Some Computational Results on a Problem Concerning Powerful ...

MATHEMATICS OF COMPUTATION VOLUME 50, NUMBER 182 APRIL 1988, PAGES 619-632

Some Computational Results on a Problem Concerning Powerful Numbers By A. J. Stephens and H. C. Williams* Abstract. Let D be a positive square-free integer and let X + Y%J~Dbe the fundamental unit in the order with Z-basis {1,Vd}. An algorithm, which is of time complexity 0(D1'4+£) for any positive e, is developed for determining whether or not D | Y. Results are presented for a computer run of this algorithm on all D < 108. The conjecture of Ankeny, Artin and Chowla is verified for all primes = 1 (mod 4) less than 109.

1. Introduction. An integer N is said to be powerful if for any prime p such that p | N we must have p2 | iV. In [4] Erdös conjectured that there do not exist three consecutive powerful numbers, and Granville [5] has shown that if this is true, then there exists an infinitude of primes p such that p2 \ 2P_1 —1. Mollin and Walsh [7] have pointed out that if there exist three consecutive powerful numbers, then there must exist some square-free D = 7 (mod 8) with X + Y\[D being the fundamental unit of Q.(\[D) such that for some odd A;, Xk is an even powerful

number and Yjt= 0 (mod D) is an odd integer, where Xk + YkV~D— (X-\-\fDY)h. If, for a given value of D, we have D \ Y', then it is a relatively easy matter to show that the least possible value for fc such that Xk is powerful and D \ Yk must be very large (see [7] for an example with D = 7). Thus, for this and other reasons, Mollin and Walsh asked the second author whether it was possible to find those values of D such that D\Y. In general, however, this seems to be a very difficult problem. Indeed, Ankeny, Artin and Chowla [1] conjectured several years ago that if (x + y,Jp)/2 is the fundamental unit of Q(y/p) when p is a prime and

p = 1 (mod 4), then p \ y. Later, Mordell [8] conjectured that if X + Y^/p is the fundamental unit of Q(yfp) when p = —1 (mod 4) and p is a prime, then p \ Y. Neither of these conjectures has been proved, but the Ankeny, Artin, Chowla (AAC) conjecture has been verified, most recently, by Soleng [13] for all p < 100028009 and Mordell's conjecture has been verified for all p < 7679299 by Beach, Williams

and Zarnke [2]. Because of the difficulty of this problem, we decided to investigate it numerically. In this paper we discuss how we found all the square-free values of D . 108 for which D | Y. We also verified the AAC conjecture for all p < 109. We describe two algorithms for conducting these numerical investigations. The first of these (the Small Step Algorithm) is basically the algorithm that has been used for previous work. We develop some general results concerning symmetric continued fractions Received March 9, 1987.

1980 MathematicsSubject Classification(1985 Revision). Primary 11R11, 11R27, 11Y16, 11Y40, 11Y65. •Research supported by NSERC of Canada Grant #A7649. ©1988 American 0025-5718/88

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Mathematical Society $1.00 + $.25 per page

A. J. STEPHENS

620

AND H. C. WILLIAMS

in order to give a slightly more compact form of this procedure than has been used in the past. We then make use of the class group infrastructure ideas of

Shanks [12] as developed and modified by Lenstra [6], Schoof [10], and Williams and Wunderlich [15] in order to derive a second algorithm (the Large Step Algorithm). The complexity of the first algorithm is 0(D1^2+E) for all e > 0; but the complexity of the second is 0(D1^4+£). We then describe the results of implementing and running both of these algorithms on an AMDAHL 5850 computer.

2. Continued Fractions and Ideals. All previous search methods for testing the AAC conjecture have made use of the properties of continued fractions. As the methods that we will employ here also involve continued fractions, we will first provide a brief review of some relevant results. Several of these can be found in standard reference works like Perron [9] or Chrystal [3], and others can be found

in [15]. We first assume that n) can be obtained by using** qo = [4>c¡] and the recursive formulas

(pi+i= 1/(0, -qt)> .

,

I,

. _

\l — U, 1, £., . . . ).

qi+i = [0,

Qi+i< o.

more efficient method of determining

these numbers has been given

by Tenner (see [15]).

If we define A_2 = 0, A_i = 1, B-2 = 1, B_i = 0 and

(2.4)

Ai+i =qi+iAi + Ai-i,

(2.5)

Bt+i=ql+xBl

(2.6)

9i+i = (-iy(Ai-i - 4>Bi-i),

+ Bl-U

(i = -1,0,1,2,...

then

(2.7)

9i+i = (-lY(Gi-i - y/DBi-i)/Q0

(i>-l),

where (2.8)

Gl-1 =Q0AJ_1-.P0BI_1

(i>-l).

**We use [q] to denote that integer such that a — 1 < [a] < a.

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),

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In [14] it is pointed out that***

(2.9)

9l+Â+1 = (G2_x- DBlx)/Q2 = (-lYQt/Q0, i

(2.10)

*r+i= n*' 3= 1

and

(2.11)

Gi-i = PlBl-i + Q%Bi-2,

(2.12)

DBl-1 = PiGi-i + QiGi-2-

Let a,ß£ Q(\fDô) and denote by [a,ß] the set {ua + vß | u, v G Z}. Let uo = (r — I + \fDo)/r, where Do (> 0) is square-free, r = 1 when D0 = 2,3 (mod 4) and r = 2 when Dr¡ = 1 (mod 4). Any order 0 of Q(y/D~ô) must have Z-basis {l,noJo}, where n 6. Z. We can denote such an order by [l,nwn]. Now Oi is the maximal order of Q(y/Dô) and if So (0 < So < 1) is the fundamental unit of Q(y/L%) and r¡ (0 < r\ < 1) is the fundamental unit of the order 0r = [l,y/Dô\,

then either r¡ = £0 or rj = Eq. Thus, if £r¡ = (x ■+y\fL%)/r and ry = x 4- \fDr]Y

(x, y,X,Y e Z), we see that if D0 \ y, then Dq \ Y and if D0 \ Y, then D0 \ 3y. It follows that if 3 -f Do, any algorithm that determines whether or not Do I Y can be used to determine whether or not Do | y. Let 0 = 0n be any order in Q(y/D) and let a be any primitive, integral ideal of 0. As mentioned in [15], we can write a in the form [Q/a, (P + \fD)/a], where a = r/g, D = (n/g)2Dç,, g = gcd(r, n). Here we have trQID — P2, and if we develop the continued fraction expansion of 0 — (P+ \/~D)/Q, we find that each of the primitive ideals

at+i = [Qi/a, (Pi + s/D)/a]

(i = 0,1,2,...

)

is equivalent to a = Oi. Also, we have

(2.13)

(Qo9m)am = (Qm-i)a,,

where by (a) we denote the principal ideal with generator a. If by L(a) we denote the least positive integer of the ideal a, then L(a) = \Q\/a. Also, if a is a primitive ideal and a does not contain any nonzero a such that both

|a| < L(a),

\á\ < L(a)

hold, then we say that a is a reduced ideal. If ax — a is a reduced ideal and if fa is a reduced ideal equivalent to a, then b = a¿ for some k. Also, if ai is a reduced

ideal, then by results given in [15],we have | 1); hence,

(2.14)

0 2). Thus, we

have Pi = Pp+i, Qi = Qv+i and (2.15)

Qo = QP,

Po = Pp (modQo).

Let p be the least positive integer for which (2.15) holds. If rj (0 < r\ < 1) is the fundamental unit of 0, by Theorem 4.5 of [15] we must have rj = 9k for some k > 1. Since, in this case, a¿ = Oi, we get k = p + 1 and by (2.5),

(2.16)

r, = 9p+i = (-ir(Ap-i-9Bp-i).

If a = [Q/a, (P + \fD)/o], define ä to be [Q/a, (P - \fD)/o\. If a = a, we say that a is an ambiguous ideal. We can now give a result which is well known in the

case a= [1,%/D]-

THEOREM 2.1.

Ifa=

[Qo/a, (P0 + \/D)/a] (Q0 > 0) is a reduced, ambiguous

ideal, then in the continued fraction expansion of (fio= (Pq + y/D)/Qo,

(2.17)

Qp_,= Qi,

(2.18)

Pp_t = Pt+1

we have

(0 1. Hence,

(2.20)

\9k+i\>\Sk\

(*>1).

If ai is a reduced, ambiguous ideal of 0n such that gcd(L(ai),n) = 1, then we can find Bp-i in (2.16) by only going up to about p/2 terms in the continued fraction expansion of —2),

we must have r = p/2 by Corollary 2.1.1. Suppose next that Q3 = Qs+i. a3+i = [Qs/a,

(Ps + s/D)/a\

In this case we have Ps+i = —Ps = [Qs+i/a,

(-Ps+1

+ y/D)/tr] = as+2.

By using the same reasoning as above, we get r¡ = 9s+i/\9s+2\. (—l)p+ Qs+iBp-i

= Bs-iG9

(mod Qs) and

It follows that

+ BsGs-i;

hence, p is odd and Qs+i-Bp-i

= qsQsBsBs-i

+ Qs+iBs_x

+ QsBsBs-2

by (2.11) and (2.1). Since Qs = Qs+i, we have

Bp-i = B2a+Bl_i by (2.5). Also, we must have p = 2s + 1. D We can use Theorem 2.2 to develop the Small Step Algorithm for determining whether or not D\Y. We assume here that D = Do is square-free and ai = Oi-

The Small Step Algorithm (SSA) (1) Put P0 = 0, Qo = 1 when D = 2,3 (mod 4) or P0 = 1, Q0 = 2 when D = l (mod 4). (2) Compute the continued fraction expansion of (Po + \/D)/Qo

PuQiiBi-i

(mod D)

(z = 1,2,3,...)


by evaluating

A. J. STEPHENS AND H. C. WILLIAMS

624

until we find the least r or s such that Pr = Pr+1

or

Qs ~ Qs+1-

(3) lfPr = Pr+1,thenD|y

if and only if D | Dr_i(Pr + Pr_2). lfQs = Qs+i,

then D | Y if and only if D | B2 + B2S_X. As mentioned above, simple variants of this algorithm have been used to obtain the known numerical results on the AAC conjecture. As D increases, however, this algorithm tends to slow down. Since we know that p = 0(D1//2+e) for any e > 0 (see, for example, Williams [14]), we see that this algorithm is of complexity 0(D1/2+£). In the following sections we will develop an algorithm, called the Large Step Algorithm, which will solve this problem in time complexity 0(D1/4+E). In order to do this, we will need to know how to find a reduced ideal which is equivalent to a given primitive ideal a. We point out here that by results in [15]

we know that if Oi = a = [Qo/a, (Po + \fD)/a] and 0o = (Po + y/D)/Q0, then the continued fraction expansion of 0 must yield some Qm such that 0 < Qm < d. When this occurs, we know that Om+i is a reduced ideal equivalent to Oi. Further,

the value of m is O (log |Qo|)3. A Regulator Algorithm and Further Results. In order to develop our next algorithm we must first mention that if a = [Q/a, (P + \/D)/a], fa = [Q'/a, (P' + \J~D)/a] are both primitive ideals of an order 0, then we can find the primitive ideal c = \Q"¡a, (P" + \fD)/o\ and an integer U such that

afa= (U)c by using the formulas below. These formulas are essentially those of Shanks [11] and can be easily derived by using the method discussed in [6] or [10].

We put G = gcd(Q/a, Q'/a) and solve (Q/a)Xl = G

(mod Q'/a)

for xi (mod Q'/a). Put

(3.1)

U = gcd(G, (P + P')/a) = gcd(Q/cr, Q'/a, (P + P')/a)

and solve x2(P + P')/o + Gy2 = U for x2, y2. Then

(3.2)

Q" = QQ'KaU2),

(3.3)

P" = P + XQ/(aU) (modQ"),

where

X s 2/2X1 (P' - P) + x2(D - P2)/Q

(mod Q'/U).

Note that when, as frequently occurs, U = G, we can put x2 = 0, y2 = 1.

If we denote the pair (P,Q) (Q > 0) by A and the pair (P',Q') (Q' > 0) by B, we use A o ß to denote the pair (P", Q") given by the formulas (3.2) and (3.3).


A PROBLEM CONCERNING POWERFUL NUMBERS

625

Also, since the ideal a which corresponds to A is the same as that corresponding to

B when and only when Q —Q' and P = P' (mod Q), we will define equality of A and B by these conditions. Let a = ai be any reduced ideal in 0 and let fa be any reduced ideal which is equivalent to o. By our remarks at the end of Section 2 we know that fa = Om for some m < p. Define ¿(a^cii) = log|0m|. Notice that, by (2.20), 6(am,ai) is a strictly increasing function of m. Also,

«5(ap+i,Oi) = log|öp+i| = -logr?. Thus we can always assume that

0 0) is a reduced ideal and oi = (1). If (U)c = a3bt,

where Ci = c is a primitive ideal of 0 and cm.|_i ~ q ~ fai is found by using the continued fraction method of reduction described at the end of Section 2, then by Theorem 5.2 of [15], we have cm+i = bk for some Á;> 1 and

O'k= (0s9't9'm+i)/U,

where (L(bi)9't)bt = (L(bt))bi, (9s)as = (L(as)), (L(ii)9'^)tm = (L(cTO))Cl. It

followsthat if we put 6¡ = Wó; thus,

A > -log(Q'¿U) > -\og(Qs„iQ't_i) > -log4D by (2.14). We have shown, then, that in (3.4) we have

(3.5)

-log4D
1).



626

Now 6(as+2,ai) thus, since ¿(aa+2,Oi)

= log|0s+2| = ¿(aa,ai) + log|Vijâ+i|;

> ê(as,ai)

+ log2 and 6 is an increasing function of s, the

lemma follows. D We are now able to present an algorithm which can be used to compute the regulator R of 0 in 0(D1/M+£) (e > 0) elementary operations. We let ai = (1), 6i = 6(oi-i,ai), Ai = (Pí,Qí), and put L = [cD1/4] where L 6 Z and c is some constant. We usually use c > 2.

Algorithm

to Compute R.

(1) Using the continued fraction algorithm, compute and store Ai, 6i for i =

0,1,2,...,

s + 1, where s = L + l. If any Qi = Qq with 0 < i < s + 1, then

R = 6i and we can exit from the algorithm.

(2) Put Bi = As, 6X*= 6S, j = 1. (3) Compute U and (P",Q") from C = (P",Q") = Aso Bj. By expanding the continued fraction of (P" + \fD)/Q",

find the least nonnegative m such

that 0 < Qm < d and put

(3.7)

A, = log(|C+il/Ea

S;+1= 6* + 6S+ Ar

(4) If C = Ai for some i such that 0 < i < s + 1, then

R = Sj+i -8t, and we can exit from the algorithm;

otherwise, we replace j by j + 1, Bj

by C, and go to (3). Proof of the Regulator Algorithm. 2I(fc-i)/2]; henCe,

From (3.6) and (2.10) we see that

|^fc| >

6k > (fc/2-2)log2 and

(3.8)

6S > (s/2 - 2) log 2 > (cL/2 - 2) log 2 > log 4D

when D is sufficiently large. (Certainly, this is so if c > 2 and D > 106.) Since by

(3.5) we have A¿ > —log4D in (3.7), we must have

s;+x>s;. Further, if we do not exit the algorithm at Step (1), then R > 6S = (D1/4 —2) log 2 - log4D by (3.8) and selection of L, we see that k = 0(R/D^4) = 0(D1'4+e) and 8*k+x> R. Thus, t = 0(D1'4+E). Let (P",Q") = As o Bt and let cm+1 = [Q'm/a, (P" + \fD)/a\.

Since cm+1 is

reduced and cm+1 ~ Oi, we must have cm+i = ay_i for some 1 < j < p + 1. Now since (3.9) holds, we must have

¿(oj_i,ai) = St+1 - R = 6t - R + 6S+ At < 6„ + log2 by (3.5). It follows from Lemma 3.1 that (Pm,Q'm) = Ai for some 0 < i < s + I. Since R = 5t*+1- 0 and

(3.13)

9k+i = (9s+i9t+i9'm+i)/U.

If we define V¿ = gcd(Qt,D), then we have V%| P¿ and Vl\Al-i by (2.11). Set License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use

628


Si-i = Ai-i/Vi.

We conclude this section with

THEOREM 3.1.

// (3.13) holds, then for some r = ±1, we have

rTkSk-i = LiH + L2K,

rTfcPfc-i = L3H + LAK (mod D),

where gcd(Tfc,D) = 1,

H = WSs-iSt-i + (D/(WZsZt))Bs-iBt-i, K = ZaSs-iBt-i

+ ZtSt-iBs-i

(mod D),

and W = gcd(V„,Vi), Z, = Vs/W, Zt = Vt/W, Qk = Q'm, G'm_x= PmB'^_x + Q'mB'm-2, Li = G'm_i/Vk, L2 = DB'm_i/(ZsZtVk) (mod D), L3 = B^_it L4 =

G'm_il(ZaZt),Tk = UWQ'¿/(VsVt). Proof. By our previous remarks and (3.11), we see that gcd(Xfc,D) = 1. We next point out that from (2.6) we get

(-l)s+k9s+i9t+i = Hi(VtVs)/W- WKiVD, where Hi = H, Ki = K (mod D); hence,

UQ"(-iy(Ak-i

- VDBk-i) = G'm_iHi(VtVs)/W+ DWKiB'm_i -sfD(WKiG'm-i+B'^xHi(VtVs)/W),

where j = s + t + m + k. Now ZsZt | Q'¿, and since c is an ideal of 0, we have

ZsZt | D - P'¿2; thus, ZsZt | P¿', and by (2.8) we deduce that ZsZt | G'/_i (i > 0). It follows immediately that L4 is an integer.

Also, since Vk = gcd(Qk,D)

and

Qk I D - Pm2, we have Vk \ Pm and Vk \ G'm_x by (2.11). Now B'fx = 0 and from (2.12) BBm_i

= PmGm_x +QmGm_2;

hence, since ZsZt [ G^, ZsZt \ G^_2 (m > 1), we get ZsZtVk | DB'm_v Thus, Lt (i = 1,2,3,4) are all integers and by (3.11) so is D/(WZsZt). D 4. The Large Step Algorithm. We are now able to present our second algorithm for determining whether or not D\Y.

The Large Step Algorithm (LSA) (1) Let L be defined as above. In the continued fraction expansion of \/D,

compute and store Ai = (P¿, Qi) and P¿_i (mod D) for i = 1,2,3,...,

s+l,

whereL + l < s < L + 2 and Qs < y/D. (If Qi > sfD, then Qi+i ),W - &d(V,V%

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NUMBERS

V" = gcd(Q',D), Z = V/W, Z' = V'/W, Li = G'^JV", L2 = DB'^_i/(ZZ'V") (mod D), L3 = B'm_x,L4 = G'm_J(ZZ'). Compute H = WEE' + (D/(WZZ'))FF', K = ZEF' + Z'E'F, E' = LiH + L2K, F' = L3H + L4K (mod D).

(4) If (P', Q') = Ai for some i such that 0 < i < s + 1, put S = (Pt/V")Bî

+ (Ql/V")Bl.2,

Y' = BiîE' -F'S

(modD);

otherwise, put V = V" and go to (3).

(5) D|F if and only if D | Y'. Proof of the Large Step Algorithm. Since Qp = 1, it is clear that the algorithm is correct when p < s + l- We will suppose, therefore, that p > s + 1. By (2.11) we have E — As-i/Vs,

where Vs = gcd(Qs,D).

Thus, by Theorem

3.1 we see that at

any stage of the execution of the algorithm the ideal [Q', P' + \/~D\ = afc+i for some

fc > 0 and Sk-i = ME', Bk-i = MF' (mod D), where M is some (undetermined by the algorithm) integer such that gcd(M, D) = 1. By the reasoning in the proof of the Regulator Algorithm, we must eventually have some (P',Q') = Ai where i is some integer such that 0 < i < s ■+1. When this occurs, we have

V —9i+i/9k+i,

where ak+i = [Q',P' + \ÍD\; hence, by (2.6),

(-l)p+iQiBp-i

= BîAk-i

- Bk-iAî.

Now S = St_! EE(Pi/V^Bi-i + (Qi/VJBi-i (mod D), gcd(Qfc/I4,D) = l,Qk = Qi, and Vjt= V¿;hence, (-l)p+t(Qk/Vk)Y = Bt-iME' - SF'M = MY'

(mod D).

Thus, Y' = MY (mod D), where gcd(M, D) = 1. D Notice that by the reasoning used in the proof of the Regulator Algorithm, this algorithm too will execute in 0(D1/4+e) elementary operations when we sort the >?¿'s in step (1) and do a binary search each time we wish to execute the search in step (4). In practice, however, it is faster to conduct this search by using hashing techniques. In our implementation we hashed on the last byte of each Qi in Ai. Because B'i'_x, Q", P" (i < m) and G'¿n_x must be computed explicitly (not just modulo D) for this algorithm, it is important for implementation purposes to know just how large these numbers can get. We answer these questions in the following two theorems.

THEOREM4.1. Let abe the least nonnegative

[Qo/a, (P0 + \/D)/a] with 0 < P0 < Qo and let m

integer such that 0 < Qm < d in the continued fraction

expansion of (Po + \fD)/Qo-

We must have \Qk\ y/D (m > 0), we see that 0 < q0 = [(P0 + v/D)/Qo] < 2. If q0 = 0, then

Pi = -Po, Qi = (D- P02)/Qo;when P0 < y/D, then 0 < Qi < y/D < Q0- If 9o = 0 and P0 > n/D, then |Qj| - (P2 - D)/Q0 < P$/Qo < Qo- If 1, 0 < Q0 - P0 < v^D, and Pi = Qo - -Po- Hence,

0 < Qi = (D - P¡)/Qo 1, we have Qo/\Qk\ > 1- □ COROLLARY4.1.1.

Under the conditions of the theorem we have |Pfc| < y/~D+

Qo (0 < k < m). Proof. Since (fa-i = [(Pjt-i + y/D)/Qk-i), we have from (2.1)

Pk = D-eQk-i, where 0 < £ < 1. Since 0 < Po < Qo and |Qk_i | < Qo, we have our result.

THEOREM 4.2.

D

Under the conditions of Theorem 4.1 we have |Gm_i| < Qo,

Bm-i < Qo/y/D. Proof. If m = 0, then Gm-i = Qo, Pm-i

= 0; if m = 1, then Gm-i =

QoQo — Po = Pi, Pm-i = Bo — I. Since, as shown in the previous theorem, we have Pi = -Po or Pi = Qo —Po, we see that the theorem holds for m — 0,1. If

m = 2 and q0 = 1, then Qo - Po < \fB and 0 < Pi < y/D; hence, 0 < Qi < (D - P2)/Qo

< \/D, which contradicts

the definition of m. Thus, if m = 2, we

have 9o = 0, Pi = -P0, Qi = (D - P02)/Qo, Gm_i = Q0 - 0. We get 0 < 9m+i < 1/Pm_2

< 1

from Theorem 2.5 of [15]. Since 2(—l)mGm_i = (9m+i + 9m+i)Qo

and 2(—l)"vDPm_i

= (9m+i —9m+i)Qo,

we have |Gm_i| < Qo, Pm-i < Qo/y/D. If m > 3 and 0m < 0, we have 0 0; thus, -tpm < 2, and we have

= 9mifim + 9mifirn > -2. Also, 2(-l)mGm_1/Q0

since 9m+i < ifim and -9m+i

= -ifim9m

< -ifim,

< 1. Further,

we get

2y/DBm-X< Qo(i>m- 4>m)= 2Q0v/D/Qm-i < 2Q0. G License or copyright restrictions may apply to redistribution; see http://www.ams.org/journal-terms-of-use

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Both the SSA and a version of the LSA were written in IBM/370 assembly language and run on an AMDAHL 5850 computer. The SSA program was run up to 107 only and succeeded in finding 8 values of D such that D\Y. These are:

46 = 2-23, 430 = 2-5-43, 1817 = 23-79, 58254 = 2-3-7-19-73, 209991= 3-69997, 1752299= 41 • 79 ■541, 3124318= 2 ■1562159,4099215= 3 • 5 • 273281. The total amount of machine time consumed by this run was 2.25 hours. Tests with the LSA revealed that our version of it became faster than the SSA at values of D somewhat in excess of 2 x 106. Experiments in finding a good value for c for the numbers in the range that we considered resulted in our using c =2.5. The LSA program was run on all values of D from 106 to 108. The three D values given above in the range between 106 and 107 were found, but no further values of D were discovered. This run took about 40 hours of CPU time. Because of the interest in the AAC conjecture, we ran our LSA program on all primes (= 1 (mod 4)) in the range 108 to 109. When D is a prime we always have V%— gcd(Q¿, D) = 1 and this allows for considerable simplification of step (3) of the LSA. This run required 31 additional CPU hours and found no prime which did not satisfy the conjecture.

Notice that since the word size for the AMDAHL 5850 is 32 bits and all of the D values which we considered were less than 109 and Q'¿ < y/D ■y/D = D, we were able to use single-precision arithmetic throughout most of the algorithm. Since none of the D values given above is a prime, we have, incidentally, verified Mordell's conjecture for all primes < 108. Finally, we remark that Mollin has proved that for each of the values of D = 7 (mod 8) in the list above, i.e., 209991 and 4099215, the corresponding X value is not powerful. Department of Computer University of Manitoba

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Winnipeg, Manitoba R3T 2N2, Canada

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Some Computational Results on a Problem Concerning Powerful ...

Some Computational Results on a Problem Concerning Powerful ...

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