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PRIMES OF THE FORM x2 + ny 2 AND THE GEOMETRY OF (CONVENIENT) NUMBERS THOMAS R. HAGEDORN Fermat’s theorem that an odd prime p can be expressed as a sum x2 + y 2 precisely when p ≡ 1 mod 4 is one of the highlights of a first course in number theory. Fermat first stated this theorem in a 1640 letter to Mersenne, and over the next two decades, he found similar statements for when a prime p could be written in the form x2 + ny 2 for n = 2, 3. Namely: p = x2 + 2y 2 ⇐⇒ p = 2 p = x2 + 3y 2 ⇐⇒ p = 3 or or p ≡ 1, 3 mod 8, p ≡ 1 mod 3. The first rigorous proofs of these results were given by Euler using Fermat’s method of descent. Euler also made similar conjectures for other values of n (see Cox [1]). When n = 5, for primes p 6= 5, he conjectured: (1) p = x2 + 5y 2 ⇐⇒ p ≡ 1, 9 mod 20 2p = x2 + 5y 2 ⇐⇒ p ≡ 3, 7 mod 20. A natural question is whether for any integer n, there are congruence conditions that precisely describe the set of primes p such that p = x2 + ny 2 , for some x, y ∈ Z (with a finite number of exceptions). We call n a convenient number,1 or numerus idoneus in Latin, if there is a finite set S, an integer N , and congruence classes c1 , . . . , ck mod N such that for all primes p 6∈ S, (2) p = x2 + ny 2 ⇐⇒ p ≡ c1 , . . . , ck mod N. Euler used convenient numbers to find large primes. When n is convenient, m is prime precisely when there is exactly one solution to the equation x2 + ny 2 = m of relatively prime positive integers (x, y). We will see that n = 1848 is convenient. Then since x = 197, y = 100 is the only relatively prime solution to the equation 18, 518, 809 = x2 + 1848y 2 , Euler could prove with minimal calculation that 18, 518, 809 is prime. Euler found the 65 convenient numbers listed in Table 1. He was astonished that after n = 1848, he could not find another convenient number despite searching upto 10, 000. To quote Weil [6, p. 229], “that such a naturally defined set of numbers should apparently be finite, as [Euler] now came to conjecture, was a novel experience to him.” In his Disquisitiones, Gauss rigorously proved that they were convenient numbers using his theory of positive definite quadratic forms. We now know due to 1This definition can be shown to be equivalent to the definitions of Euler and Gauss. 1 2 T. R. HAGEDORN Weinberger [7] that there that there are at most 66 convenient numbers. The 66th convenient number, if it exists, must be greater than 109 . In this paper, we present an alternative proof that these 65 numbers are convenient using Minkowski’s geometry of numbers. This proof uses quadratic reciprocity, but doesn’t use Gauss’s theory of binary quadratic forms or genus theory, making it suitable as a supplement at the end of a first course in number theory. When n = 1, this application is well-known, giving Fermat’s theorem on p = x2 + y 2 , and is often presented as the first application of Minkowski’s theory. For other values of n, however, we have not seen this application documented in the literature. As an example, for n = 93, we can easily show: Proposition 1. For a prime p, p = x2 + 93y 2 for some x, y ∈ Z ⇐⇒ p ≡ 1, 25, 49, 97, 109, 121, 133, 157, 169, 193, 205, 253, 289, 349, 361 mod 372. We can similarly prove statements of type (2) for all 65 numbers in Table 1. Conversely, the method of proof can be combined with the methods of genus theory to establish criteria for n to be a convenient number [3]. We can also prove some similar statements for non-convenient numbers. For example, if n = 14, Euler made the following conjecture for all primes p 6= 2, 7, p = x2 + 14y 2 (3) ⇐⇒ p ≡ 1, 9, 15, 23, 25, 39 mod 56. or p = 2x2 + 7y 2 This theorem is the strongest possible true statement for n = 14 that uses only congruences. One can show by elementary means [4] that there is no choice of modulus N for which a statement of type (2) is true when n = 14. We note that both of the forms in (3) represent primes is each of the specified congruence classes and that no prime can be represented by both forms. For instance, 113, 281 are both primes p ≡ 1 mod 56. But 281 = 152 +14(2)2 , 113 = 2(5)2 +7(3)2 , and neither prime is represented by the other form. Euler also conjectured that for a prime p, (4) 3p = x2 + 14y 2 ⇐⇒ p ≡ 3, 5, 13, 19, 27, 45 mod 56. Convenient Numbers n h(−4n) 1, 2, 3, 4, 7 1 5, 6, 8, 9, 10, 12, 13, 15, 16, 18, 22, 25, 28, 37, 58 2 21, 24, 30, 33, 40, 42, 45, 48, 57, 60, 70, 72, 78, 85, 88, 93, 4 102, 112, 130, 133, 177, 190, 232, 253 105, 120, 165, 168, 210, 240, 273, 280, 312, 330, 345, 357, 8 385, 408, 462, 520, 760 840, 1320, 1365, 1848 16 Table 1. Table of the 65 known convenient numbers and the class numbers h(−4n) of the group of binary quadratic forms with discriminant −4n. GEOMETRY OF NUMBERS 3 We can use the geometry of numbers to prove both (3) and (4). The subject of convenient numbers and genus theory is a rich one. For more information on these areas and the complete answer to when we can express a prime p in the form p = x2 + ny 2 , we encourage the reader to consult [1]. 1. Quadratic Reciprocity The law of quadratic reciprocity, discovered by Euler and Legendre and first proved by Gauss [2], gives an efficient way to determine the squares mod p, for an odd prime p. The Legendre symbol is defined for numbers a with (a, p) = 1 by ( 1, if a is a non-zero square mod p, a = p −1, if a is a not a square mod p. 3 2 Thus, 11 = 1 as 52 ≡ 3 mod 11, whereas 11 = −1 as none of the squares 12 , 22 , . . . , 102 modulo 11 equals 2. The following properties of the Legendre symbol are easy to prove. b a = . Lemma 1. 1. If a ≡ b mod p, then p p −1 2. = (−1)(p − 1)/2 . p ab a b 3. Let a, b be relatively prime to p. Then = . p p p −1 By part (2) of the lemma, 11 2 = −1 and −1 is not a square mod 11. By part −1 −2 (3) of the lemma, 11 = 11 11 = (−1)(−1) = 1 and −2 is a square mod 11, as −2 ≡ 9 ≡ 32 mod 11 demonstrates. The law of quadratic reciprocity makes the calculation of ap quite easy for large a, p. Proposition 2 (The Law of Quadratic Reciprocity [2]). Let p, q be two odd primes. p−1 q−1 p q = (−1) 2 · 2 . 1. q p ( p2 −1 1, if p ≡ 1, 7 mod 8, 2 2. = (−1) 8 = p −1, if p ≡ 3, 5 mod 8, Example. We can now easily determine if 2, 31 are squares mod 71. Since 71 ≡ 7 mod 8, 2 is a square. Trial and error shows that 122 ≡ 2 mod 71. Now 31−1 71−1 31 71 71 9 = (−1) 2 2 =− =− = −(+1) = −1, 71 31 31 31 so 31 is not a square mod 71. The first equality follows from part (2) of the proposition and the third equality follows from part (1) of Lemma 1. The following lemma follows quickly from the law of quadratic reciprocity. Lemma 2. Let p be an odd prime and n a positive integer with (n, p) = 1. Then −n 2 2 p = 1 ⇐⇒ p|x + ny , for some relatively prime integers x, y. 4 T. R. HAGEDORN Proof. If −n = 1, then −n ≡ x2 mod p for some x ∈ Z. So p|x2 + n = x2 + n · 12 . p 2 Conversely, if p|x2 +ny 2 , then x2 ≡ −ny mod p. If y 6≡ 0 mod p, then y is invertible −n 2 and (x/y) ≡ −n mod p. So p = 1. In fact, this case must occur as y ≡ 0 mod p implies x ≡ 0 mod p and then p would be a common divisor of x, y, contradicting the given. The following proposition is a crucial one that establishes the congruence criteria for a prime p. Proposition 3. Let n ≥ 1, and p be an odd prime with p - n such that p = x2 + ny 2 . p 1. −n = 1 and p q = 1 for every odd prime q|n. 2. If n ≡ 0, 1 mod 4, then p ≡ 1 mod 4. 3. If 8|n, then p ≡ 1 mod 8. Proof. Part (1) follows immediately from reduction mod q and from Lemma 2 as x, y must be relatively prime. Since x, y cannot both be even, x2 + ny 2 ≡ a + bn mod 4, where a, b = 0, 1. If n ≡ 0, 1 mod 4, then p ≡ a + bn ≡ 0, 1 mod 4. Since p is odd, p ≡ 1 mod 4. Part (3) follows similarly by working mod 8. 2. Minkowski’s Geometry of Numbers and Convenient Numbers An amazing discovery of 19th -century mathematics was Minkowski’s simple use of geometry to prove theorems in about numbers. We will apply his methods to reprove Gauss’s result that the 65 numbers in Table 1 found by Euler are all convenient number. We begin by recalling some geometric definitions. We say a subset R of R2 is symmetric if −x ∈ R whenever x ∈ R. We say R is convex if for all points x, y ∈ R, the line segment l connecting x, y is contained in R. In particular, the midpoint (x + y)/2 of l is an element of R. Lastly, a lattice L in R2 is the set of integral linear combinations av + bw, a, b ∈ Z, for two linearly independent vectors v, w ∈ R2 . A fundamental domain for L is the region of R2 defined by F = { rv + sw| 0 ≤ r, s < 1, r, s ∈ R}. F has the property that one can translate every point x ∈ R2 by an element of the lattice L to a point in F. In other words, x = l + y, for some l ∈ L, y ∈ F. From linear algebra, the area of F equals | det M |, where M is the matrix whose rows are the vectors v, w. If v, w ∈ Z2 , then L is a subgroup of Z2 , and the area of F equals its index [Z2 : L]. The following simple geometric proposition is the foundation for Minkowski’s Geometry of Numbers. Proposition 4 (Minkowski). Let L be a lattice in R2 whose fundamental domain has area V . If R is a convex, symmetric subset of R2 with Area(R) > 4V , then R contains a non-zero point of L. Proof. Consider the lattice 2L. Since 2L has index 4 in L, its fundamental domain has area 4V . Now translate every point of R by a vector in 2L into the fundamental GEOMETRY OF NUMBERS 5 domain for 2L. Since Area(R) > 4V , there must be two points x, y ∈ R whose translates coincide. Hence x = y + 2z, for some non-zero element z ∈ L. Now if y ∈ R, then −y ∈ R since R is symmetric. And if −y, y + 2z ∈ R, then the midpoint of the line connecting them [−y + (y + 2z)]/2 = z must be in R since R is convex. Hence R contains a non-zero point of L. It is well-known that the Geometry of Numbers can be used to prove Fermat’s theorem that an odd prime p can be expressed as p = x2 + y 2 precisely when p ≡ 1 mod 4 [5]. We illustrate the method for n = 3 to prove: Proposition 5. Let p 6= 3 be a prime. Then p = x2 + 3y 2 for some integers x, y if and only if p ≡ 1 mod 3. Proof. (⇒) We know p 6= 2. Then by Proposition 3, −3 = 1. By quadratic recip procity, p−1 3 −1 3 −1 p −3 = = (−1) 2 = . p p p p p 3 p Hence 3 = 1 and p ≡ 1 mod 3. (⇐) Reversing the above implications, we have −3 2 p = 1 and there exists an integer u such that u ≡ −3 mod p. Now consider the lattice L ⊂ R2 defined by L = {(a, b) ∈ Z2 |a ≡ ub mod p}. L is generated by the vectors v = (u, 1), w = (0, p) and it is straight-forward to see that L has index p in Z2 . Consequently, the area of its fundamental domain √ has area p. Let Rk be the ellipse defined by x2 + 3y 2 = k. Then Rk has area πk/ 3 > 1.8k. Choose k = 2.3p. Then Area Rk > 4p, and Rk ∩ L contains a non-zero point (a, b). Now a2 + 3b2 ≡ (ub)2 + 3b2 ≡ −3b2 + 3b2 ≡ 0 mod p, so p|a2 + 3b2 . Now since (a, b) ∈ Rk , we have a2 + 3b2 < 2.3p, so a2 + 3b2 = p or 2p. We now show that a2 + 3b2 = 2p cannot occur. Suppose a2 + 3b2 = 2p, then a2 ≡ 2p ≡ 2 mod 3, which gives a contradiction. Hence a2 + 3b2 = p. We can prove Proposition 1 and similar statements for all 65 convenient numbers by slight modifications of this argument. Analyzing the proof, we see there are three steps: 1. Use Proposition 3, to determine congruence criteria for when p = x2 + ny 2 . 2. For a prime p satisfying the congruence criteria in (1), Minkowski’s Geometry of Numbers shows that x2 + ny 2 = dp has a solution (a, b) for some positive integer √ def d ≤ Bn = [4 n/π]. 3∗ For each d satisfying 1 < d ≤ Bn , use congruence conditions to show that either the equation x2 + ny 2 = dp has no solution. When n = 3, we saw that B3 = 2 and equation x2 + 3y 2 = dp had a solution for d = 1 or 2. Wefurther showed that the equation x2 + 3y 2 = 2p could not have a solution since 32 = −1. It is simple to generalize this argument with the following useful lemma: 6 T. R. HAGEDORN Lemma 3. Let q - d be an odd prime factor of n. If pq = 1 and dq = −1, then the equation x2 + ny 2 = dp has no solution in integers. Proof. Since pq = 1, q 6= p. If a2 + nb2 = dp, then a2 ≡ dp 6≡ 0 mod q. So dp q = 1, but this contradicts the hypothesis. Hence, there is no solution to x2 + ny 2 = dp. Let us now try our method to prove a proposition similar to Proposition 5 for n = 93. We will run into one new situation. Proof of Proposition 1. (⇒) follows immediately using Proposition 3. We now show (⇐). Since B93 = 12, following the proof of Proposition 5, using n = 93 in place of n = 3, we find that there is a positive integer d ≤ 12 such that the equation x2 + 93y 2 = dp has a solution for some positive integers a, b. We will now show that d = 1 has this property by considering the other possible choices for d. The choices d = 2, 3, 5, 6, 7, 8, 11 can be eliminated by Lemma 3 as d3 = −1 or p p d = −1. We note that = 31 3 31 = 1 by the given congruence conditions for p. That leaves the choices d = 4, 9, 10, 12. If d = 4, then a2 + b2 ≡ 0 mod 4, which forces a, b to both be even. Thus (a/2)2 + 93(b/2)2 = p and the original equation x2 + 93y 2 = p has a solution. So x2 + 93y 2 = 4p has a solution precisely when x2 + 93y 2 = p has a solution. The same situation occurs when d = 9, 12. Hence, we are left to analyze the case d = 10. If x2 + 93y 2 = 10p, then reducing modulo 4, x2 + y 2 ≡ 2 mod 4 and one sees that x, y must both be odd. So x2 , y 2 ≡ 1 mod 8 and x2 + 93y 2 ≡ x2 + 5y 2 ≡ 6 mod 8. But as 10p ≡ 2 mod 8, x2 + 93y 2 6≡ 10p mod 8 and x2 + 93y 2 = 10p cannot have a solution. Summarizing, we have seen that every choice of d 6= 1 either leads to a contradiction or shows that the equation x2 +93y 2 = p has a solution. Thus, the equation x2 +93y 2 = p has a solution. In light of the new wrinkle caused by the cases d = 4, 9 in the above proof, we need to replace step 3∗ above by the following: 3. For each d satisfying 1 < d ≤ Bn , use congruence conditions to show that either the equation x2 + ny 2 = dp has no solution or that a solution of x2 + ny 2 = dp leads to a solution of x2 + ny 2 = p. Unfortunately, the case d = 10 in the above proof shows that the congruence arguments needed in Step 2 can be more involved than simply using Lemma 3. There are two additional congruence arguments we need. As Lemma 3 only applies to odd primes, we need to establish some congruence arguments for the prime 2. As one often sees when working with the even prime 2, we will have congruences modulo powers of 2. The second new congruence argument we will need is best illustrated in the proof that n = 57 is a convenient number. The new methods appear when considering the choices d = 6, 7. Proposition 6. Let p be a prime. Then p = x2 + 57y 2 ⇐⇒ p ≡ 1, 25, 49, 61, 73, 85, 121, 157, 169 mod 228. GEOMETRY OF NUMBERS 7 Proof. (⇒) follows from Proposition 3. (⇐) Formally following the proof of Proposition 1, we find that x2 + 57y 2 = dp has a solution for some positive integer d ≤ 9. There four distinct situations to consider. d i. The cases d = 2, 3, 5, 8 cannot occur by Lemma 3 as either d3 = −1 or 31 = −1. 2 ii. d = 4, 9. As in the proof of Proposition 1, one can show the equation x + 57y 2 = dp has a solution precisely when x2 + 57y 2 = p has a solution. iii. d = 6. If x2 + 57y 2 = 6p, then 3|x. Let x = 3z. Then 3z 2 + 19y 2 = 2p. Reducing this equation modulo 3, we have y 2 ≡ 2p ≡ 2, giving a contradiction. So when d = 6, there is no solution. iv. d = 7. As p 6= 7, 7p is squarefree and if x2 + 57y 2 = 7p then (x, y) = 1. Then −57 −57 = 1 by Lemma 2. But quadratic reciprocity shows = −1, which 7 7 would give a contradiction. So d = 7 cannot occur. As every choice of d 6= 1 either leads to a contradiction or shows that the equation x2 + 57y 2 = p has a solution, the proposition is proved. Generalizing the new arguments we made in the cases d = 6, 7 in the previous proof, we have the following lemma. Lemma 4. Let p be an odd prime. (a). Suppose that q 6= p is an odd prime such that dp q|(n, d), q 2 - d, and q 2 - n. If x2 +ny 2 = dp has a solution, then q 2 |n and n/q = q q . (b). Assume x2 + ny 2 = dp has a solution and q = 6 p is an odd prime divisor of d −n 2 with q - d, q - n. Then q = 1. We summarize in the following proposition all the needed congruence arguments, including the additional ones for the prime 2 and for d = 9 (whose proofs we omit). Proposition 7. Given n, assume that the prime p satisfies the congruence conditions listed in Proposition 3. If (n, d) satisfies one of the following conditions, then either the equation x2 + ny 2 = dp has no solution or if it has a solution, then the equation x2 + ny 2 = d0 p has a solution for some positive integer d0 < d. a. There is an odd prime factor q of n with q - d with dq = −1. = −1. b. There is an odd prime factor q of d with q - n, q 2 - d and −n q d c. There exists an odd prime q|(n, d) with q 2 - d, q 2 - n such that n/q = 6 q q . k d. There is an odd integer k such that 2 is the highest power of 2 dividing (n, d) and either 2k+1 |n or 2k+1 |d. e. n ≡ 3 mod 4 and d = 2. f. n ≡ 1 mod 4 and 4|d. g. n ≡ 7 mod 8 or n ≡ 0, 8, 12 mod 16, and d = 4. h. 4|n and d ≡ 3 mod 4. i. 8|n and d = 5. j. n ≡ 48 mod 64 and d ≡ 16 mod 64. k. n ≡ 18 mod 27 and d = 9. Using this proposition when analyzing the choices of d that arise in the proofs of statements of type (2), we can easily prove the following. Corollary 1. The 65 values in Table 1 are convenient numbers. 8 T. R. HAGEDORN 3. Applications to x2 + ny 2 = pq and Non-Convenient Numbers Minkowski’s Geometry of Numbers can be used in a similar way as in the previous section to prove theorems about the expressibility of specific composite numbers by a quadratic form. For example, we can prove the second statement in (1) using the same methods used to prove Proposition 5. By slightly modifying the method of proof, we can also prove the following similar statement. Proposition 8. Let p, q ≡ 3, 7 (mod 20) be odd primes. Then pq = x2 + 5y 2 for some integers x, y. Proof. The congruence conditions give −5 = −5 = 1. So there exists solutions p q 2 2 to the equations x ≡ −5 (mod p), x ≡ −5 (mod q). By the Chinese Remainder Theorem, there is an integer u such that u2 ≡ −5 mod pq. Now consider the lattice L ⊂ R2 defined by L = {(a, b) ∈ Z2 | a ≡ ub mod pq}. As L has index pq in Z2 , its fundamental domain √ has area pq. Let Rk be the ellipse 2 2 defined by x + 5y = k. Then Rk has area πk/ 5 > 1.4k. Choose k = 2.9pq. Then Area Rk > 4pq, and Rk ∩ L contains a non-zero point (a, b). Now a2 + 5b2 ≡ (ub)2 + 5b2 ≡ −5b2 + 5b2 ≡ 0 mod pq, so pq|a2 + 5b2 . Now since (a, b) ∈ Rk , we have a2 + 5b2 < 2.9p, so a2 + 5b2 = pq or 2pq. Now if a2 + 5b2 = 2pq, then regarding the equation mod 5 shows that 2pq ≡ ±2 (mod 5) is a square, giving a contradiction. Hence a2 + 5b2 = pq. It is straightforward to see that one can prove statements similar to Proposition 8 and (1) for any convenient number n. But the usefulness of this method is not limited solely to convenient numbers. It can prove a limited number of similar statements for other numbers. For example, as n = 14 is not a convenient number, we cannot prove a statement regarding x2 +14y 2 = p. Nevertheless, we can reprove the following conjecture of Euler. Proposition 9. Let p be a prime. Then 3p = x2 + 14y 2 for some integers x, y ⇐⇒ p ≡ 3, 5, 13, 19, 27, 45 mod 56. Proof. (a)(⇒) We can p 6= 2, 3, 7. Then any solution (x, y) must be relatively assume 3p = = 1 and quadratic reciprocity gives p ≡ 3, 5 (mod 8), prime. Then −14 p 7 p ≡ 3, 5, 6 (mod 7), which gives the desired congruence. (⇐) By quadratic reciprocity, the congruence conditions give −14 = 1. So there exists an integer u such that p 2 u ≡ −14 mod p. Then as in the proof of Proposition 5, we use u to construct a lattice L, an ellipse Rk , and choose k such that Minkowski’s Theorem shows that there is a solution (a, b) with a2 + 14b2 = dp, for some d = 1, 2, 3, 4. If d = 1, 2, 4, then p 2 a ≡ dp (mod 7) and 7 = 1, contradicting the congruence conditions on p. Thus d = 3 and a2 + 14b2 = 3p. Does Proposition 9 have a generalization similar to Proposition 8? Yes, Proposition 9 is but a specific case of the following more general proposition. GEOMETRY OF NUMBERS 9 Proposition 10. Let p, q be odd primes with p, q ≡ 3, 5, 13, 19, 27, 45 mod 56. Then pq = x2 + 14y 2 for some integers x, y. However, the methods of this paper are too weak to prove Proposition 10! In order to prove it, we must use the more powerful methods of Gauss’s theory of genera. We now illustrate the problem one encounters when one tries to generalize the proof of Proposition 9 to prove Proposition 10. Assume p, q are given as in the hypothesis of Propostion 10. Following the proof of Proposition 8, one can show that there exists a solution to x2 + 14y 2 = pqd, for some positive integer d ≤ 4. If d = 4, then x, y would there would both be even, and p q 3 be a solution to x2 + 14y 2 = pq. If d = 3, then 3pq = 1, but = = 7 7 7 7 = −1, giving a contradiction. But when we consider the case d = 2, we cannot obtain a contradiction as 2 is a square mod 7. As a result, we know that there is a solution to the equation x2 + 14y 2 = pq or the equation x2 + 14y 2 = 2pq, but we cannot definitely say that either equation can be solved! Using genus theory one can show that the both equations can always be solved. We now sketch a proof of both this claim and Proposition 10, and refer the reader to [1] for background material. The class group of binary quadratic forms with discriminant −68 consists of 4 forms divided up into two genera. The principal genus of forms with discriminant −56 consists of e = x2 + 14y 2 , the identity element for the group, and f = 2x2 + 7y 2 ; the other genus consists of the two forms g = 3x2 + 2xy + 5y 2 , h = 3x2 − 2xy + 5y 2 . By genus theory, one can show that primes p, q satisfying the hypotheses in Proposition 10 are represented by either g or h. However, as g, h represent exactly the same values in Z (since one can replace (x, y) by (−x, y)), we can assume g represents p and h represents q. Gauss’s law of composition for quadratic forms then shows that pq is represented by the form e = g ◦ h. Hence x2 + 14y 2 = pq has a solution, proving Proposition 10. Similarly, f represents the number 2, so h = f ◦ g represents 2p. Since g represents q, we have that 2pq is represented by g ◦ h = e, and x2 + 14y 2 = 2pq has a solution as well. References 2 2 [1] D. Cox, Primes of the Form x + ny : Fermat, Class Field Theory, and Complex Multiplication, Wiley, 1997. [2] C. F. Gauss, (1801) Disquisitiones Aritmeticae, trans. by Arthur A. Clarke, Yale University Press, 1965. [3] T. Hagedorn, Lower Bounds for the 66th Convenient Number, in preparation. [4] B. Spearman, K. S. Williams, Representing Primes by Binary Quadratic Forms, American Mathematical Monthly, 99(5) (1992), pp. 423-426 [5] I. Stewart, D. Tall, Algebraic Number Theory and Fermat’s Last Theorem, 3rd Ed., A.K. Peters, Natick, MA 2001. [6] A. Weil, Number Theory: An Approach Through History, Birkhäuser, Boston, Basel, and Stuttgart, 1984. [7] P.J. Weinberger, Exponents of the class groups of complex quadratic fields, Acta Arith. 22 (1973), pp. 117–124. E-mail address: [email protected] Department of Mathematics and Statistics, The College of New Jersey, P.O. Box 7718, Ewing, NJ 08628