Canad. Math. Bull. Vol. 46 (1), 2003 pp. 71 79 The Number o Fields Generated by the Square Root o Values o a Given Polynomial Pamela Cutter, Andrew Granville, and Thomas J. Tucker Abstract. The abc-conjecture is applied to various questions involving the number o distinct ields Q ( (n) ), as we vary over integers n. 1 Introduction We irst investigate the ollowing problem: Conjecture 1 Suppose that (x) Z[x] has degree 2 and no repeated roots. Then there are N distinct quadratic ields amongst (1.1) Q ( (1) ), Q ( (2) ), Q ( (3) ),..., Q ( (N) ). Conjecture 1 is actually untrue or linear polynomials. Indeed an elementary argument gives: Theorem 1A Let (x) = ax + b with integers 0 b < a < N and gcd(a, b) = 1. There are ( 6 ( π 2 1 1 ) ) 1c(a) p 2 N + O( an log a) p a distinct quadratic ields amongst those in (1.1) where c(a) = m M a 1/m 2 and M a is the set o integers m, coprime to a, such that there is no l < m with l 2 m 2 (mod a). By urther elementary arguments one can prove conjecture 1 or quadratic polynomials: Theorem 1B Suppose that (x) Z[x] has degree 2 and no repeated roots. Then there are N + O(N/ log N) distinct quadratic ields amongst Q ( (1) ), Q ( (2) ), Q ( (3) ),..., Q ( (N) ). For higher degree polynomials we have proved the conjecture assuming the abcconjecture, using several o the ideas rom [3]: Received by the editors March 26, 2001; revised August 20, 2001. The second author is supported, in part, by the National Science Foundation. AMS subject classiication: 11N32, 11D41. c Canadian Mathematical Society 2003. 71
72 Pamela Cutter, Andrew Granville, and Thomas J. Tucker The abc-conjecture (Oesterlé, Masser, Szpiro) positive integers satisying a + b = c then Fix ε > 0. I a, b, c are coprime (1.2) c ε N(a, b, c) 1+ε, where N(a, b, c) is the product o the distinct primes dividing abc. A special case o Theorem 1 in [3] states that i we assume the abc-conjecture then or (x) Z[x] without repeated roots with gcd{ (n) : n Z} = 1, there exists a constant c > 0 such that there are c N positive integers n N or which (n) is squareree. (Note that this result can be proved unconditionally i has degree 2 using the sieve o Eratosthenes; and was proved unconditionally by Hooley [5] or o degree 3 by deeper arguments. Perhaps it might be possible to use the techniques o [5] to prove Conjecture 1 or degree 3 polynomials unconditionally). The main result in this paper is that Conjecture 1 ollows rom the abc-conjecture (a weaker consequence o the abc-conjecture was given in Corollary 2 o [3]). Theorem 1C I the abc-conjecture is true then Conjecture 1 is also true. A key component in the proo o Theorem 1C is the ollowing result (which may be o independent interest) on integral points on (x) c (y), which is proved using Siegel s Theorem. Theorem 2 Suppose that (x) Z[x] has at least three distinct roots. For any ixed c > 1 there are at most initely many integral points (a, b) on the curve (x) c (y) = 0. A harder though perhaps more important problem is to determine, or a given (x) Z[x] without repeated roots, an estimate or A (D), the number o squareree integers d D such that there exists some integer n with (n) = dm 2 or some integer m. From Theorem 1C we deduce that A (D) D 1/ deg( ) i the abc-conjecture holds. To get an upper bound we use Corollary 1 rom [3] (which is deduced rom (26) o Elkies paper [2]): Lemma 1 Assume that the abc-conjecture is true. Suppose that (x) Z[x] has no repeated roots. Then p n deg( ) 1 o(1). primes p (n) Now, note that in the above case ( ) 2 ( p = primes p (n) primes p dm p) 2 (dm) 2 = d (n) D n deg( ), and combining this with Lemma 1 shows that that n D 1/(deg( ) 2)+o(1). Recalling the deinition o the quantity A (D) given above, we have now proved: Theorem 3A roots. Then Assume the abc-conjecture. Suppose that (x) Z[x] has no repeated D 1/ deg( ) A (D) D 1/(deg( ) 2)+o(1). In particular A (D) D 1/2+o(1) i deg( ) 4.
Quadratic Fields 73 In act we believe that A (D) = D 1/ deg( )+o(1) i deg( ) 2, and we will give a heuristic to justiy this or deg( ) 3 in Section 7. For linear polynomials we prove the ollowing, in Section 6. Theorem 3B I (x) = ax + b with (a, b) = 1 then A (D) 2 α(a) 6 π 2 D p a p 2(p + 1), where α(a) equals 1 i 8 a, equals 0 i 4 a but not 8, and equals 1 i 4 does not divide a. For quadratic polynomials we believe that A (D) = D 1+o(1) (which might be provable). More precisely we believe that i is reducible then A (D) c D, and i is irreducible then A (D) c D/ log D or some constants c > 0. It is easy to see that such a dichotomy exists rom the ollowing examples: Example 1 Consider (x) = x 2 1. Then (x) = dm 2 precisely when x 2 dm 2 = 1; in other words, when we have a solution to Pell s equation. Pell s equation always has an integer solution, so A (D) (6/π 2 )D. Example 2 Consider (x) = x 2 3. Then (x) = dm 2 is equivalent to x 2 dm 2 = 3; that is, 3 is represented by the principal binary quadratic orm o discriminant 4d. Now 3 is represented by a binary quadratic orm o discriminant 4d i and only i d is a square mod 3, so that d 0 or 1 (mod 3). I that is the case then 3 is represented by a binary quadratic orm rom the principal genus, o discriminant 4d, i and only i 3 is a square mod p or every prime p dividing d, so that p = 3 or p ±1 (mod 12). In other words we have d 0 or 1 (mod 3), with no prime actors congruent to ±5 (mod 12): By the undamental lemma o the sieve [4] there are D/ log D such integers d D. For such d we know that 3 is represented by some orm rom the principal genus, but it is a relatively deep problem to determine which orm(s). However, Cohen and Lenstra conjectured that there is just one class o quadratic orms in each genus or over 75% o real quadratic discriminants, and we expect this to be true or the restricted class o discriminants considered here. Thus we surmise that A x2 3(D) D/ log D, and a more detailed analysis o the Cohen-Lenstra heuristic leads to the guess that A x2 3(D) D/ log D. 2 Algebraic Preliminaries Lemma 2.1 Suppose that (x) Z[x] has no repeated roots. Then there exists a constant B = B such that or all prime powers q, there are no more than B values o a (mod q) or which (a) 0 (mod q). Proo Let K be the splitting ield or over Q, let P be a prime in K lying over a prime p with ramiication index e and with N P = p g, let α 1,..., α n be the roots o, and let p r be the highest power o p dividing the leading coeicient o. We
74 Pamela Cutter, Andrew Granville, and Thomas J. Tucker will show that or any m 0, the number values o a (mod p m ) or which (a) 0 (mod p m ) is at most n B,p := max p g(er+v P,i) P p where v P,i = k i v P(α k α i ). I p does not divide the discriminant D o then r = 0 and each v P,i = 0, so that B,p n. Thereore B := max{n, max p D B,p }. Suppose (a) 0 (mod p m ). Select i such that v P (a α i ) = max k ( vp (a α k ) ), so that v P (α i α k ) min ( v P (a α k ), v P (a α i ) ) = v P (a α k ). Thereore ( ) em v P (a) = er + v P (a α k ) er + v P (a α i ) + v P (α i α k ). k i k i=1 In other words, v P (a α i ) em er v P,i, so that there are at most (NP) er+v P,i such values o a (mod p m ). Summing over i then inishes the proo. By the Chinese Remainder Theorem we immediately deduce: Corollary 2.2 Suppose that (x) Z[x] has no repeated roots. Then there exists a constant B = B such that there are no more than B ω(d) values o a (mod d) or which (a) 0 (mod d), where ω(d) denotes the number o distinct prime actors o d. We will now prove a couple lemmas about the nonexistence o various types o low degree actors or certain (x) Q[x]. Lemma 2.3 Suppose that (x) Q[x] has at least two distinct roots. I c is neither zero nor a root o unity then (x) c (y) has no linear actor in Q[x, y]. Proo I (x) c (y) has a linear actor x ay b, with a, b in Q, then the linear map L(y) = ay + b has the property that ( L(y) ) = c (y). This implies that L permutes the roots o, which means that some power L k o L ixes all the roots o. Since there are at least two distinct roots, L k must be the identity, but this is impossible since then (y) = ( L k (y) ) = c k (y) and c is not a root o unity. Lemma 2.4 Suppose that (x) Q[x] has at least three distinct roots. I c is neither zero nor a root o unity then (x) c (y) has no quadratic actor o the orm x 2 ay 2 + 2dx + ey + h Q[x, y]. Proo Let g(x) = (x d) so that g(x) cg(y) has a actor o the orm x 2 ay 2 +ly+q over Q. Deine P(x, y) by (x 2 ay 2 + ly + q)p(x, y) = g(x) cg(y), and let P k (x, y) be the degree k part o P(x, y). Assuming without loss o generality that g is monic o degree n, we ind that x 2 ay 2 divides x n cy n so that n is even,
Quadratic Fields 75 say 2m, and c = a m, giving P 2m 2 (x, y) = (x 2m a m y 2m )/(x 2 ay 2 ). The degree 2m 1 terms in the equation above then give (x 2 ay 2 )P 2m 3 (x, y) + ly (x2m a m y 2m ) (x 2 ay 2 ) or some integer u. Replacing x to x and subtracting, gives = u(x 2m 1 a m y 2m 1 ) (x 2 ay 2 ) ( P 2m 3 (x, y) P 2m 3 ( x, y) ) = 2ux 2m 1, so that x 2 ay 2 divides 2ux 2m 1, which can only happen i u = 0. But then x 2 ay 2 divides lyp 2m 2 (x, y), which can only happen i l = 0. We have proved that our quadratic actor must be o the orm x 2 ay 2 + q. In act each P 2 j 1 (x, y) = 0 or, i not, select the largest j or which P 2 j 1 (x, y) 0 and then (x 2 ay 2 )P 2 j 1 (x, y) = v(x 2 j+1 a m y 2 j+1 ), which is impossible. We deduce that g(x) a m g(y) has no terms o odd degree, so that g(x) can be written as G(x 2 ) or some G(t) Q[t]. Letting X = x 2 and Y = y 2, we deduce that X ay + q is a actor o G(X) a m G(Y ), and so G has no more than one distinct root by Lemma 2.3. But then g, and so, can have no more than two distinct roots, contradicting the hypothesis. Remark It is plausible, ollowing the two previous results, that (x) c (y) has no actor o degree k when c is not a root o unity, and has more than k distinct roots. However we do not see how to generalize the proos above. 3 Integer Points on (x) c (y) We will use the ollowing amous theorem due to Siegel, oten reerred to as Siegel s theorem, in what ollows. This will involve introducing the idea o points at ininity, which we now explain. Let h(x, y) be a polynomial in two variables. We will denote the highest degree part o h(x, y) (which is obtained by homogenizing h(x, y) and then setting the homogenizing variable to 0) as h(x, y). The linear actors o h(x, y) correspond to the points at ininity or the curve h(x, y) = 0. Siegel s Theorem (see [6], or Section 2 o [1] or a contemporary account) Let h(x, y) be an absolutely irreducible polynomial with coeicients in a number ield. I the curve h(x, y) = 0 has more than two distinct Q-points at ininity, then there are at most initely many integer points (a, b) on the curve h(x, y) = 0. Using this we can proceed to the proo o Theorem 2: Proo o Theorem 2 Suppose (x) c (y) actors over Q into absolutely irreducible actors as r i=1 h i(x, y), each o which has degree 2 by Lemma 2.3. We will show that each curve h i (x, y) = 0 has at most initely many integer points. Since r h i=1 i (x, y) = x n cy n, and x n cy n has distinct roots, each h i (x, y) has deg h i distinct linear actors and so h i (x, y) = 0 has deg h i distinct points at ininity. Thereore, by Siegel s theorem (as stated above) we deduce that there are at most initely many integral points on h i (x, y) = 0 whenever deg h i 3.
76 Pamela Cutter, Andrew Granville, and Thomas J. Tucker We are let with the case where deg h i = 2. We may assume that h i (x, y) = 0 can be written with rational coeicients, else h i (x, y) = 0 has at most initely many rational solutions (corresponding to its intersection with its conjugate curves). Now h i (x, y) divides x n cy n = ξ n =1 (x ξαy), where α is the positive real n-th root o c, so that h i (x, y) = (x ξ 1 αy)(x ξ 2 αy), or two distinct n-th roots o unity, ξ 1, ξ 2 : I ξ 1 is real then ξ 2 is real, since h i (x, y) is real, and thus, since they are distinct real roots o unity, they must be 1 and 1. Thereore h i (x, y) = x 2 ay 2 where a is rational, which is impossible by Lemma 2.4. I ξ 1 = ξ is not real then ξ 2 = ξ, since h i (x, y) is real. Moreover, the coeicients α(ξ + ξ) and α 2 are rational, and so ξ 2 + ξ 2 = ( α(ξ + ξ) ) 2 /α 2 2 is rational. Thereore ξ 2 generates a ield o degree at most two over the rationals, so that ξ 2 is a primitive k-th root unity where k = 1, 2, 3, 4 or 6. Now k = 1 is impossible as ξ is not real, the case k = 2 gives h i (x, y) = x 2 ay 2 or some rational a, which is impossible by Lemma 2.4. The cases k = 3, 4, 6 give h i (x, y) = x 2 2baxy+4ba 2 y 2 or b = 1, 2, 3, respectively, where a is rational. Making the change o variables x = X + aby we ind that h i (x, y) = X 2 + Ay 2 + cx + dy + e where A = b(4 b)a 2 is a positive rational number. By urther changes o variables to remove the linear terms, and scaling up to make solutions integers, we ind that integer points on h i (x, y) = 0 correspond to integer points (u, v) on an ellipse o the orm u 2 +Av 2 = N; evidently u N and v N/A so there are initely many. 4 Large Squares Dividing (n) Fix an ɛ > 0 and a suiciently large integer y N 1/3. We will show that i N is suiciently large then (4.1) #{n N : m 2 (n) or some m > y} ɛn, unconditionally i deg( ) 2, and assuming the abc-conjecture otherwise. It ollows immediately rom Theorem 8 o [3] and the discussion preceding it, that i we assume that the abc-conjecture is true, then or any ixed c > 0 there are o(n) integers n N such that (n) is divisible by the square o a prime > cn. O course i (x) has degree 1 or 2, then (n) N 2 or all n N, so that (n) cannot be divisible by the square o a prime N. The number o integers n N or which (n) is divisible by the square o some prime in the range y < p < cn is, using Lemma 2.1, y<p<cn ( N ) B p 2 + 1 N y + N log N. Finally we must consider those n or which (n) is divisible by the square o some integer m > y, all o whose prime actors are y. For each n we select the smallest such m and we claim that this is y 2 : or i not select any prime actor p o m so that p y and let M := m/p which is > y 2 /y = y and such that M 2 divides m 2 which
Quadratic Fields 77 divides (n), contradicting the minimality o m. Thus using Corollary 2.2, and since ω(m) log m/ log log m so that B ω(m) = m o(1), we have that the number o such n is B ω(m) y<m y 2 ( N ) m 2 + 1 y o(1) (N/y + y 2 ) N/y 1 o(1). The result (4.1) ollows rom combining the last three paragraphs, so long as y was chosen suiciently large. 5 Proo o Theorems 1B and 1C We want to determine the number o distinct squareree integers d or which there exists some integer n N such that (n) = dm 2 or some integer m. Evidently there are no more than N such values o d. To get a lower bound we remove all cases where m > y (where y is as deined in section 4), as well as all those d or which there is more than one such pair m, n with m y and n N. In other words, using (4.1), our quantity is (1 ɛ)n m 1,m 2 y #{n 1 n 2 N : (n 1 )/m 2 1 = (n 2 )/m 2 2 is squareree}. Note that i m 1 = m 2 then we are asking or solutions to (n 1 ) = (n 2 ) with n 1 n 2. However or any non-constant polynomial, (n) is monotone increasing or n suiciently large, so there are at most initely many such pairs n 1, n 2. Otherwise, assuming without loss o generality that m 1 > m 2, each pair (n 1, n 2 ) gives rise to an integer point on the curve (x) c (z) = 0 with c = (m 1 /m 2 ) 2 > 1. By Theorem 2 there are c, 1 such points when deg( ) 3. Thereore or ixed but large y, m 1,m 2 y #{n 1 n 2 N : (n 1 )/m 2 1 = (n 2 )/m 2 2 is squareree} 1, which completes the proo o Theorem 1C or deg( ) 3. When deg( ) = 2 we can get more uniorm bounds. In that case i (x) = ax 2 + bx + c, we can complete the square to get 4a (x) = X 2 +, where X = 2ax + b and = (b 2 4ac). Then, i we ix positive integers m 1, m 2 y, take an integral solution (n 1, n 2 ) to (n 1 )/m 2 1 = (n 2 )/m 2 2, and let r j = 2an j + b, we obtain an integral solution (r 1, r 2 ) to (m 2 r 1 m 1 r 2 )(m 2 r 1 + m 1 r 2 ) = (m 2 1 m 2 2). This has τ ( (m 2 1 m 2 2) ) = y o(1) solutions, where τ(.) is the number o divisors o an integer. Theorem 1B then ollows rom the proo in Section 4 with y = log 2 N.
78 Pamela Cutter, Andrew Granville, and Thomas J. Tucker 6 Linear Polynomials Proo o Theorem 1A We wish to ind the number o squareree integers d or which there exists an integer m such that dm 2 = an + b or some n N. For those d q (mod a), we select m to be the smallest positive integer or which qm 2 b (mod a) i such an m exists. Thus we need to determine m M a #{squareree d (an + b)/m 2 : d b/m 2 (mod a)}. By elementary sieve theory we have that the number o squareree integers d q (mod a), when (a, q) = 1, is, writing d = ar + q, r x g 2 ar+q µ(g) = = g ax+q g ax+q,(g,a)=1 = x p a µ(g)#{r x : g 2 ar + q} µ(g) ( x/g 2 + O(1) ) (1 1 ) p 2 + O( ax). Summing up over m M a gives the result. Proo o Theorem 3B I (x) is a linear polynomial, say ax + b with (a, b) = 1, then we are interested in the proportion o squareree integers d D or which there exists some integer m such that dm 2 b (mod a). In other words d belongs to one o a certain set o congruence classes mod a; the number o such congruence classes being φ(a)/2 w(a) where w(a) denotes the number o distinct odd prime actors o a, plus 2 i 8 divides a, or plus 1 i 4 divides a but not 8. By the estimate in the proo o Theorem 1A each o these arithmetic progressions contains (D/a) p a (1 1/p2 ) + O( D) such integers d D, and so we obtain the result. 7 Heuristic The argument preceding Lemma 1 in the introduction tells us that i (n) = dm 2 with d D then n D 1/(deg( ) 2)+o(1), assuming the abc-conjecture. I n D 1/ deg( ) then (n) n deg( ) D, so that d D. Thereore we need to explore urther or D 1/ deg( ) n D 1/(deg( ) 2)+o(1). For N = cd 1/ deg( ) 2 j with j = 0, 1, 2,..., J we consider N < n < 2N, so that (n) N deg( ). Then m 2 = values o n (mod m 2 ), or which (n) 0 (mod m 2 ) by Corollary 2.2. Thereore the number o such N < n < 2N is B ω(m) (2N/m 2 + 1). For the heuristic assume the term 1 is irrelevant at least on average, and recall that B ω(m) N o(1). Thereore #{n N : m 2 (n) or some m M} m M N1+o(1) /m 2 D/N deg( )/2 1 o(1), where (n) /d N deg( ) /D, and obviously m 2 N deg( ). Now there are B ω(m)
Quadratic Fields 79 M is the interval N deg( )/2 /D 1/2 m N deg( )/2. Summing over all such N we get D 1/ deg( )+o(1), or deg( ) 3, as required. Added in proo: In March 2002, we received a preprint by Bjorn Poonen, Squareree values o multivariable polynomials, in which he proves our Theorems 1A, 1B, 1C; his proos are slightly dierent rom the ones in this paper. Reerences [1] E. Bombieri, Eective Diophantine Approximation on G m. Ann. Scuola Norm. Pisa Cl. Sci (4) 20(1993), 61 89. [2] N. Elkies, ABC implies Mordell. Internat. Math. Res. Notices 7(1991), 99 109. [3] A. Granville, ABC means we can count squarerees. Internat. Math. Res. Notices 19(1998), 991 1009. [4] H. Halberstam and H.-E. Richert, Sieve Methods. Academic Press, London-New York-San Francisco, 1974. [5] C. Hooley, On the power ree values o polynomials. Mathematika 14(1967), 21 26. [6] C. L. Siegel, Über einege Anwendungen diophantischer Approximaten. Abh. Preuss. Akad. Wiss. Phys. Math. Kl. 1(1929), 209 266. Department o Mathematics University o Georgia Athens, Georgia 30602 USA email: pcutter@kzoo.edu andrew@math.uga.edu ttucker@math.uga.edu