#A88 INTEGERS 16 (2016) ON RECIPROCAL SUMS FORMED BY SOLUTIONS OF PELL S EQUATION

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1 #A88 INTEGERS 6 (06) ON RECIPROCAL SUMS FORMED BY SOLUTIONS OF PELL S EQUATION Carsten Elsner Fachhochschule für die Wirtschaft University of Alied Sciences Hannover Germany carsten.elsner@fhdw.de Received: 8//5 Revised: 6//6 Acceted: /4/6 Published: /9/6 Abstract Let (X n Y n ) n denote the ositive integer solutions of Pell s equation X DY = or X DY =. We introduce the Dirichlet series X (s) = P n= /Xs n and Y (s) = P n= /Y n s for <(s) > 0 and rove that both functions do not satisfy any nontrivial algebraic di erential equation. For any ositive integers s and s the two numbers X (s ) and Y (s ) are algebraically indeendent over a transcendental field extension of Q whereas the three numbers X () Y () and P n= /(X ny n ) are linearly deendent over Q. From the transcendence of Y () and the corresonding alternating series we obtain an alication to the Archimedean cattle roblem. Irrationality results for series of the form P n= ( )n+ /X n P n= ( )n+ /Y n and P n= ( )n+ /X n Y n are obtained by a theorem of R. André-Jeannin.. A Summary on the Solutions of Pell s Equation Let D denote a ositive integer which is not a erfect square. In this aer we investigate ositive integer solutions (X Y ) of Pell s equations X DY = and X DY = rovided that the Pell-Minus-equation X DY = is solvable in integers. Let (X n Y n ) N : Xn DYn = ± ^ n =... be the set of all integer solutions which are given recursively by X n + Y n D = X + Y D n (n =... ) if X DY = () and X n + Y n D = X + Y D n (n =... ) if X DY = () where (X Y ) is the rimitive solution of the corresonding equation with smallest coordinates X and Y. Let k be the length of the eriod of the continued fraction

2 INTEGERS: 6 (06) exansion of D and let m /q m denote the m-th convergent of D. First we treat the integer solutions of X DY =. It is well-known [8 7] that (X n Y n ) = ( mk q mk ) (3) holds for n = 3... with m = n when k is even; in articular we have (X Y ) = ( k q k ). When k is odd all solutions X n and Y n are given by (3) for m = n. Then (X Y ) = ( k q k ). Pell s equation X DY = is solvable by integers if and only if k is odd. Then we obtain all solutions X n and Y n from (3) for m = n where (X Y ) = ( k q k ). Exanding the right-hand sides of () and () using the Binomial theorem and rearranging the terms we find the reresentations X n = Y n = X 0 ale ale m 0 (mod ) X ale ale m (mod ) m D / Y X m (4) m D ( )/ Y X m (5) where m {n n } deending on whether the solutions of X DY = or X DY = are under consideration. This gives the formulas X n = m + m (6) Y n = D m m (7) where m {n n } and := X + Y D := X Y D. Then we obtain = X DY = ±. Since > it is clear that <. We have 0 < < and = (if X DY = ) and < < 0 and = (if X DY = ). Let s s be integers. In this aer we investigate the series X n= Xn s X n= Yn s X n= X n Y n X n= (X n Y n ) and the corresonding alternating sums. The first and second sum can be considered as Dirichlet series which code the algebraic structure of Pell s equations in analytic terms. We shall show in Section that these series do not satisfy any nontrivial algebraic di erential equation similar to the Riemann Zeta function or the Fibonacci

3 INTEGERS: 6 (06) 3 Zeta function. Discrete values of these series can be investigated by dee methods of transcendence theory. In articular we are interested in algebraic indeendence (and deendence) results between two or three numbers reresented by these series (Sections 4 to 6).. Hyertranscendental Series In this aer we investigate the two Zeta functions X (s) := Y (s) := X X s n= n X Y s n= n <(s) > 0 <(s) > 0 associated with all ositive integer solutions (X n Y n ) n of Pell s equation X DY = ±. Both functions X (s) and Y (s) can be extended meromorhically to the comlex lane by the method described in [6] for the Fibonacci Zeta function. Let D N be not a erfect square and let (X n (D) Y n (D) ) with < X (D) < X (D) <... and ale Y (D) < Y (D) <... denote all integer solutions of Pell s equation X the sets DY =. Moreover we introduce M (D) X := X(D) X (D)... and M (D) Y := Y (D) Y (D).... Lemma.. Every ositive integer k divides infinitely many elements of the set M (D) Y.. There is a sequence ( m ) m of distinct rimes such that every rime m divides at least one element from the set M (D) X. Proof.. For every given k N Pell s equation X Dk Y = has infinitely many solutions X (Dk ) n Y (Dk ) n N because Dk > is not a erfect square by the condition on D. Therefore for n =... we have X (Dk ) n D ky (Dk ) n = which shows that ky (Dk ) n M (D) Y (n =... ). With ky (Dk ) n 0 (mod k) for n the first art of the lemma is roven.. We construct the rimes m of the sequence ( m ) m recursively.

4 INTEGERS: 6 (06) 4 Construction of : Let D 0 := D. The equation X D 0 Y = has infinitely many ositive integer solutions. We know from X (D0) > that there is a (smallest) rime divisor of X (D0) M (D) X. Construction of m+ : Let us assume that we have already defined distinct rimes... m such that every µ ( ale µ ale m) divides at least one element from the set M (D) X. Set D m := D( m ). D is not a erfect square. Therefore Pell s equation X by ositive integers. We consider the articular solution X (Dm) D m Y (Dm) D m Y = can be solved X (Dm) Y (Dm) i.e. =. (8) Assume that there is a rime µ from the set {... m } dividing X (Dm). With D m 0 (mod µ ) it follows from (8) that 0 0 Y (Dm) (mod µ ) a contradiction. This shows that no rime from the set {... m } divides X (Dm). Thus from X (Dm) > we conclude that there is a (smallest) rime divisor m+ of X (Dm) satisfying Equation (8) can be transformed into m m. (9) X (Dm) D m Y (Dm) = which roves that X (Dm) belongs to the set M (D) X. By X(Dm) 0 (mod m+ ) and (9) we comlete the roof of the lemma. Remark. In general the sequence ( m ) m does not increase monotonously. In what follows we aly a theorem due to A. Reich [9]. In order to state this result we firstly introduce the set D of ordinary Dirichlet series f(s) = P n= a nn s with integer coe cients a n satisfying two conditions:.) The abscissa of absolute convergence of P n= a nn s is finite. We denote this abscissa by a (f)..) The set of all divisors of indices n corresonding to nonvanishing coe cients a n contains infinitely many rime numbers. Next let be a nonnegative integer. For a meromorhic function f set f [ ] (s) := f(s) f 0 (s)... f ( ) (s).

5 INTEGERS: 6 (06) 5 Lemma ([9] [] A. Reich ). Let f D h 0 < h < < h m be any real numbers 0... m be any nonnegative integers and let 0 > a (f) h 0. Put k := P m j=0 ( j + ). If : C k! C is a continuous function such that the di erencedi erential equation f [ 0] (s + h 0 ) f [ ] (s + h )... f [ m] (s + h m ) = 0 holds for all s with <(s) > 0 then vanishes identically. A function is said to be hyertranscendental if it satisfies no nontrivial algebraic di erential equation. We may consider the functions X (s) and Y (s) as ordinary Dirichlet series of the form P n= a nn s having the (finite) abscissas 0( X (s)) = 0( Y (s)) = 0 of absolute convergence i.e. they converge absolutely for = <(s) > 0. Moreover by Lemma we know that for both functions the set of all divisors of indices n with a n 6= 0 contains infinitely many rime numbers since a n = for n M (D) (D) X or n M Y resectively and a n = 0 otherwise. On this situation we aly Reich s theorem (Lemma ). Theorem. Let h 0 < h < < h m be any real numbers 0... m be any nonnegative integers and let 0 > h 0. Put k := P m j=0 ( j + ). If : C k! C is a continuous function satisfying or [ 0] X [ 0] Y (s + h 0) [ ] (s + h X )... [ m] (s + h X m) = 0 (s + h 0) [ ] (s + h )... [ m] (s + h m ) = 0 for all s with <(s) > 0 then vanishes identically. Y Corollary. Let (X n Y n ) n be the sequence of all ositive integer solutions of Pell s equation X DY =. Then both the functions X (s) and Y (s) are hyertranscendental. Y 3. On the Irrationality of Some Recirocal Sums It follows from (6) and (7) that every sequence (X n ) n (Y n ) n and (X n Y n ) n satisfies a linear three-term recurrence formula of the form Z n+ + az n+ + bz n = 0. (0) In articular both sequences (X n ) n and (Y n ) n satisfy the same formula. In the sequel we comute the coe cients a b in (0) for every recirocal sum in the set n X X X o X n Y n X n Y n n= n= n=

6 INTEGERS: 6 (06) 6 and show that a b are integers with b = ±. Case. (X n Y n ) satisfies X DY =. We study the recurrence for (X n ) n and (Y n ) n simultaneously. The characteristic olynomial of the recurrence is P ( ) = + a + b = ( )( ) = ( + ) + = X +. Thus for both sequences (X n ) n and (Y n ) n (0) can be written as The sequence (X n Y n ) n Hence we have Z n+ = X Z n+ Z n. () is given by X n Y n = 4 D n + n n n = 4 D ( ) n 4 D ( ) n. P ( ) = ( )( ) = ( + ) + = X + DY + and Z n+ = X + DY Z n+ Z n. () Case. (X n Y n ) satisfies X DY =. The characteristic olynomial of the recurrence for the sequences (X n ) n and (Y n ) n is P ( ) = ( )( ) = X + DY + since X n = n + n = ( ) n + ( ) n Y n = D n n = D ( ) n D ( ) n. Therefore (0) reduces to Moreover for the sequence (X n Y n ) n X n Y n = Z n+ = X + DY Z n+ Z n. (3) = we obtain the formula 4 D n + n 4 D ( 4 ) n 4 D ( 4 ) n. n n

7 INTEGERS: 6 (06) 7 Hence we have and P ( ) = ( 4 )( = 4 ) = ( ) X 4 + 6DX Y + D Y 4 + Z n+ = X 4 + 6DX Y + D Y 4 Z n+ Z n. (4) On the above sequences and their recurrences () () (3) and (4) we may aly a result of André-Jeannin [5 Théorème] from which we derive that the recirocal sums of these sequences as well as the corresonding alternating sums are irrational. In order to state André-Jeannin s result we consider a sequence (w n ) nz of integers satisfying a linear three-term recurrence formula w n = rw n sw n (n Z) with nonvanishing constant integer coe cients r s and arbitrary initial values w 0 w Z. We assume that w n 6= 0 for n and that the discriminant := r 4s is ositive. The characteristic equation X rx + s = 0 of the recurrence has two real zeros such that > and >. Then there are two real constants C C with w n = C n n + C for n 0. Lemma 3 ([5] Théorème R. André-Jeannin). Let x be a nonvanishing integer and let the following conditions be satisfied; (i) s = ± C C 6= 0 (ii) x < C C x <. Then the series P n= xn /w n has a real irrational limit. We aly this lemma with s = r {X (X + DY ) (X 4 + 6DX Y D Y 4 )} > 0 x { +} C C x ale /4D < <. Theorem. Let (X n Y n ) denote ositive integer solutions of either Pell s equation X DY = or of Pell s equation X DY =. Then any number in the set n X n= X n is irrational. X n= Y n X n= X n Y n X n= ( ) n+ X n X n= ( ) n+ Y n X n= ( ) n+ o X n Y n +

8 INTEGERS: 6 (06) 8 4. An Algebraic Indeendence Result for Two Recirocal Sums Formed by Powers of Solutions of Pell s Equation Let K and E be the comlete ellitic integrals of the first and the second kind defined by K = K(k) = E = E(k) = Z 0 Z 0 dt ( t )( k t ) r k t t dt with k C \ {0} [ [ ) where the branch of each integrand is chosen so that it tends to as t! 0. Furthermore let K 0 = K 0 (k) := K(k 0 ) where k + (k 0 ) =. In this section we shall rove the following theorem. Theorem 3. Let (X n Y n ) n be the sequence of all ositive integer solutions either of Pell s equation X DY = or of Pell s equation X DY =. Let the modulus k be given by X Y D = e K(k 0 )/K(k) if X n DY n = (5) and (X Y D) = e K(k 0 )/K(k) if Xn DYn =. (6) Then for any ositive integers s and s the two numbers X (s ) = X n= X s n and Y (s ) = X n= Y s n are algebraically indeendent over Q(E/). Remark. It can be shown that the modulus k with 0 < k < is uniquely determined either by (5) or by (6). Remark 3. It follows from [3 Table ] for q = X Y D or q = (X Y D) according to (5) or (6) that E = q + 0q 3q 3 + 8q 4 96q q 6 960q q 8 + O(q 9 ).

9 INTEGERS: 6 (06) Prearation of the Proof Here we introduce several quantities which will be used for the roof of Theorem 3. The Jacobian ellitic function w = sn z = sn (z k) with modulus k may be defined by Z w dt z = ( t )( k t ) and 0 cn z = sn z dn z = k sn z ns z = sn z nc z = cn z nd z = dn z. We cite the subsequent lemmas 4-7 from []. For some slightly modified roofs see [0 ch. 3.]. Lemma 4. The coe cients c j of the exansion ns z = z + X c j z j j=0 are given by c 0 = 3 ( + k ) c = 5 ( k + k 4 ) c = 89 ( + k )( k )( k ) Xj (j )(j + 3)c j = 3 c i c j i (j 3). i= Lemma 5. The coe cients d j of the exansion X ( k )nd z = k + d j z j j= are given by d = k ( k ) d = 3 k ( k )( k ) Xj j(j )d d j = 6d d j 3d d i d j i (j 3). i=

10 INTEGERS: 6 (06) 0 Lemma 6. The coe cients e j of the exansion ( k )(nc z ) = X e j z j j= are given by e = k e = 3 ( k )( k ) Xj j(j )e e j = 6e e j + 3e e i e j i (j 3). i= Lemma 7. The coe cients f j of the exansion X dn z = + f j z j j= are given by f = k f = 3 k ( + k ) Xj j(j )f f j = 6f f j 3f f i f j i (j 3). i= All the coe cients c j d j e j f j introduced above are olynomials in k with rational coe cients. The recurrence formulas in the receding lemmas allow us to comute the degree of each such olynomial. Lemma 8. We have for j that deg c j = deg d j = j + deg e j = deg f j = j. Proof. We restrict the arguments on the degree of e j. The roofs are similar for the remaining olynomials. Let µ(e j ) be the sign of the leading coe cient of e j. We show the simultaneous identities deg e j = j and µ(e j ) = ( ) j by induction on j. We have deg e = µ(e ) = deg e = 4 and µ(e ) = by Lemma 6. Now we fix some j 3 and

11 INTEGERS: 6 (06) assume the identities to be true for... j. When i runs from to j we obtain ste by ste that µ(e i e j i ) = ( ) i ( ) j i = ( ) j deg(e i e j i ) = i + (j i ) = j Xj µ 3e e i e j i = ( ) j = ( ) j i= Xj deg 3e e i e j i i= = + (j ) = j µ(6e e j ) = ( ) j deg(6e e j ) = 4 + (j ) = j + Xj µ 6e e j + 3e e i e j i = µ(6e e j ) = ( ) j (7) i= Xj deg 6e e j + 3e e i e j i i= = deg(6e e j ) = j +. (8) For the left-hand side of the recurrence formula in Lemma 6 we know that µ j(j )e = and deg j(j )e =. Therefore we obtain by (7) and (8) the desired identities µ(e j ) = ( ) j = ( ) j deg e j = (j + ) = j. The q-series A j+ (q) = C j+ (q) = X n= X n= n j+ q n q n B j+(q) = n j+ q n q n D j+(q) = X ( ) n+ n j+ q n q n X ( ) n+ n j+ q n n= n= q n are generated from the Fourier exansions of the Jacobian ellitic functions; see the formulas (i) (ii) (iii) and (iv) in [3 Table 5]. Combining each Fourier series with the ower series exansion of the corresonding ellitic function given by Lemmas 4-7 we obtain in Lemma 9 below the following closed form exressions of these q-series in terms of K/ E/ and k (cf. [3 Table ] and [4]). To state these exressions we additionally need the series exansions of the circular functions sec z and csc z

12 INTEGERS: 6 (06) given by csc z = z + X sec z = X j=0 j=0 a j z j a j = ( )j (j + ) j+ B j+ (j + )! b j z j b j = ( )j (j + ) j+ ( j+ )B j+ (j + )! (9). (0) Here B = /6 B 4 = /30 B 6 = /4... are the Bernoulli numbers. Lemma 9. Let q = e K0 /K. Then we have K 3 E K + (k ) = 4A (q) K K K j+c j = a j ( ) j j+3 (j)! A j+(q) (j ) E = + 8B (q) j+e j = b j + ( ) j j+3 (j)! B j+(q) (j ) K K E = 8C (q) K K K j+f j = ( ) j j+3 E K + j+d j = ( ) j j+3 (j)! C j+(q) (j ) (k ) = 8D (q) (j)! D j+(q) (j ). As already mentioned above for each q C with 0 < q < we can choose the modulus k such that q = e K0 /K. The following Lemma 0 can be obtained from Nesterenko s theorem [7 Theorem 4.] on Ramanujan s functions P (q ) := 4A (q) Q(q ) := + 40A 3 (q) R(q ) := 504A 5 (q)

13 INTEGERS: 6 (06) 3 which can be exressed in terms of K/ E/ and k using the identities for A A 3 and A 5 in Lemma 9. By Q we denote the set of all comlex algebraic numbers over Q. Lemma 0 ([3] Lemma ). If q = e K0 /K Q with 0 < q < then K/ E/ and k are algebraically indeendent over Q. Lemma ([4] Satz 7 Algebraic indeendence criterion). Let K be a field satisfying Q K C. Let x... x n C be algebraically indeendent over K and let y... y k C with ale k < n satisfy the systems of equations f j (x... x n y... y k ) = 0 ( ale j ale k) where f j (X... X n Y... Y k ) K[X... X n Y... Y k ] for ale j ale k. Assume det (x... x n y... y k ) 6= i aleijalek Then the numbers y... y k are algebraically indeendent over K(x k+... x n ). This indeendence criterion is a slightly generalized version of Lemma in [3]: Let x... x n C be algebraically indeendent over Q and let y... y n C satisfy the system of equations f j (x... x n y... y n ) = 0 for j =... n where f j (X... X n Y... Y n ) belongs to Q[X... X n Y... Y n ] for ale j ale n. If the Jacobian determinant of f... f n with resect to X... X n does not vanish for X i = x i Y i = y i (i =... n) then the numbers y... y n are algebraically indeendent over Q. For Theorem 3 the system of equations in Lemma will take the form f = X (s ) y and f = Y (s ) y where X (s ) Y (s ) Q[x x x 3 ] (cf. Section 4.). Relacing x i by variables X i (i = 3) we X(s i Y (s i Finally by (s)... s (s) we denote the elementary symmetric functions of 3... (s ) defined by X i(s) = ( ) i r ri ( ale i ale s ) aler < <r iales and let 0(s) = for every s. Let U n = n n V n = n + n

14 INTEGERS: 6 (06) 4 for n =.... From [] we get the following identities. Case. Let s N and (X n Y n ) n be all ositive integer solutions of X DY =. Then with A j+ = A j+ ( ) and B j+ = B j+ ( ) for = X Y D we have X (s) = s X Y (s) = n= V s n = ( )s s (s )! = s D s ( ) s s D s (s )! X n= Xs s (s)b + U s n Xs s (s)a + j= j= s j (s)b j+ () s j (s)a j+. () Case. Let s N and (X n Y n ) n be all ositive integer solutions of X DY =. Then with C j+ = C j+ ( ) and D j+ = D j+ ( ) we have X (s) = s X = n= V s n s (s )! Xs s (s)c + j= s j (s)c j+ (3) Y (s) = s D s ( ) s X n= = ( )s s D s (s )! U s n Xs s (s)d + j= s j (s)d j+. (4) The formulas (-4) rely on identities for series on hyerbolic functions found by I.J. Zucker [4]. 4.. Proof of Theorem 3 In (-4) we exress the q-series A j+ B j+ C j+ and D j+ in terms of x = k x = K/ and x 3 = E/ using the identities in Lemma 9. Case. X DY =. In what follows the trile (x x x 3 ) indicates that the variables X i are finally relaced by the values x i (i = 3).

15 INTEGERS: 6 (06) 5 X (s) = X x x x 3 X x x x 3 = + Y (s) = Y x x x 3 Y x x x 3 = s h (s )! (4x x 3 ) (s )! 8 Xs j= s (s )! Xs j= s j (s) ( )s j (j)! j+3 (x ) j+ e j b j i s j (s) ( )s j (j)! j+3 (x ) j+ e 0 j (5) s h (s )! (s )! Xs j= x 3 s j (s) ( )s j (j)!(j + ) j+ (x ) j+ e j i (6) s D s h (s )! ( ) s (s )! 4 Xs j= x x 3 (x )(x ) s j (s) ( )j (j)! j+3 a j (x ) j+ c j i s D s h (s )! ( ) s x (x ) (s )! Xs j= s D s (s )! Xs j= s j (s) ( )j (j)! j+3 (x ) j+ c 0 j h (s )! ( ) s 6 i (7) 3x 3 + (x )(x ) s j (s) ( )j (j)!(j + ) j+ (x ) j+ c j i. (8) Here c 0 j and e0 j are the derivatives of the olynomials c j = c j (k) and e j = e j (k) resectively with resect to their variable x = k.

16 INTEGERS: 6 (06) 6 Case. X DY =. X (s) = + s h (s )! ( ) s (x x x 3 ) (s )! Xs s j (s) ( )j (j)! i j+3 (x ) j+ f j X x x x 3 = s Xs (s )! j= s j (s) ( )j (j)! j+3 (x ) j+ f 0 j X x x x 3 = + Y (s) = + s h (s )! ( ) s (x x 3 ) (s )! Xs j= s j (s) ( )j (j)!(j + ) j+ (x ) j+ f j i (30) s D s h (s )! (s )! Xs j= x x 3 + (x )x s j (s) ( )s j (j)! j+3 (x ) j+ d j Y x x x 3 = Y x x x 3 = + s D s h (s )! x (x ) (s )! 4 Xs j= s j (s) ( )s j (j)! j+3 (x ) j+ d 0 j s D s h (s )! (s )! Xs j= x 3 + (x )(x ) i (3) s j (s) ( )s j (j)!(j + ) j+ (x ) j+ d j i. (3) d 0 j and f 0 j are the derivatives of the olynomials d j = d j (k) and f j = f j (k) resectively with resect to their variable x = k. Since and are algebraic numbers of degree two we know by Lemma 0 that the three numbers x x x 3 are algebraically indeendent over Q. For Theorem 3 it is enough to show by Lemma with y = X (s ) and y = Y (s ) that the

17 INTEGERS: 6 (06) 7 Jacobian determinant = (x X x 3 ) X X (s (x X x 3 Y (s Y X(s Y (s X (s Y (s ) Q(x x 3 does not vanish identically. In what follows let ( ) denote the leading coe cient of the olynomial (x X x 3 ) with resect to its variable X. We are forced to treat the cases for s = or s = searately. Case. X DY = Case.. s > s >. We have from (5-8) that deg X ( ) = s + s and D s ( ) s ( ) = (s )(s ) s c s e 0 s s c 0 s e s. (33) Let us assume that s c s e 0 s s c 0 s e s 0 (34) where the left-hand side is a olynomial in k with rational coe cients. The olynomials c j (k) and e j (k) never vanish identically. Therefore there is an interval [k 0 k ] [0 ] with c s (t)e s (t) 6= 0 for k 0 ale t ale k. Then by (34) we have e 0 s s (t) e s (t) = s c 0 s c s (t) (t) (k 0 ale t ale k ). Integrating both sides with resect to t from k 0 to k we find that e s s (k) c s s (k) = const k (35) is a constant function for k 0 ale k ale k. Consequently this identity holds for 0 ale k ale. But from Lemma 8 we obtain deg k e s s (k) = s s s and deg k c s s (k) = s s which is inconsistent with (35). Hence s c s (k)e 0 s (k) s c 0 s (k)e s (k) does not vanish identically. Since k is a transcendental number it follows from (33) that ( ) 6= 0. The three numbers x x x 3 are algebraically indeendent over Q. Thus the olynomial (X X X 3 ) does not vanish at X i = x i (i = 3) because its leading coe cient with resect to X does not vanish. Case.. s > s =. We have Y (s ) = Y () = D 6 x x 3 (x )(x ) (36)

18 INTEGERS: 6 (06) 8 which yields (x X x 3 ) X (s ) Dx 3 3 D(x )(X ) + 3 x (X X (s Here deg X ( ) = + s and with x = k we obtain ( ) = D 3(s ) ks e s (k )e 0 s. From ks e s (k) (k )e 0 s (k) 0 we obtain as in Case. that e s (k) ( k ) s is a constant function for 0 ale k ale. But again by using Lemma 8 we have deg k e s (k) = s and deg k ( k ) s = s a contradiction. Thus we find (x x x 3 ) 6= 0 by the same arguments as in Case.. Case.3. s = s >. We have and therefore X (s ) = X () = x x 3 (37) (x x x 3 ) = x Y (s We conclude that deg X ( ) = s 4 and ( ) = x 3D s ( ) s c 0 s s 6= 0. The nonvanishing of ( ) follows from deg k (c 0 s (k)) = s 3 and from the algebraic indeendence of x and x 3 over Q. As in Case. it turns out that (x x x 3 ) 6= 0. Case.4. s = s =. From (36) and (37) we obtain (x x x 3 ) = 8 3 Dx x x 3 6= 0. Case. X DY =. The arguments are essentially the same as in Case using the formulas (9-3). Therefore we omit the details and confine ourselves to the necessary identities.

19 INTEGERS: 6 (06) 9 Case.. s > s >. deg X ( ) = s + s ( ) = D s ( ) s (s )(s ) s d s fs 0 s d 0 s f s. Assuming that ( ) vanishes identically we find that which is imossible since deg k by Lemma 8. f s s (k) d s s (k) = const k f s s (k) = s s s and deg k d s s (k) = s s Case.. s > s =. We have Y () = D(x )x 3 + D (x )(x ) deg X ( ) = + s ( ) = D( )s s Assuming that ( ) vanishes identically we find that which is imossible since by Lemma 8. f s (k) ( k ) = const s k (k )f 0 s s kf s. deg k f s (k) = s and deg k ( k ) s = s Case.3. s = s >. We have X () = x With deg X ( ) = s + 5 and x x 3 hence (x x x 3 ) = (x 3 4x Y (s ( ) = Ds s d0 s 6= 0 by deg k (d 0 s (k)) = s 3 it follows again that (x x x 3 ) 6= 0. Case.4. s = s =. Here This comletes the roof of Theorem 3. (x x x 3 ) = 8Dx x (x 3 x ) 6= 0.

20 INTEGERS: 6 (06) 0 5. Auxiliary Results for Sets of Three Recirocal Sums In the following we need beside A j+ (q) B j+ (q) C j+ (q) and D j+ (q) three further q-series (cf. [3 Sec. ]). L (q) = M (q) = J 0 (q) = X n= X n= X n= nq n + q n ( ) n+ nq n + q n ( ) n+ q n / q n. The numbers = X +Y D and = X Y D were already defined in Section. 5.. Series Formed by Solutions of X DY = We have and X =. For s = we obtain by () and () the identities Y n= n X X n= n = 4D = 4 X n= X n= ( n + n ) = 4 X ( n n ) = 4D ( ) It follows by (6) and (7) for m = n that V n= n X U n= n X n Y n = 4 D n n = 4B ( ) (38) = 4DA ( ). (39) where = ( ) =. Similarly to the roof of (39) we obtain the formula X n= (X n Y n ) = 6D X n= ( n n ) = 6DA ( ). (40) The corresonding alternating series can be exressed in terms of q-series too: X ( ) n+ n= X n X ( ) n+ n= Y n X ( ) n+ n= = 4M ( ) (4) = 4DL ( ) (4) X n Y n = 6DL ( ). (43)

21 INTEGERS: 6 (06) 5.. Series Formed by Solutions of X DY = We have =. For s = we obtain by (3) and (4) the identities X X n= n = 4 X n= ( n + n ) = 4 X n= V n = 4C ( ) (44) and X Y n= n = 4D X n= ( n n ) = 4D ( ) X U n= n = 4DD ( ). (45) It follows by (6) and (7) for m = n that X n Y n = 4 D 4n 4n = 4 D ( ) n ( ) n where = ( ) =. Thus X n= (X n Y n ) = 6D X n= Similarly we derive from [] the identity ( 4n 4n ) = 6DC ( 4 ). (46) X Y n= n = X D n n = D n= X n= U n = DJ 0 ( ). (47) 6. An Algebraic Indeendence Result for Three Recirocal Sums Formed by Solutions of Pell s Equation All the statements in this section are going back to algebraic indeendence and deendence results on certain q-series roven in [3]. The three identities A (q) B (q) = 4A (q ) (48) C (q) D (q) = 4C (q ) (49) L (q) M (q) = 4M (q ) (50) can easily be roven by using the underlying q-series. Now let q be an algebraic number over Q with 0 < q <. As articular cases of [3 Theorem ] we know that in each set A (q) B (q) C (q) D (q) and L (q) M (q) the two elements are algebraically indeendent numbers over Q. Assume that A (q) and A (q ) are algebraically deendent over Q. Substituting the exression for A (q ) from (48) into the relation between A (q) and A (q ) we obtain a nontrivial relation

22 INTEGERS: 6 (06) between A (q) and B (q). The contradiction shows that A (q) and A (q ) are algebraically indeendent over Q. It can be roven analogously that B (q) and A (q ) are algebraically indeendent over Q. In a similar vein (49) and (50) are alied to rove the following lemma. Lemma. Let q be an algebraic number over Q satisfying 0 < q <. Then there are the following statements.. Any two numbers in the set are algebraically indeendent over Q.. Any two numbers in the set are algebraically indeendent over Q. 3. Any two numbers in the set are algebraically indeendent over Q. A (q) B (q) A (q ) C (q) D (q) C (q ) L (q) M (q) L (q ) The real number := X Y D is a quadratic surd satisfying 0 < <. With (38) (39) (40) and (44) (45) (46) and (4) (4) (43) resectively we deduce the following theorem alying Lemma. Theorem 4.. Let (X n Y n ) denote ositive integer solutions either of Pell s equation X DY = or of Pell s equation X DY =. Then any two numbers in the set n X X X o Xn Yn X n Y n n= n= are algebraically indeendent over Q. The three numbers in this set are linearly deendent over Q.. Let (X n Y n ) denote ositive integer solutions of Pell s equation X DY =. Then any two numbers in the set n X ( ) n+ n= X n n= X ( ) n+ n= Y n X ( ) n+ o n= X n Y n are algebraically indeendent over Q. The three numbers in this set are linearly deendent over Q.

23 INTEGERS: 6 (06) 3 The statements on the linear deendence of three numbers in each of the sets in Theorem 4 follow easily from Y n D X n = ± X ny n (n =... ). But there are nontrivial identities for recirocal sums formed by solutions of Pell s equation. Let q be any comlex number satisfying 0 < q <. We cite two identities from Theorem 3 in [3]: A (q) + B (q) = L (q) (5) C (q) + D (q) = J 0 (q). (5) Case : Let (X n Y n ) be ositive integer solutions of X from (5) (with q = ) (38) (39) and (4) the formula DY =. Then we have X Y n= n X + D X n= n X ( ) n+ =. n= Y n Case : Let (X n Y n ) be ositive integer solutions of X DY =. Here (5) (with q = ) (44) (45) and (47) rove X Y n= n X + D X n= n X =. Y n n= 7. An Alication to the Cattle Problem of Archimedes Since many authors have treated the details of the famous cattle roblem (cf. [] [3]) we omit all introductory exlanations and restrict to the diohantine equations and their solutions. Let wæ : number of white bulls w ß : number of white cows bæ : number of black bulls b ß : number of black cows dæ : number of daled bulls d ß : number of daled cows yæ : number of yellow bulls y ß : number of yellow cows

24 INTEGERS: 6 (06) 4 There are seven linear and two quadratic restrictions. given by wæ = bæ = dæ = w ß = b ß = d ß = y ß = The two quadratic equations are + bæ + yæ dæ + yæ wæ + yæ b 4 ß + bæ 4 + d 5 ß + dæ 5 + y 6 ß + yæ 6 + w 7 ß + wæ. The linear equations are wæ + bæ (53) yæ + dæ 4 (54) where denotes the set of erfect squares whereas 4 is the set of triangular numbers. There are ositive integers X Y solving Pell s equation X Y =. (55) Since X is an odd number let m be defined by X = m +. Moreover set k := Y. Then a solution of the cattle roblem is given by w ß = k b ß = k y ß = k d ß = k wæ = k bæ = k yæ = k dæ = k.

25 INTEGERS: 6 (06) 5 The restrictions (53) and (54) are fulfilled by wæ + bæ = k = ( Y ) yæ + dæ = m(m + ) k =. It is obvious that the cattle roblem has infinitely many solutions which can be comuted e ectively by solving Pell s equation (55). By C [n] := w æ [n] + b [n] æ + d [n] æ + y æ [n] + w [n] ß + b[n] ß + d[n] ß + y[n] ß we denote the number of cattles of the nth herd satisfying the Archimedean restrictions where C [] < C [] <.... One knows that the smallest herd consists of C [] = cattles where C [] is a number formed by digits. From Theorem 4 (55) and from we derive the following result. C [n] = k = Y Theorem 5. The numbers are transcendental. X n= C [n] and X ( ) n+ C [n] n= References [] C. Elsner Sh.Shimomura and I. Shiokawa Algebraic relations for recirocal sums of binary recurrences Seminar on Mathematical Sciences: Diohantine Analysis and Related Fields Keio University Yokohama 35 (006) [] C. Elsner Sh. Shimomura and I. Shiokawa Algebraic relations for recirocal sums of Fibonacci numbers Acta Arith. 30. (007) [3] C. Elsner Sh. Shimomura and I. Shiokawa Algebraic indeendence results for the sixteen families of q-series Ramanujan J. no. 3 (00) [4] C. Elsner Varianten eines Kriteriums zum Nachweis algebraischer Unabhängigkeiten Forschungsberichte der FHDW Hannover 6 (0) - ISSN: Technische Informationsbibliothek Hannover Signatur RS 853 (06). [5] R. André-Jeannin Irrationalité de la somme des inverses de certaines suites récurrentes C.R.Acad. Sci. Paris 308 Série I (989) [6] L. Navas Analytic continuation of the Fibonacci Dirichlet series Fibonacci Q. 39 (00)

26 INTEGERS: 6 (06) 6 [7] Yu.V. Nesterenko Algebraic Indeendence Narosa Publishing House 009. [8] O. Perron Die Lehre von den Kettenbrüchen Chelsea Publishing Comany New York 950. [9] A. Reich Über Dirichletsche Reihen und holomorhe Di erentialgleichungen Analysis 4 (984) [0] M. Stein Algebraic Indeendence Results for Recirocal Sums of Fibonacci and Lucas Numbers PhD Thesis Gottfried Wilhelm Leibniz Universität Hannover 0; htt://edok0.tib.uni-hannover.de/edoks/e0dh/ df [] J. Steuding The Fibonacci Zeta-Function is hyertranscendental CUBO 0 no. 3 (008) [] I. Vardi Archimedes Cattle Problem Am.Math. Monthly 05 (998) [3] A. Winans Archimedes Cattle Problem and Pell s Equation kobotis.net/math/uc/000/rojects/winans.df [4] I.J. Zucker The summation of series of hyerbolic functions SIAM J.Math.Anal. 0 (979) 9-06.