Errata to An Invitation to Modern Number Theory

Additional problems available here.

• Chapter 1: 
  • page 4: d should be 2087 and not 2807 (although amazingly both values work!)
  • page 5: in equations (1.1), (1.2) and 2^k ≤ x: replace x with n.
  • page 9: Exercise 1.2.17: should refer to Lemma 6.4.9, not 6.4.3.
  • page 12: the left hand side of equation (1.18) should be N+k choose k, and not N+k choose k-1.
  • page 15: Definition 1.3.5: it should be blackboard N (for natural numbers).
  • page 16: after Definition 1.4.6 replace the empty set with the set containing the identity element.
     
• Chapter 2:
  • page 30: after (2.3) it should be the number of primes at most n (not at most x).
  • page 34: modify (2.16) and the text around it to: We write F(x) ~ G(x) if F(x) - G(x)  = o(F(x)), and say F and G are of the same
    order. Add to Exercise 2.2.4:  Let F(x) = x²/2 and G(x) = Σ_{n ≤ x} n. Prove F(x) ~ G(x).

• Chapter 3:
  • page 56: insert a factor of (-1)ⁿ on the right hand side of the last line of (3.36).
  • page 60: Exercise 3.2.19: the hint for the first statement should read (cosθ + 1)² and not (cosθ - 1)²; better yet, consider 2(cosθ + 1)².
  • page 79: Section 3.3.4 (Imprimitive Characters): The definition was slightly off, and should read: "By definition a character χ modulo m is periodic. If (n,m) > 1 then χ(n) = 0, and if (n,m) = 1 then |χ(n)| = 1 (and hence χ(n) \neq 0 for n relatively prime to m). If for n relatively prime to m the character χ has period equal to m, χ is a called a primitive character, otherwise it is imprimitive. For example, consider the character modulo 6 given by χ(1) = 1, χ(5) = -1; as 2 and 3 are not relatively prime to 6, χ(0) = χ(2) = χ(3) = χ(4) = 0. Let ψ be the character modulo 3 given by ψ(0) = 0, ψ(1) = 1 and ψ(2) = -1. For n relatively prime to 6, χ(n) = ψ(n). Thus χ is induced from a character modulo 3."
     
• Chapter 4:
  • page 95: Exercise 4.4.1 should read: Show that while (4.56) has integral solutions, (4.57) does not.
  • page 98: Exercise 4.4.13 should read: Let F(x₁,x₂) = x₁² + x₂². Prove the number of solutions N_F(p) of F(x₁,x₂) ≡ 0 mod p satisfies |N_F(p) - p| ≤ C(F) p^1/2 for some constant C(F). Note that although F(x₁,x₂) is not absolutely irreducible over C (it equals (x₁+ix₂)(x₁-ix₂)), it is over Q and Z.

• Chapter 5:
  • page 133: after Exercise 5.7.4 it should be AB^+/-1 and not AB^+/-.
  • pages 134 - 135: in equations (5.81) through (5.84) a₀ should be replaced with |a₀|.
  • page 145: before Lemma 6.3.6 the reference should be to section 6.5, not section 6.3.6.

• Chapter 7:
  • page 158: In equation (7.2) the subscripts N₀+1 and N₀ should be replaced with -N₀+1 and -N₀.

• Chapter 8:
  • page 214: the variance and standard deviation in section 8.4.2 of X are both 1. This means the variance on page 215 should be 2N, not 2N/4. The first line of (8.73) should have (2N)^2N, not 2^2N.  Equation (8.74): It should be exp(-k² / N) and not exp(-2 k² / N). To see this, one should take the logarithm of the product in the denominator and Taylor expand. Equation (8.75) is correct. The fraction in (8.75) should be written as 2 / Sqrt(2π 2N); the exponential term is exp(-(2k)² / 2(2N)). This looks like a Gaussian with mean 0 and variance 2N except for the fact that we have a 2 in the numerator of the normalization constant. This 2 is because Prob(S_2N = m) is zero if m is odd, and thus if we spread this out to be over 2k and 2k+1, the factor of 2 is basically replaced with 1.
  • The exposition of the proof of the Central Limit Theorem on pages 214 - 215 has been expanded, and can be downloaded HERE.

• Chapter 11:
  • page 265: replace all the text from equation (11.41) to the end of the proof with: Therefore g_x0(x) is bounded everywhere, say by B. As
    g_x0 is a bounded function, it is square-integrable, and thus the Riemann-Lebesgue Lemma (see Exercise11.2.2) implies that its Fourier coefficients tend to zero. This completes the proof, as i ∫^1/2_1/2 g_x0(x) sin((2N+1)πx)dx = Im(∫^1/2_1/2 g_x0(x) exp(2π i (2N+1)x) dx); thus our integral is just the imaginary part of the 2N+1^st Fourier coefficient, which tends to zero as N →∞. Hence as N →∞, S_N(x₀) converges (pointwise) to f(x₀).

• Chapter 13:
  • page 321: Should say we are approximate F_N on the major arc M_a,q. The integral should be over just the major arc M_a,q and not all the major arcs. In (13.74) replace Q³ with Q.
  • page 322: equation (13.78): should equal -1 otherwise (not 0).
  • page 324: equation (13.90): should equal N log N (not N).

• Chapter 14:
  • page 353: equation (14.111): c_p(N) instead of c_q(N).

• Chapter 15:
  • page 362: it should be Prob(A)dA and not Prob(A) in equation (15.5).
  • page 363: Exercise 15.1.5 should only be on even probability distributions, p(-x) = p(x).

• Chapter 16:
  • page 395: carriage return missing in equation (16.2.4) so hard to read: expression equals 1.

• Chapter 16:
  • page 395: carriage return missing in equation (16.2.4) so hard to read: expression equals 1.

• Appendix A:
  • Exercise A.5.9:  the problem on summing elements in the Cantor Set was not in the Monthly, but rather in the Mathematics Magazine (problem Q785, proposed by Jeffrey Shallit, Mathematics Magazine vol 64, no 5 (1991), page 351).

• Appendix B:
  • page 458: replace "this subgroup is disjoint from SO(2)" with "these matrices are disjoint from those in SO(2)".

• Appendix D:
  • page 475: should be Chapters 13 and 14 near the end of the page, not sections.