数理解析研究所講究録別冊 = RIMS Kokyuroku Bessa (2012), B32:

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1 Imaginary quadratic fields whose ex Titleequal to two, II (Algebraic Number 010) Author(s) SHIMIZU, Kenichi Citation 数理解析研究所講究録別冊 = RIMS Kokyuroku Bessa (01), B3: Issue Date URL Right Type Departmental Bulletin Paper Textversion publisher Kyoto University

2 RIMS Kôkyûroku Bessatsu B3 (01), Imaginary quadratic fields whose exponents are less than or equal to two, II By Kenichi Shimizu Abstract Shimizu [1] gave some necessary conditions for e D, where e D is the exponent of the ideal class group of an imaginary quadratic field Q( D). In this paper we mainly consider some relations between prime-producing polynomials and the condition e D. First we give a generalization of Mollin s result. Next we consider the inverse of Mollin s result when d (mod ) and t D = 3, and give some relations between invariants of Q( D). 1. Introduction Given a square-free integer d > 0, we define D by { d if d 1, (mod ) D := d if d 3 (mod ), and call D the discriminant of the imaginary quadratic field K D = Q( D). We denote by h D the class number of K D, and e D the exponent of the ideal class group of K D that is the least positive integer n such that a n are principal for all ideals a. We denote by t D the number of different prime factors of D. We call a rational prime q a split prime if (q) = qq (q q ) and a ramified prime if (q) = q for prime ideals q and q in K D. Let q D denote the least split prime. We define f D (x) by { x + d if d 1, (mod ) f D (x) := x + x + (1 + d)/ if d 3 (mod ), Received April 1, 011. Revised November 19, Mathematics Subject Classification(s): Primary 11R11, Secondary 11R9 Key Words: imaginary quadratic field, exponent, prime-producing polynomial Supported by JAPAN SUPPORT s-357@gaia.eonet.ne.jp c 01 Research Institute for Mathematical Sciences, Kyoto University. All rights reserved.

3 56 Kenichi Shimizu then we have that prime divisors of f D (x) are split primes or ramified primes (Lemma 3.). Further for every divisor e of d, we define q D (e) by e + d/e if d (mod ) e + d/e q D(e) := if d 1 (mod ) e + d/e if d 3 (mod ), and denote by q D the minimum of q D (e) for all divisors e of d. We have that q D (e) and q D are split primes (see Shimizu [1]). Let b be any divisor of d and a = d/b. We assume b > 1. Let p be the least prime divisor of b, then we define I b by {x 0 x p 1} if d (mod ) and b is a prime {x 0 < x p 1} if d (mod ) and b is not a prime I b := {x 0 x p 1} if d 1 (mod ) {x 0 x p 1 p } if d 3 (mod ), d where x are integers. We define quadratic polynomials f D,b (x) by ax + b if d (mod ) f D,b (x) := ax + ax + a + b if d 1 (mod ) ax + ax + a + b if d 3 (mod ). In Lemma 3.3, we show that the value of f D,b (x) in I b is a split prime or a product of split primes. We denote by ν(n) the number of (not necessarily different) prime factors of an integer n, and define the Ono number p D as follows, { max{ν(f D (x)) x are integers in 0 x D/ 1 if d 1, 3 p D := 1 if d = 1, 3. In this paper we assume d 1, 3 through all sections. For these invariants, we pose the following conjecture: Conjecture 1.1. (i) e D. (ii) p D = t D. The following conditions (i) (vi) are equivalent.

4 Imaginary quadratic fields whose exponents are less than or equal to two, II 57 (iii) For every divisor b of d, f D,b (x) takes only prime values for all integers x with I b. (We call this condition GEP-property in this paper.) (iii) For the largest prime divisor b of d, f D,b (x) takes only prime values for all integers x with I b. (We call this condition EP-property in this paper.) (iv) q D = q D. (v) q D > D/. (vi) f D (x) = qd for an integer x. (Only when d 1, 3 (mod )) Similar conjecture was given by Shimizu [1], two conditions are improved in this paper as follows. In [1], we did not have the condition (iii), and we wrote the condition q D > R D instead of (v), where R D is D or D/ when d (mod ) or d 1, 3 (mod ), respectively. We here note that some relations have been known between those conditions of Conjecture 1.1. G.Rabinowitsch [10] and F.G.Frobenius [1] showed independently that (i) (iii) when t D = 1. M.D.Hendy [] essentially showed that (i) (iii) when t D =. H.Möller [5] proved that (i) (iv) and (iv) (v). R.A.Mollin [6][7][8] showed that (i) (ii) and (ii) (iii). Shimizu [1] showed that (i) (vi) and (iv) (vi) when d 1, 3 (mod ), and showed that (i) and (iv) are equivalent when d (mod ). X.Guo and H.Qin [3] showed that (i) (ii) when t D = 3 under the Extended Riemann Hypothesis. In Section 3 we show that (i) (iii) (Theorem 3.1), which is a generalization of Mollin s result. In Section, we show that (v) (iv) when d 1, (mod ) (Theorem.). In Section 5, only when t D = 3 and d (mod ), we show that (iii) (i) (Theorem 5.1) and that conditions in Conjecture 1.1 are equivalent except (vi) (Theorem 5.). Consequently by Theorem 5. we obtain the same result as X.Guo and H.Qin [3] without the Extended Riemann Hypothesis in the case of d (mod ). In Section 6, we show that (iii) (v) when t D = 3 and d 1 (mod ) (Theorem 6.1).. Prime-producing polynomials In this section we sketch the history of prime-producing polynomials. In 177, L.Euler discovered that the quadratic polynomial x + x + 1 takes only prime values for all integers with 0 x 39. Euler also noted that the quadratic polynomial x + x + A takes only prime values with 0 x A, in the cases of A =, 3, 5, 11, 17, 1. In 191, F.G.Frobenius [1] and G.Rabinowitsch [10] independently showed that the above fact is related to the class number of imaginary quadratic field Q( 1 A). They proved:

5 58 Kenichi Shimizu Theorem.1. The following (1) and () are equivalent. (1) The quadratic polynomial x +x+a (A ) takes only prime values for all integers with 0 x A. () Q( 1 A) has class number one. In the same paper, Frobenius referred to prime-producing polynomials related to imaginary quadratic fields with h D = as follows. Theorem.. Let p be an odd prime number. (i) If Q( p) has class number two, then x + p takes only prime values for all integers with 0 x < p. (ii) If p 1 (mod ) and Q( p) has class number two, then x x + (p + 1)/ takes only prime values for all integers with 0 x < (p + 1)/. In 197, M.D.Hendy [] gave a necessary and sufficient condition for prime-producing polynomials related to imaginary quadratic fields with h D = as follows. Theorem.3. Let p and q be odd prime numbers. (i) If Q( p) has class number two if and only if x + p is prime for all integers with 0 x p/. (ii) If p 1 (mod ) and Q( p) has class number two if and only if x +x+(p+1)/ is prime for all integers with 0 x ( p 1)/. (iii) If pq 3 (mod ), p < q and Q( pq) has class number two if and only if px + px + (p + q)/ is prime for all integers with 0 x pq/1 1/. In 1995, R.A.Mollin [7][8] generalized these results to imaginary quadratic fields whose class numbers are more than two. He considered imaginary quadratic fields with h D = td 1. It is known that h D = t D is equivalent to e D. From now on we use e D instead of h D = td 1. Mollin proved the following. Theorem.. (Mollin [6], [7] p.110) If e D, then the equality p D = t D holds. Using f D,b (x), we state Mollin s result. Theorem.5. (Mollin [7] p , [8]) Let b be the largest prime divisor of d and a=d/b. If e D, then f D,b (x) takes only prime values for all integers x with I b. Theorem.5 is proved by using Theorem..

6 Imaginary quadratic fields whose exponents are less than or equal to two, II A generalization of Mollin s result In this section we give a generalization of Theorem.5. Though we have taken b the largest prime divisor of d in Theorem.5, we can take b any divisor of d as follows. Theorem 3.1. Let b > 1 be any divisor of d, a = d/b and p the least prime divisor of b. We assume a > 1 when d 3 (mod ) and d is not a prime. If e D, then the quadratic polynomial f D,b (x) takes only prime values for all integers x with I b. When d 3 (mod ) and d is not a prime, if a = 1, then there are counter examples. For example, if d = 15, then f D,b (x) = x + x + does not take prime values x in I b. If b is the largest prime divisor of d, then Theorem 3.1 gives Theorem.5. For the proof of Theorem 3.1, we give the following lemmas. Lemma 3.. Prime divisors of f D (x) are split primes or ramified primes. Conversely all split primes and all ramified primes divide f D (x) for some integers x. Proof. At first, we consider in the case of d 1, (mod ). Let p be a prime which does not divide D, then we have that p is odd and that p f D (x) x + d 0 (mod p), X D (mod p) is solvable, p is a split prime. If p D, then p is a ramified prime. When d (mod ) we have. When d 1 (mod ) we have p D p D p f D (0) { p f D (0) for an odd prime p, p f D (1) for p =. Second, we consider in the case of d 3 (mod ). For an odd prime p we have p f D (x) x + x d = (x + 1) + d (x + 1) d (mod p), X D (mod p) is solvable, p is a split prime. 0 (mod p), For p =, when d 7 (mod 8) we have that is a split prime and f D (x) for all x. When d 3 (mod 8) we have that is neither a split prime nor a ramified prime, and f D (x) for all x. Thus we complete the proof.

7 60 Kenichi Shimizu Lemma 3.3. In I b, the value of the quadratic polynomial f D,b (x) is a split prime or a product of split primes. Proof. At first, we state a relation of f D (x) and f D,b (x). When d (mod ) we have f D (ax) = a x + d = a(ax + b) = af D,b (x). ( When d 1 (mod ) we have f D (ax + a) = a x + a x + a + d = a ax + ax + a + b ) = af D,b (x). When d 3 (mod ), since f D (x) = x + x d f D (ax + a 1 ) = (ax + a) + d ( = a ax + ax + a + b ) = af D,b (x). = (x + 1) + d, we have Therefore by Lemma 3., we get that prime divisors of f D,b (x) are split primes or ramified primes. Next we show that no ramified primes divide f D,b (x). Assuming that f D,b (x) is divided by a prime divisor p of d, we derive a contradiction. It holds that a and b have no common prime divisors since d is square-free. Let p be the least prime factor of b. When d (mod ) we have f D,b (x) = ax +b. By p (ax +b), p a immediately implies p b, which is a contradiction. If p b, then p ax. Since x I b, namely, 0 < x < p p, we get p a, which is a contradiction. When d 1 (mod ) we have f D,b (x) = ax + ax + a + b. We show that the values of f D,b (x) are odd. By d = ab 1 (mod ), it holds that a b 1 (mod ) or a b 3 (mod ). Hence we get a + b (mod ), and so the values of f D,b (x) are odd. As we assume p f D,b (x), p a implies p (a + b). Hence we have p b, which is a contradiction. On the other hand, from p f D,b (x) we have p (ax + ax + a + b). If p b, then we have ax + ax + a + b ax + ax + a a(x + 1) 0 (mod p ). Hence we have p (x + 1) since p a. By x I b, namely, 0 x p/ 1, we have 0 < x + 1 p 1 p 1, which is contradict to p (x + 1).

8 Imaginary quadratic fields whose exponents are less than or equal to two, II 61 Finally, when d 3 (mod ) we have f D,b (x) = ax + ax + a + b. As we assume p f D,b (x), p a implies p b, which is a contradiction. If p b, then ax + ax + a + b = a(x + 1) + b a(x + 1) 0 (mod p ). Hence we get p a(x + 1). Thus we have p (x + 1) since p a. From x I b, namely, 0 x p/ 1/ p/(d), we have 0 < x + 1 p p d < p p, which is contradict to p (x + 1). Thus we complete the proof. Möller gave an lower bound for q D. Theorem 3.. (Möller [5]) If e D, then q D > D or q D > D/ when d or d 1, 3 (mod ), respectively. Using Lemma 3.3 and Theorem 3., we prove Theorem 3.1. Proof. (Theorem 3.1) We prove this theorem by two steps. First, assume that b is a prime divisor of d, second, assume that b is not a prime. Step I: Let b be a prime divisor of d. When d (mod ) we have f D (ax) = af D,b (x). By Theorem. we have p D = t D, and ν(a) = t D 1 since b is a prime divisor of d. Hence we get that f D,b (x) takes only prime values for all integers x with 0 ax D/ 1 = d 1, which corresponds to 0 x d/a 1/a = b 1/a, that is, 0 x b 1. Therefore f D,b (x) takes only prime values for all integers x with I b. Next, when d 1 (mod ) we have f D (ax + a) = af D,b (x). Since p D = t D and ν(a) = t D 1, we get that f D,b (x) takes only prime values for all integers x with 0 ax + a D/ 1 = d 1, which corresponds to 0 x d/a 1/a 1/, that is, 0 x b/ 1. Therefore f D,b (x) takes only prime values for all integers x with I b. Last, when d 3 (mod ) we have f D (ax + a 1 ) = af D,b(x). From p D = t D and ν(a) = t D 1, we get that f D,b (x) takes only prime values for all integers x with 0 ax+(a 1)/ D/ 1 = d/ 1, which corresponds to 0 x d/(a) 1/ 1/(a), that is, 0 x b/ 1/ b/(d). Therefore f D,b (x) takes only prime values for all integers x with I b.

9 6 Kenichi Shimizu Step II: Let b be not a prime and p the least prime divisor of b, then it holds ap < ab = d. We assume that there is an integer x such that f D,b (x) is not a prime in I b, and let f D,b (x) = q 1 q r (q i is a split prime, 1 i r and r ) by Lemma 3.3. Then we have f D,b (x) qd. At first, when d (mod ) we have f D,b (x) = ax + b qd. Since e D, we have qd > D = d by Theorem 3.. Hence we get d < q D ax + b. On the other hand, since x < p and ap < d, we get ax + b < ap + b < d + b. Therefore we have d < d + b, which is a contradiction. Hence f D,b (x) takes only prime values for all integers x with I b. Second, when d 1 (mod ) we have f D,b (x) = ax + ax + (a + b)/ q D. Since x p/ 1, ap < d and a + b 1 + d, we have ax + ax + a + b ( p ) ( p ) a 1 + a 1 ( ) p = a p + 1 = ap ap d < d d = d d + ap a d Hence ax + ax + (a + b)/ d. By Theorem 3. we have qd > D/ = d, therefore we get d < qd ax + ax + a + b d, which is a contradiction. Finally, when d 3 (mod ) we have f D,b (x) = ax + ax + (a + b)/ qd. By Theorem 3., we have qd > D/ = d/. Since x p/ 1/ p/(d) and ap < d, we get ax + ax + a + b = a(x + 1) + b a ( p p ) b + d ( = ap 1 1 ) + b d < d ( 1 1 d + 1 ) d + b = d d + b < d 16 + b.

10 Imaginary quadratic fields whose exponents are less than or equal to two, II 63 In this case, since we assume a > 1, we have a 3. Thus we have b d/3. Therefore d 16 + b d 16 + d 1 = 7 8 d < d. Hence we get d < q D ax + ax + a + b < d, which is a contradiction. Thus we complete the proof of Theorem On the condition q D > D/ In this section we show Theorem. below. For the proof of Theorem. we give the following lemma. Lemma.1. For an odd split prime q there are two integers x in 0 x < q such that q f D (x). If d 1, (mod ), then they are positive, and one is even and the other is odd. If d 3 (mod ), then they are non-negative, and both even or both odd. Proof. By Lemma 3., there are integers x such that q f D (x). When d 1, (mod ), since q f D (0) we may put x 0 the least positive integers such that q f D (x 0 ). Since f D (q x 0 ) = (q x 0 ) + d = q(q x 0 ) + f D (x 0 ), we get that q f D (x 0 ) implies q f D (q x 0 ). Then we have that two integers x 0 and q x 0 are different, for x 0 = q x 0 implies x 0 = q, which is impossible since q is odd. Thus we get 0 < x 0 < q/ < q x 0 < q. Therefore there are two integers x in 0 < x < q such that q f D (x), and one of them is even and the other odd since q is odd. When d 3 (mod ), let x 0 be the least non-negative integers such that q f D (x 0 ). We have: f D (q 1 x 0 ) = (q 1 x 0 ) + (q 1 x 0 ) + (1 + d)/ = q{q 1 x 0 } + x 0 + x 0 + (1 + d)/ = q{q 1 x 0 } + f D (x 0 ), hence we get that q f D (x 0 ) implies q f D (q 1 x 0 ). Assuming x 0 = q 1 x 0, we have x 0 = (q 1)/ and f D (x 0 ) = (q +d)/. Thus by q f D (x 0 ) we get q d, which is a contradiction. Thus we obtain x 0 < q 1 x 0, that is, 0 x 0 < (q 1)/ < q 1 x 0 < q. Hence there are two integers x in 0 x < q such that q f D (x). Further we obtain that both x 0 and q 1 x 0 are even or both odd since q is odd.

11 6 Kenichi Shimizu Theorem.. Suppose d 1, (mod ). If q D > D/, then q D = q D. Proof. When d 1, (mod ), we have that q D is odd, and that there are two integers x in 0 < x < q D such that q D f D (x) by Lemma.1. First, we prove the theorem when d (mod ). Assume that x is even, then we have that x + d is even, hence we put x + d = q D c (c 1). If c is divided by a split prime, then x + d qd. By q D > x we get qd + d > x +d qd, hence we obtain q D +d > q D, and so q D < D/. Therefore q D > D/ implies that c = 1 or the divisors of c are only ramified primes. Since x is even, it holds x + d (mod ). Hence we have (x + d), and so c is odd. Then an odd ramified prime p divides c exactly, because p c implies p d, which is impossible since d is square-free. Thus we get c d, and so c x. Putting x = cx 1, we have c x 1 + d = q D c, hence we get cx 1 + d/(c) = q D. Since x 1 1, we have q D c + d/(c) = q D (c) q D. On the other hand, we have q D q D since q D is the least split prime. Hence we get q D = q D. Second, we show this theorem when d 1 (mod ). Assume that x is odd, then x + d is even, hence we put x + d = q D c (c 1). If c is divided by a split prime, then by the same way as d (mod ), we get q D < d = D/. Therefore q D > D/ implies that c = 1 or the divisors of c are only ramified primes. Since x is odd, it holds x + d (mod ). Hence we have (x + d), and so c is odd. We get that an odd ramified prime divides c exactly by the same reason as d (mod ). Hence we get c d. Therefore we obtain c x. Putting x = cx 1, we have cx 1 + d/c = q D. Since x 1 1, we have q D (c + d/c)/ = q D (c) q D. On the other hand, we have q D q D. Thus we get q D = q D. We conjecture that q D > D/ implies q D = q D when d 3 (mod ), but we can not prove it yet. From Theorem. we have two corollaries as follows. Corollary.3. When d (mod ), it holds that the conditions in Conjecture 1.1 (i), (iv) and (v) are equivalent. Proof. Theorem 3. says that e D implies q D > D > D/. And Theorem. says that q D > D/ implies q D = q D, On the other hand, q D = q D implies q D > D/3, because by q D = q D there is a divisor e of d such that q D = e + d/e, hence q D = e + d/e > d = D > D/3. We have the property such that the ideal class group of K D is generated by split ideals or ramified ideals which norms are less than D/3. Therefore by qd > D/3 the ideal class group is generated by only ramified prime ideals, and so a is principal for any ideal a of K D. Thus we obtain e D. Hence we show that q D = q D implies e D. Therefore (i), (iv) and (v) are equivalent.

12 Imaginary quadratic fields whose exponents are less than or equal to two, II 65 Corollary.. When d 1 (mod ), it holds that the conditions in Conjecture 1.1 (iv), (v) and (vi) are equivalent. Proof. Shimizu [1] proved that q D = q D is equivalent to f D(x) = qd for a integer x. Further we have that q D = q D implies q D > D/, because by q D = q D there is a divisor e of d such that q D = (e + d/e)/, hence q D > d = D/. On the other hand we have proved that q D > D/ implies q D = q D in Theorem.. Therefore (iv), (v) and (vi) are equivalent. 5. Relations between prime-producing polynomials and e D In this section, we consider the inverse of Theorem.5, that is, whether it holds that EP-property implies e D. But it does not hold when d 1, 3 (mod ). There are three counter examples with d < 100, 000, 000. d prime factors d mod h D t D t D 1 p D These imaginary quadratic fields satisfy EP-property, but do not satisfy e D. On the other hand, when d (mod ) we conjecture that EP-property implies e D, and only when t D = 3 we can prove it as below. Furthermore we generally conjecture that GEP-property implies e D, but we can not prove it yet. As we have described in Section, Rabinowitsch and Frobenius showed when t D = 1 that EP-property holds if and only if e D, and Hendy obtained the similar result when t D =. From now on we consider the problem on EP-property when t D = 3. We show the following theorem. Theorem 5.1. When d = p 1 p (mod ) (p 1 < p ), the quadratic polynomial p 1 x + p takes prime values for all integers with 0 < x p 1, then it holds e D. For the proof of Theorem 5.1 we show the following lemma and theorem. Lemma 5.. For any split prime q, there are integers x such that q f D,b (x).

13 66 Kenichi Shimizu Proof. There is an integer x 0 such that q f D (x 0 ) by Lemma 3., then it holds q f D (x 0 + yq) for every integer y. When d (mod ), we consider the equation x 0 + yq = ax, that is, ax qy = x 0 in which x and y are integers. Since gcd(a, q) = 1, there are solutions of this equation. Let (x 1, y 1 ) be a solution, then we have x 0 +y 1 q = ax 1. Since f D (x 0 +y 1 q) = f D (ax 1 ) = a(ax 1 + b), we obtain q a(ax 1 + b), and so q (ax 1 + b). Hence we have q f D,b (x 1 ). When d 1 (mod ), we consider x 0 + yq = ax + a. Since gcd(a, q) = 1, there are solutions of this equation. Let (x 1, y 1 ) be a solution, then we have x 0 + y 1 q = ax 1 + a. Since f D (x 0 + y 1 q) = f D (ax 1 + a) = a{ax 1 + ax 1 + (a + b)/}, we obtain q a{ax 1 + ax 1 + (a + b)/}, and so q {ax 1 + ax 1 + (a + b)/}. Hence we have q f D,b (x 1 ). When d 3 (mod ), we consider x 0 + yq = ax + (a 1)/. Since gcd(a, q) = 1, there are solutions of this equation. Let (x 1, y 1 ) be a solution, then we have x 0 + y 1 q = ax 1 + (a 1)/. Since f D (x 0 + y 1 q) = f D (ax 1 + (a 1)/) = a{ax 1 + ax 1 + (a + b)/}, we obtain q {ax 1 + ax 1 + (a + b)/}. Hence we have q f D,b (x 1 ). Using Lemma 5. we prove the following. Theorem 5.3. Suppose d (mod ). Let b be the largest prime divisor of d and a=d/b. We assume b > D/16. If the quadratic polynomial f D,b (x) takes prime values for all integers x with I b, then it holds e D. Proof. In this case, we have f D,b (x) = ax + b and I b = {x 0 < x b 1}. From Lemma 5., there are integers x such that q D (ax + b). Suppose that x 1 is the least positive integer such x. If 0 < x 1 b 1, namely, x I b, then by Lemma 3.3 we have that ax 1 + b is a split prime. Since q D is the least split prime, we get q D = f D,b (1), thus q D = a + b > d > D/. Now suppose x 1 b. Since x 1 is the least positive integer such that q D (ax + b), we have q D / > x 1. By x 1 b, we have q D / > b. Since b > D/16 we get q D / > D/16, namely, q D > D/. Therefore we obtain e D by Corollary.3. Using Theorem 5.3 we prove Theorem 5.1. Proof. (Theorem 5.1) Let a = p 1 and b = p, then we have f D,b (x) = p 1 x + p and I b = {x 0 < x p 1}. In this case the condition b > D/16 holds. Thus we complete the proof.

14 Imaginary quadratic fields whose exponents are less than or equal to two, II 67 Theorem 5.. When d (mod ) and t D = 3, it holds that the conditions in Conjecture 1.1 (i), (ii), (iii), (iii), (iv) and (v) are equivalent. Proof. We have already shown that (i), (iv) and (v) are equivalent in Corollary.3. Further by Theorem 5.1 EP-property implies e D when t D = 3. On the other hand, it holds that e D implies p D = t D and that p D = t D implies EP-property by Mollin [6][7][8]. Furthermore by Theorem 3.1 e D implies GEP-property, and it is trivial that GEP-property implies EP-property. Thus we complete the proof. X.Guo and H.Qin [3] showed under the Extended Riemann Hypothesis that e D is equivalent to p D = 3 when t D = 3. By Theorem 5. we have obtained without the Extended Riemann Hypothesis that e D is equivalent to p D = 3 and t D = 3 when d (mod ). 6. Relations between prime-producing polynomials and q D > D/ We want to prove that EP-property implies e D when t D = 3 and d 1 (mod ), too. But we cannot prove it yet. In this section we show that EP-property implies q D > D/ when t D = 3 and d 1 (mod ). We prove the following theorem. Theorem 6.1. Suppose d = p 1 p 1 (mod ) (p 1 < p ). If the quadratic polynomial p 1 x + p 1 x + (p 1 + p )/ takes prime values for all integers with 0 x p / 1, then it holds q D > D/. For the proof of Theorem 6.1 we show the following. Theorem 6.. Suppose d 1 (mod ). Let b be the largest prime divisor of d and a = d/b. We assume b > D/. If the quadratic polynomial f D,b (x) takes prime values for all integers x with I b, then it holds q D > D/. Proof. In this case, we have f D,b (x) = ax +ax+(a+b)/ and I b = {x 0 x b/ 1}. From Lemma 5., there are integers x such that q D {ax + ax + (a + b)/}. Suppose that x 1 is the least non-negative integer such x. If 0 x 1 b/ 1, namely, x 1 I b, then we have that f D,b (x) = ax 1 + ax 1 + (a + b)/ is a split prime. Since q D is the least split prime, we get q D = f D,b (0). Hence we have q D = (a + b)/ > d = D/.

15 68 Kenichi Shimizu Next, suppose x 1 > b/ 1. Since (q D 1)/ > x 1, we have (q D 1)/ > b/ 1, and so q D b. By b > D/, we have q D > D/. Thus we complete the proof. From Theorem 6. we have the proof of Theorem 6.1. Proof. (Theorem 6.1) Let a = p 1 and b = p, then we have f D,b (x) = p 1 x + p 1 x+(p 1 +p )/ and I b = {x 0 x p / 1}. In this case, the condition b > D/ holds. Thus from Theorem 6. we complete the proof. Theorem 6.3. When d 1 (mod ) and t D = 3, we have the following relations between conditions in Conjecture 1.1. (i) (ii) (iii) (iv) (v) (vi) Proof. Theorem. and.5 say that (i) (ii) (iii), and Theorem 6.1 says that (iii) (v) when t D = 3. Further the equivalence between (iv), (v) and (vi) is given by Corollary.. Thus we complete the proof. Acknowledgment The author would like to thanks the referee for his valuable comments and suggestions. References [1] F. G. Frobenius: Über quadratische Formen, die viele Primzahlen darstellen, Sitzungsber. d. Königl. Akad. d. Wiss. zu Berlin, (191) [] H. Gu, D. Gu and Y. Liu : Ono invariants of imaginary quadratic fields with class number three, J. Number Theory 17, 6-71 (007) [3] X. Guo and H. Qin : Imaginary quadratic fields with Ono number 3, Comm. in Algebra, Vol.38, Issue 1, 30-3 (010) [] M. D. Hendy : Prime quadratics associated with complex quadratic fields of class number two. Proc. Amer. Math. Soc.3, (197) [5] H. Möller : Verallgemeinerung eines Satzes von Rabinowitsch über imaginär-quadratische Zahlkörper, J. Reine Angew. Math. 85, (1976). [6] R. A. Mollin : Orders in quadratic fields III, Proc. Japan Acad. 70A, (199).

16 Imaginary quadratic fields whose exponents are less than or equal to two, II 69 [7] R. A. Mollin : Quadratics, CRC Press, Boca Raton (1995). [8] R. A. Mollin : A completely general Rabinowitsch criterion for complex quadratic fields, Canad. Math. Vol. 39(1), (1996). [9] T. Ono : Arithmetic of algebraic groups and its applications, St. Paul s International Exchange Series Occasional Papers VI, St. Paul s University, Tokyo (1986). [10] G. Rabinowitsch : Eindeutigkeit der Zerlegung in Primzahlfaktoren in quadratischen Zahlkörpern, J. Reine Angew. Math. 1, (1913). [11] F. Sairaiji and K. Shimizu : A note on Ono s numbers associated to imaginary quadratic fields, Proc. Japan Acad. 77A, 9-31 (001). [1] F. Sairaiji and K. Shimizu : An inequality between class numbers and Ono s numbers associated to imaginary quadratic fields, Proc. Japan Acad. 78A, (00). [13] R. Sasaki : On a lower bound for the class number of an imaginary quadratic field, Proc. Japan Acad. 6A, (1986). [1] K. Shimizu: Imaginary quadratic fields whose exponents are less than or equal to two, Math. J. Okayama Univ. Vol. 50, (008)