299 = 3 × 102 − 7 102 = 15 × 7 − 3 7 = 2 × 3 + 1 Appendix A: Solutions and Hints to Exercises 259 Then 1 = 7 − 2 × 3 = 7 − 2 × (15 × 7 − 102) = 2 × 102 − 29 × 7 = 2 × 102 − 29 × (3 × 102 − 299) = 29 × 299 − 85 × 102 = 29 × 299 − 85 × (700 − 2 × 299) = 199 × 299 − 85 × 700 = 199 × (999 − 700) − 85 × 700 = 199 × 999 − 284 × 700 1.8 As in Lemma 1.20. 1.9 As in Theorem√ 1.19, using the result of the last exercise to derive unique fac- torisation in Z[ −2].√ √ 1.10 Given α and β in Z[ 2], write α/β = x + y 2 with x, y ∈ Q, and√ let m and n be the closest integers to x and y respectively. Put κ = m + n 2, and ( / − ) =|( − )2 − ( − )2|≤ 1 then N α β κ x m 2 y n 2 . The result follows as in Lemma 1.20. √ √ 1.11 We can take α = −7 and β = 2. Then the imaginary part of α/β is 7 , and √ √ √ 2 7 > Z[ − ] Z[ − ] this is at least 2 1 from any point of 7√, since points in 7 have imaginary parts which are integer multiples of 7. We will see later that the problem is that we are working with the wrong ring! ( 2 ) = ( 3 ) = ( 5 ) = ( 7 ) = ( 11 ) =− 1.12 We find 1009 1009 1009 1009 1, but 1009 1(ei- ther by using known properties of Legendre symbols or using Euler’s formula (p−1)/2 ≡ ( a ) (1009−1)/4 = 252 ≡ a p (mod p)). Compute 11 11 469 (mod 1009). Then x = 469 is a solution (and so is x =−469 ≡ 540 (mod 1009)). 1.13 1. The elements except for those where f (s) = s may be partitioned into pairs {s, f (s)}.SoS is the union of FixS( f ) with all the pairs {s, f (s)}, and so |S| is the sum of |FixS( f )| and 2r, where r is the number of pairs. 2. This is rather fiddly! First take (x, y, z) with x < y − z. Then f (x, y, z) = (X, Y, Z), where X = x + 2z, Y = z and Z = y − x − z. Then f ( f (x, y, z)) = f (X, Y, Z). But clearly X = x + 2z > 2z = 2Y , and as X > 2Y , f (X, Y, Z) = (X − 2Y, X − Y + Z, Y ) = (x, y, z). The other two cases are similar. 3. Suppose that f (x, y, z) = (x, y, z). It is easy to see that (x, y, z) must satisfy y − z < x < 2y, as the other two possibilities get mapped to triples with different conditions. Then f (x, y, z) = (2y − x, y, x − y + z) = (x, y, z),sox = y. Then (x, x, z) ∈ S means that x2 + 4xz = p,so p = x(x+4z).Butp is prime, so x = 1 and then z = k (where p = 4k+1). So the unique fixed point is (1, 1, k). By the first part, |S| is odd. 4. It is clear that f is another involution on S.As|S| is odd (by the last part), f has a fixed point (by the first part). 5. If (x, y, z) is a fixed point of f , then (x, z, y) = (x, y, z),soy = z.So there is a point of S of the form (x, y, y), and so x2 + 4y2 = p, and thus p is the sum of two squares. 1.14 For Step 1, we have already found x = 469 in Exercise 1.12. The Euclidean algorithm for 1009 and 469 is: 260 Appendix A: Solutions and Hints to Exercises 1009 = 2 × 469 + 71 469 = 6 × 71 + 43 71 = 1 × 43 + 28 43 = 1 × 28 + 15 28 = 1 × 15 + 13 15 = 1 × 13 + 2 13 = 6 × 2 + 1 2 = 2 × 1 + 0 and the algorithm√ gives 1009√= 282 + 152. 1.15 Write x3 = (y + −2)(y − −2).Ifx were even, we would have y2 + 2 ≡ 0 (mod 8), which is not possible, so x is odd, and then y will also be odd. √ √ √ If α is a common√ divisor of y + −2√ and y − −2,√ then α|2 −2, and√ so N(α)√|N(2 −2) = 8. So α is a unit, or −2|α.Butif −2|α, and α|y+ −2, then −2|y, and then y would be even. So this is not possible, and α is a unit. √ √ Then we have√ y + −2 = uβ3, y − −2 = vγ3 for units u and v. The only units in Z[ −2] are ±1, so we need √ √ y + −2 =±(a + b −2)3. Expanding, and taking the imaginary parts gives 1 =±(3a2b − 2√b3),or b(3a2 −√2b2) =±1. This is solved with b =±1, a =±1, so y + −2 = ±(1 ± −2)3, and we can read off y =±5, and then it is easy to recover x = 3.
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