Complete solutions to a family of cubic Diophantine equations

JOURNAL

OF NUMBER

THEORY

%i,

235-250

(1990)

Complete Solutions Cubic Diophantine

to a Family of Equations

EMERY THOMAS Deparimeni

of Mathematics, University Berkeley, California 94720 Communicated

of Caljfornio,

by H. Zassenhaus

Received April 5, 1989

The following equation

theorem is proved: If n 2 1.365 x lo’, then the Diophantine 2 - (n - 1) xzy - (n + 2)

has only the “trivial”

.y? - y’ = + 1

(*I

solutions (k LO), (0, * I), ( + 1. f 1).

Moreover, we show that for 0 3. then (*) has only trivial solutions. ((6 1990 Academic Press. Inc.

Since the mid-1960’s, with the work of Baker, one is frequently able to find all the solutions to a given Diophantine equation (of Thue type), at least if the coefficients are reasonable in size (see [ 1,4, 10, 11, 13, 161 for examples). On the other hand, it seems quite difficult to find all the solutions to a parametrized family of equations, since the data shift from one equation to the next. In this paper we consider an infinite family of Diophantine equations (the norm-form equation for the “simplest cubic?-see Shanks [12]) and show that for a sufficiently large value of the parameter one can find all the solutions. The equation we consider is x3-(n-l)x2y-(n+2).xy2-y3=

+_l,

n >, 0.

(*)

Note that for all n this equation has the solutions (0, f I),

(h 1, Oh

(+I,

Tl);

235 0022-314X/90 Ml!34,‘2-X

$3.00

Copyright ( 1990 by Academic Press. Inc. All rights of reproduction m any form reserved.

236

EMERYTHOMAS

we call these the trivial solutions.

For n = 0, 1, and 3 it is known that (*) solutions; see Section 3. Our result is

has non-trivial THEOREM

1. Let n be an integer with n 2 1.365 x 10’. Then the equation x3-(n-l)x*y-(n+2)xy*-y3=

+l,

n 2 0,

has only the trivial solutions given above.

In this section we outline a proof of the Theorem, with some details put over to the next section. In Section 3 we find all solution to (*) with n between 0 and 103. The cases n = 0, 1 are already known [2,8, 151, and so we now assume n 3 2. Suppose, then, that (p, q) is a non-trivial solution to (*); since ( -p, -9) is also a solution, we may assume that q > 0. Since n 3 2, one readily verifies that (*) has no non-trivial solution with q = 1; hence, we now assume q 2 2. For an integer n, set F,(x)=x3-(n-1)x*-(n+2)x-1,

11)

and denote the roots of F,,(x) = 0 by &, > 2, > R,. Remark 1. We regard the roots as functions of n and so sometimes will write A,(n), i= 0, 1, 2. Also, we usually will write simply R ( = A(n)) for b(n).

By Lagrange (see Remark 2 at end of section) we have: If (p, q) is a non-trivial solution to (*), then p/q is a principal convergent for precisely one of the roots &.

(2)

In this case we say that (p, q) has type i (i=O, 1,2). The family (*) has the following striking property: If (p, q) is a solution of type i, then (q, - (p + q)) is a solution with type i+ 1 (with i+ 1 taken mod 3).

(3)

We prove this in the next section. Thus, to prove Theorem 1 it suffices to show PROPOSITION

1. If n > 1.365 x 107, then (*) has no non-trivial solution of

type 0.

Suppose that (p, q) is a non-trivial solution of type 0. We will show that then one must have n < 1.365 x 107.

CUBIC

DIOPHANTINE

237

EQUATIONS

Let K, = Q(n), where F,,(i) = 0. As is well known, (p, q) is a solution to (*) o p - q2 is a unit in K,. Set i=o,

Yi= P-d;,

1,2.

(4)

Thus, (p,q) is a solution to (*)~y~y,~~= f.1. The field K, is Galois with action I, -+O i., --)O i.,, where &=A,

A, = - l/(J. + l),

i,,=

-(I

+ l/3,).

(5)

As is easily shown (e.g., [14]), A,, and iI generate the group of units for the order Z[n] in K,. Thus there are integers A, B (with the choice of signs below A and B will be positive) such that yi= o’(PQf),

i=o,

1,2.

(6)

(We will say that the integers A and B are associated to the solution (p, q).) We will need some estimates for I, q, and the ‘J~s. For n B 2, (a)

n
q>, [$I]

(b)

y(;+II(L+l))-~
a$-2),

q>,2

(7) (c)

q(A+ 1+ l/j-l-&<

(4

hl < l/q2jv2.

ly21
1 + l/j-)+$

We give the proofs in the next section. We now estimate the exponent B. Define a number k, = k&n, q) by B = k0 log q/log i.

(8)

We will prove: LEMMA 1. Let (p, q) be a non-trivial solution to (*) of type 0, with n 2 2, and let A, B be the associated exponents. Then

O
k,(n,q)<(l+log(l+2)/logq)N,, where N,=(l

+ l/Rlog(/2+

1)).

(9) (10)

238

EMERY

THOMAS

Set 6, = 1,/J.,, 6, = A/&, b2 = (2 - A,)/(2 -A,). In the next section we show (cf. [3, 131)

rA0 w*-

l= + C(4 -4M~-&)l

Yoh,.

(11)

where w= _+C(&-&)/(A-A,)1

yo/yI. (12)

Let 6, = - l/6*. Then, 6,Adf+83=W,

Set A = A log16,) -B log16,1 + log16,1. We will show 0 < IAl < l/qY

(13)

From (5) we obtain s,=(n+l)*/&

bi= -1Z/@+l),

63 = (2 + 1)/d.

Therefore, A = C log@ + 1) - D log(A) = C log( 1 + l/n) - E log(A),

(14)

where C=2A+B+l, D=2B+A+l, A=f(2C-D-l), B=f(2DC-l),E=D-C=B-A.Asin(6)wesaythatCandDareassociaredto the solution (p, q). Now D-C=B-A>0 by (9); hence D>C>O (by (9) and (14)). We now give a lower bound for D. By (13) and (14), (C log( 1 + l/n) - Elog(l)l

< l/q314

and so c log( 1 + l/J) - E < l/q%4 log J.. log 1

(15)

We wish to find the least values of C and E (both positive integers) that satisfy (15). Clearly, we minimize C by taking E = 1. In this case C3 Co, where Co is the largest integer such that co log( 1 + l/n) < log 1. (We use here the fact that l/n > log( 1 + l/1) > l/1 - l/22* > l/q3,14.) Hence Co + 12 Co, where Cb is the least integer such that Co/,? 2 log I. Thus

Since D 2 C + 1, we obtain D 2 A log 1.

(16)

CUBIC

DIOPHANTINE

239

EQUATIONS

By (8), (14), and (16) we have B 2 D/3 3 A log A/3,

q 2 2” log i.i3ko.

We now find an upper bound for D. For this we need a preliminary

(17) result.

LEMMA 2. Let f(x) and g(y) be continuous, real-valued, monotonically decreasing functions, with f(x) defined for x > 0 and g( y) for y > 1. Suppose that for all such x and y, f(x) > 1 and g(y) > 1. Suppose, finally, that the equation

g(f (x)) - x = 0 has a unique root, x0, and that there is x, )x0 such that g(f(x,)) x’ and y’ are numbers such that O

Y’ >f(x’),

then x’ 6 x0 and y’ b f (x0). The proof is given in Section 2. We apply the lemma as follows. We now take n > 4. Set f,(x) = Ai tog1./3-y,

g,(y)

= (1 + log(i

+ 2mg

Y) N,

h,,(x) = g,( f,(x)) - x = (1 + 3x log(A + 2)/A(log i)‘) N, - x,

(18)

N, = 1 + l//I log(l + 1). Note that f,,(x), g,(y) and h,(x) satisfy the hypotheses of Lemma 2, with .x0 (the unique root of h,(x) = 0) given by .x,=k,(n)=(N,j~‘-310g(I.+2)/Alog~)2))’. BY @h (lo),

(19)

(17), and (18),

and so by Lemma 2 we have k,(n, q) < k,(n). Note that k,(n) is monotonically decreasing, as a function of n; define an absolute constant, k,, independent of n and q, by k, = k,( 10’). Using (19), we find k, d 1 f E,

where

E= 2.5 x lo-‘.

(20)

240

EMERYTHOMAS

Finally, if n 2 lo’, we have k,(n, q) < k,(n) < k,, and so by (8), (17), and (20) we obtain If n > 107,

then

B < (1 + E) log q/log 1, q 2 ABi(’ +‘I > Ai.10gA/3(’+‘). (21)

To find an upper bound for D we use a recent result of Mignotte and Waldschmidt [9]. Let a be algebraic of degree d over Q, with conjugates O~GI,.... G~CI and minimal polynomial c,xd+ ... + cd, c0 > 0. Set log c,+ i

log Max(1, la,al)

i=l

Suppose now that a, and a2 are real, positive algebraic numbers with exact degrees d, and dz. Let d denote the degree over Q of the field Q(a,, a*). Let b,, 6, be positive rational integers such that b, log aI # b2 log a2.

Define A4 = Max { b, , b2} and choose two positive real numbers a, and a2 satisfying ai2 l,

Ui 2 h(ai) +

log

2,

ai 2 (2e/d) IlOg ai1

for i= 1 and 2. Finally, define W= 512d4a,a2(5.5

w, = 500d4a, a2(3 + log kf)2.

+ log M)*,

We then have [9, Corollary

(22)

1.1 and subsequent remark]:

(i) Jb, log a, - b2 log a21 2 eeW. rf (log a,1 2 1 and log A42 15, then

THEOREM (Mignotte-Waldschmidt).

(ii)

lb, log a1 -b,

log a21 > epW1.

We apply the theorem to our situation aI=&

az = 1 + l/A,

by taking (for n 2 2),

d= d, = dz = 3,

b,=E,

b2=C;

thus, M= C. (It readily follows from (14) that C> E.) We show in Section 2 that A # 0 in (14). Moreover, h(a,) = h(a,) = 3-‘(log

1+ log(1 + l/A)) = 3-l log(il+

1).

CUBIC

DIOPHANTINE

241

EQUATIONS

Therefore, we may take a, = (2e/3) log 1,

a2 = log(R + 1)/3 + log 2.

Suppose now that n 2 10’; then by (17) log D 3 18. Also, log LY,> 1, and so W, = 4500(log(R + 1) + 3 log 2) x (2e) x log 1 x (3 + log C)‘.

Furthermore,

W, < W’,, where w; = 4500k,(log

2)’ (3 + log D)’

and k3 = 2e(log( 10’ + 1) + 3 log 2)/log( 107).

Hence, by the above theorem of Mignotte

and Waldschmidt,

(Al >e-Wl>e.-wi. On the other hand, by (13) and (21) l/11 < l/q9.4
+fl,

and so 3BlogL/(l

+E )< w;.

Consequently, taking k, = 45OOk,( 1 + E). using the fact that B3 D/3, and dividing through by log 1, we obtain D < k, log /I( 3 + log 0)‘.

(We find that k4 = 27620.77...). Set H,(x)

= x - k, log L( 3 + log x)‘.

Thus we have shown: If D is associated with the non-trivial solution type 0, for (*) with fixed n, then H,(D) < 0.

(p, q) of

Note that for n fixed the function H,(x), with x 2 1, has a single zero--call this M,. Also, if x > M,, then H:(x) ~0. Therefore, D < M,. But we have shown in (16) that D >, I log A, and so by (7a), nlogn
242

EMERY

THOMAS

Therefore, H,(n log n) < 0, since H,(x) Consequently,

has no zero between n log n and D.

n log n , lo’, then n < k;(3 + log(n log n))2

(**)

where ki = k4 log(10’ + l)/log(lO’) = 2.762 ... x 104. But by (**), n < 1.365 . 107, as claimed. This completes the proof of Proposition 1 and hence of Theorem 1. (I am indebted to N. Tzanakis for his useful comments on the proof of Proposition 1.) Remark 2. As noted in (7), A is close to n and so 2, is close to 0 and A2 is close to - 1. Using this and (4) one readily shows that if (p, q) is a non-trivial solution to (*) with q > 2, then

where p/q is closest to the root li. Thus, p/q is a principal convergent for li, as claimed in (2). Note Lagrange [6], Legendre [7], and Petho [lo] for further discussion on this topic. Remark 3. Suppose that n is negative, say n = -m, m > 0. Then (p, q) is a solution to

x3-(n-l)x2y-(n+2)xY2-yY3=

+l

if, and only if, (q, p) is a solution to l)xy2--‘=

x3-(m-2)x2y--(m+

_+l.

Thus, there is no loss of generality in taking n > 0, in (*).

2.

PROOFS

Proof of (3). We are to show two things: (q, -(p + q)) is a solution, and it has type (i+ 1). Since (p, q) is a solution we see by (4) and (5) that (P-4~)(P+4+Pa)(4+(P+q)l)=

+l(A+

1).

CUBIC

DIOPHANTINE

EQUATIONS

243

Let

Clearing fractions, we obtain (using (5))

and so t = + 1. This proves that (q, - (p + q)) is again a solution. Suppose first that (p, q) has type 1; we are to show that (q, - (p + q)) has type 2. One readily shows (Eq. (7a)) that for all n > 0, A > 1, and hence - 1 < h 1 < 0. Since p/q is a principal convergent for A,, we must have p d 0 and )pJ < q. Thus, - q/( p + q) < - 1, and so - q/( p + q) can only be a principal convergent for A,-i.e., (q, -(p + q)) has type 2. The other two cases are similar and are left to the reader. Proof’ of (7). One checks that F,,(n) < 0 and that for s > PI, F,(s) > 0; hence I = n + l/a, where CI> 1. Setting x = n + l/u and solving for Q using (1 ), we find that CL> $I. Thus, the first two principal convergents for R are n/l and (a,n+l)/a,, with a,>&]. Hence, since q>2,

since A c (n + 1). This proves (7a). Since p/q is a principal convergent for 1, and by Remark 2,

I I z-1

< 1/2q2;

thus (7b, c) follow at once by (4). For (7d) note that IY~I = I h and so by (a), (b), and (c), [y( < l/q’i’. Proof of Lemma

1. Set ci=log(yil, R=log(IZ+

Taking

l)log(l)+

Y21, We leave the details to the reader. i- 1, 2; and let (log(A,I)*.

absolute values and logs on both sides of (6), we obtain c i = A log@ + 1) + B log( Ilb\ ) c2= -A log(II,,)+

Blog(A).

244

EMERY

THOMAS

Solving for A and B, we obtain

A = Cc,log(a)- c2hdl&l)l/~ B = Cc, log( (&I ) + c2 log(A + 1)1/R.

By (7), O
and logl;l,l

< 4 log A. Since R > 0, we have A > 0,

A) = (cl + c2) logll,l

+ c2 log(l+

1) - c1 log 2

> (Cl + c2) log I&( -I- Cl log(1 + l/A) = (2c, + CJ logJ1,l > 0

and so B-A > 0; i.e., B > A > 0, which proves (9). For (lo), note that log(A) log(A + 1) < R, so that l+log(l+l/4

B
log A

1 [
log@ + 1)

1+

log 1

1

Ilog(;l+

1)

1

Now by (7c), lyzl < q(n + 2) and so cz < log q + log(il + 2). Thus, B<

logq+log(~+2) [

I[

1+

log I

1

Alog(A+

1)

1

= h log q/log 1,

where h = ( 1 + log(l + 2)/lag q) NA, NA = (1 + l/A log(;l + 1)). This proves (10). Proof of (11). Solving for p and q in (4) (taking i = 0, 1) we find (cf. Steiner [ 131)

Substituting

these into the expression y2 = p - q&, we obtain (23)

Now let 60 =1,/A,, (6) imply (11). Proofof(13).

6, = A/A,, S2 = (A - A,)/(A - A*); we see that (23) and

By (12), k&y sg1 = I-6,

+ WI - (6,1 (1 - w&l.

CUBIC DIOPHANTINE

Consequently, have shown

-A log16,( + B log[6,1=

IAl = llwll -wIUI

EQUATIONS

245

log(6,( + log( 1 - MI/&), and so we

G llogll - leu

I I.

(24)

We now give an estimate for jw(/(6,1. LEMMA

3. 6, = --A*( = (A + 1)/n); (A, - &)/(,I-

These follow at once from the definitions LEMMA

4.

jb,) = l/k

and from (5 ).

IyoIy, ( G l/q3J3.

Proof By (7), (yO( dl/q’I’, and Iy,(>qA+q/(A+ q > 4(jV- 2) and q > 2. Thus, the result follows.

l)-

1/2q>qA.

since

As a corollary, we obtain IwP31 = IY~IYJ

h31=

MY I %I = h/h VW + 1)I

d l/q93(A + 1) d l/(qV

-I- 1).

Consequently, (A( = (log(1 - - lu’/~,ll d Iw/&l .

I

1 I 1 - lW3l

I

= , w,63, -, _ 1 G l/q3j”4,

which proves one half of (13). To complete the proof of (13), we show Let n be the linear form given in (14). Then n # 0.

(25)

Proof. If n = 0, then by (14), (1. + 1)’ = 3,O. But this is impossible since 1 and ;C+ 1 are independent units. Thus, n # 0. Proof‘ of Lemma 2. Let h(x) = g(f(x)) -x. For any x with O 1, g(f(x)) > 1, and so g(f(x)) > x, i.e., h(x) > 0. Since x0 is the unique number such that h(x,) = 0, and since there is x1 > x0 with h(x,) x0, then g(f(x)) x0 and show that this leads to a contradiction. By what we have shown above, if x’> x0, then x’> g(f(x’)). But y’>f(x’), by hypothesis, and so (since g is monotonic decreasing) g(~)‘) < g(f(x’)). Therefore,

246

EMERYTHOMAS

x’ > g(f(x’)) > g(y’). But this is a contradiction, and so x’ ,< x0, as asserted, Since f is also monotonic decreasing, we have

which completes the proof of the lemma.

3. SOLUTIONS In this section we find all solutions to (*), with 0 < n < 103. The cases n = 0 and n = 1 are already known [2,8]; we repeat these results below. (See also [lo, 15, 161.) Thus, from now on we assume that 26nG 103. Our result is THEOREM 2. For Odn < lo3 the only non-trivial solutions to (*) with q > 0 are asfollows:

n = 0:

(5,4), (-4,9),

(-9,

n= 1:

(2, I), (-3,2),

(-1, 3)

n=3:

(7.2), (-2,

9), (-9,

5), (-2,

11, (1, I), (- 42)

7).

We prove the theorem in two steps. First, we find all solutions (p, q) to (*), 2 < n 6 103, with 2 < q 6 lo’-this gives the results in the table above. We then show that (*) has no solutions with q> lo5 (and 2 10. By (19), k,( 10) = 1.2204, .. .. and so by (17), since 2 = A(n) > 10, q>lO

~O~og(10)/3h(~O)

>

105.

Thus, we must have 2 < n d 9. As noted in (2), p/q is a principal convergent for the root il =A(n). By the usual continued fraction process (e.g., see [ 16]), for a fixed n we test each principal convergent pk/qk, with qk < 105, to see whether it is a solution to (*). We find that the only such solutions, for 2 d qk d 105, are those listed in Theorem 2. We leave the details to the reader. The proof of Theorem 2 is then complete when we show: PROPOSITION

2.

There are no solutions to (*) with q > 105, 2
The proof will follow at once from the following three results. (Proofs given at end of section.) Recall the number k,(n, q), defined in (8).

247

CUBIC DIOPHANTINEEQUATIONS LEMMA

5. If q > lo5 and 2
LEMMA 6. Suppose that C is associated to the solution (p, q), where 2 105. Then

c < 1.281 x 108. For each n between 2 and lo3 we now describe a finite algorithm for proving Proposition 2. Fix n, let p’ = log( 1 + l/A)/log I, and let ,u be a rational (decimal) approximation to $, so that 1~- $1 < 10p4’. Moreover, let { Pk/Qk} be the continued fraction expansion for p and (Pi/Q;} that for $. We say that a principal convergent Pk/Qk is proper if lb - P,JQkl < l/(2 + E) Q:,

where

E =

lo-‘.

(26)

Note that if in (26) we replace 2 + E simply by 2, then (26) is always satisfied either by Pk/Qk or Pk + ,/Qk + , ; see [S]. Thus, one can expect that proper convergents occur frequently. We show: 7. For a fixed value of n, 2
1.281 x lo8 f Q, 6 1016

(27)

then, for that value of n, (*) has no solution (p, q) with q > 105.

We now have a finite algorithm for proving Proposition 2. Take the continued fraction expansion of p, where ,u is a decimal approximation to log( 1 + l/A)/log I, as above. For each n between 2 and lo3 one finds that there is a proper convergent that satisfies (27), thus proving Proposition 2. (By changing the various constants that occur, the same proof readily handles the cases n = 0 and n = 1.) We are left with Lemmas 5,6 and 7 to prove. Proof of Lemma 5. As noted above, k,( 10) = 1.22..., so if n > 10, then kO(n, q) 4 k,( 10) -=z1.267, which proves the lemma in this case. On the other hand, if 2 6 n < 9, then by (lo),

kdn, 4) < (1 + log(W)

+ 2Vog 4) Nj.c,,

d 1 + log(A(9) + 2)/lag q) N;,,, = 1.2664... < 1.267. This completes the proof.

248

EMERY

THOMAS

Proof of Lemma 6. We again use the Mignotte-Waldschmidt Theorem [9], as in Section 1, but now taking W in (22) instead of W,. Thus, W= 512 x 34x (2e/3) x log(A) x (log(A + 1)/3 + log 2) x (5.5 + log C)‘. Hence, w < w = k,(log A)( 5.5 + log C)“, where k, = 9612e(log(A( 103) + 1) + 3 log 2). (We find that k, = 1.730... x lo’.) 2
By Mignotte-Waldschmidt

for each n,

(Al 2ewW>eeW’, where n is given in (14). As before, now taking E’ = 0.267, we see that 3B log A/( 1 + E’) < k,(log i1)(5.5 + log C)‘. Since C < 38, by (9) and (14) we obtain C - k6(5.5 + log C)* < 0, where k, = kS x 1.267 = 2.192... x 105. Let x1 denote the root of the equation x - k6( 5.5 + log x)’ = 0,

x > 10’.

We find that x, = 1.2808... x lo’, and so C< 1.281 x 108, as claimed. This completes the proof of the lemma. We need one more fact to prove Lemma 7. Let (C, E) be the exponents associated to the solution (p, q). Then by (15) and (17), since C < D, Ilog( 1 + l/A)/log A- E/C/ < 1/Cq314 log A< l/C,IC’k0~4 log 1. Since A 2 2.65 and C > 1, it is easily seen that /ZClk0A4log A 3 2c, and so Jlog( 1 + l/n)/log

A - E/Cl < 1/2C2. Consequently,

E/C is a principal

convergent for log( 1 + 1/1)/lag ,I = ,u’.

(28)

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DIOPHANTINE

249

EQUATIONS

Proof’ of Lemma 7. Let Pj/Qj be a proper Qi < 1016. Since 1~- $1 < 1O-~4oone sees that

convergent

for p with

IP’-P,/Qjl < 1/2Qt, and hence P,/Qj is also a principal convergent for p’ (see [S] ). Suppose now that P,,/Q, is the proper convergent given in Lemma 7 (so that (27) is satisfied). Let (p, q) be a solution to (*) and let (C, E) be the pair of exponents associated with (p, q) as in (14). By (28), E/C is a principal convergent for p’, say E/C= Pi/Q;. Since lE/C - ~‘1 < l/q3A4(log A)C by (15 ), it follows by a standard argument from continued fractions (e.g., [16]) that

l/(Q;+Q;+,

) < l/q%4 log A,

which implies q312410gjW
(29)

Since C = Qh we must have Qb < Q,, by (27) and Lemma 6. But P,/Q, is also a principal convergent for p’, and so Q;, , < Q, < lOI by (27). Consequently, qY

log A.< 2 x 1016.

But q > 10’ and 1. > A(2) > 2.65, which implies q3A4 log A > 1Or5 x (2.65)4 log(2.65) > 2 x 1016,

which is a contradiction.

Thus, there cannot be a solution

to (*) with

q I=-lo5 and 2 d n < 103. This completes the proof of Lemma 7.

Remark 4. By an argument similar to that given for Proposition can easily show

2, one

Suppose that lo3
250

EMERYTHOMAS

Conjecture.

The Diophantine

equation

X3-((n-l)xZy-(n+2)xy2-y3=

has non-trivial

+1,

iZ20

solutions if, and only if, n = 0, 1 or 3.

ACKNOWLEDGMENT Part of the research for this paper was done while the author was a visiting fellow at the Research School of Physical Sciences, Australian National University, Canberra.

Note added

in proof:

(12/14/89): M. Mignotte has proved the Conjecture.

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in Belgium