1_750_naika.dvi


EUROPEAN JOURNAL OF PURE AND APPLIED MATHEMATICS

Vol. 3, No. 6, 2010, 924-947

ISSN 1307-5543 – www.ejpam.com

SPECIAL ISSUE ON COMPLEX ANALYSIS: THEORY AND APPLICATIONS

DEDICATED TO PROFESSOR HARI M. SRIVASTAVA,

ON THE OCCASION OF HIS 70TH
BIRTHDAY

Some New Modular Equations of Degree Four and Their

Explicit Evaluations

M. S. Mahadeva Naika1,∗, K. Sushan Bairy1, M. Manjunatha2

1 Department of Mathematics, Bangalore University, Central College Campus, Bangalore-560 001,

INDIA
2 Department of Mathematics, P E S College of Science, Mandya, INDIA

Abstract. In this paper, we derive several new modular equations of degree 4 by using Ramanujan’s

modular equations. We also establish several general formulas for explicit evaluations of h4,n.

2000 Mathematics Subject Classifications: Primary 33D10, 11F27

Key Words and Phrases: Modular equation, Theta-function

1. Introduction, Definitions and Notations

In Chapter 16 of his second notebook [7], Ramanujan develops the theory of theta-

function and his theta-function is defined by

f (a, b) :=

∞
∑

n=−∞
an(n+1)/2 bn(n−1)/2 (1)

= (−a; ab)∞(−b; ab)∞(ab; ab)∞.

Following Ramanujan, we define

ϕ(q) := f (q,q) =

∞
∑

n=−∞
qn2

= (−q; q2)2∞(q
2; q2)∞, (2)

∗Corresponding author.

Email addresses: msmnaika�rediffmail.
om (M. Naika), ksbairy�rediffmail.
om (K. Bairy),mmanjunathapes�gmail.
om (M. Manjunatha)

http://www.ejpam.com 924 c© 2010 EJPAM All rights reserved.



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 925

ψ(q) := f (q,q3) =

∞
∑

n=0

qn(n+1)/2 =
(q2; q2)∞
(q; q2)∞

,

f (−q) := f (−q,−q2) =

∞
∑

n=−∞
(−1)nqn(3n−1)/2 = (q; q)∞,

χ(q) := (−q; q2)∞,

where

(a; q)∞ =
∞
∏

n=0

(1− aqn).

The ordinary hypergeometric series 2F1(a, b; c; x) is defined by

2F1(a, b; c; x) =

∞
∑

n=0

(a)n(b)n

(c)nn!
xn, | x |< 1,

where

(a)0 = 1, (a)n = a(a+ 1)(a+ 2)...(a+ n− 1), for n≥ 1.

Let

Z(r) := Z(r; x) := 2F1

�

1

r
,

r − 1

r
; 1; x

�

and

qr := qr(x) := exp

 

−π csc

�π

r

�

2F1(
1

r
, r−1

r
; 1; 1− x)

2F1(
1

r
, r−1

r
; 1; x)

!

,

where r = 2,3,4,6 and 0< x < 1.

Let n denote a fixed natural number, and assume that

n
2F1

�

1

r
, r−1

r
; 1; 1−α

�

2F1

�

1

r
, r−1

r
; 1;α

� =
2F1

�

1

r
, r−1

r
; 1; 1− β

�

2F1

�

1

r
, r−1

r
; 1;β

� , (3)

where r = 2,3,4 or 6. Then a modular equation of degree n in the theory of elliptic functions

of signature r is a relation between α and β induced by (3). The Ramanujan-Weber class

invariant is defined as

Gm/n = (α(1−α))−1/24 and Gmn = (β(1− β))−1/24, (4)

where β is of degree n over α.

In [9], J. Yi introduced two parameterization hk,n as follows

hk,n =
ϕ(e−π

p
n/k)

k1/4ϕ(e−π
p

nk)
. (5)

Yi established some properties and several explicit evaluations of hk,n for different real values

of n and k. In [6], M. S. Mahadeva Naika and S. Chandan Kumar have established several



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 926

new modular equations of degree 2. They have also established general formula for explicit

evaluation for h2,n. For more detail, one can see [8].

In Section 2, we collect some results which are useful to prove our main results. In Section

3, we establish the modular equations of the ratios of the theta-function ϕ(q). In Section 4,

we establish several general formulas for explicit evaluations of h4,n using modular equations

established in Section 3.

2. Preliminary Results

In this section, we collect some results which are useful to prove our main results.

Lemma 1. [3, Entry 10 (i), (v) pp.122] For 0< x < 1,

ϕ(e−y ) =
p

z, (6)

ϕ(e−4y ) =
1

2

p
z
�

1+ (1− x)1/4
�

, (7)

where z = 2F1

�

1

2
, 1

2
; 1; x

�

and y = π
2F1

�

1

2
, 1

2
; 1; 1− x

�

2F1

�

1

2
, 1

2
; 1; x

� .

Lemma 2. We have

1. [3, Eq. (24.21), p.215] If β is of degree 2 over α, then

β =

�

1−p1−α
1+
p

1−α

�2

. (8)

2. [3, Entry 5 (ii), p.230] If β is of degree 3 over α, then

�

αβ
�1/4

+
�

(1−α)(1− β)�1/4
= 1. (9)

3. [3, Eq. (24.22), p.215] If β is of degree 4 over α, then

β =

�

1− 4
p

1−α
1+

4
p

1−α

�4

. (10)

4. [3, Entry 13(i), p. 280] If β is of degree 5 over α, then

�

αβ
�1/2

+
�

(1−α)(1− β)�1/2
+ 2
�

αβ(1−α)(1− β)�1/6
= 1. (11)

5. [3, Entry 19(i), p. 314] If β is of degree 7 over α, then

�

αβ
�1/8

+
�

(1−α)(1− β)�1/8
= 1. (12)



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 927

6. [1, Entry 11.3.3, p. 275] If β has degree 8 over α, then

(1− 4
p

1−α)(1− 4
p

β) = 2
p

2
8
p

β(1−α). (13)

7. [3, Entry 3(x),(xi), p. 352] If β has degree 9 over α, then

�

β

α

�1/8

+

�

1− β
1−α

�1/8

−
�

β(1− β)
α(1−α)

�1/8

=
p

m, (14)

�

α

β

�1/8

+

�

1−α
1− β

�1/8

−
�

α(1−α)
β(1− β)

�1/8

=
3p
m

. (15)

8. [4, pp. 387–388] Let

U = 1−
p

αβ −
p

(1−α)(1− β), (16)

V = 64
�
p

αβ +
p

(1−α)(1− β)−
p

αβ(1−α)(1− β)
�

, (17)

and

W = 32
p

αβ(1−α)(1− β). (18)

If β has degree 13 over α, then

p
U(U3 + 8W ) =

p
W (11U2+ V ). (19)

9. [1, Entry 17.3.26, pp. 391–392] If β has degree 17 over α, then

m=

�

β

α

�1/4

+

�

1− β
1−α

�1/4

+

�

β(1− β)
α(1−α)

�1/4

− 2

�

β(1− β)
α(1−α)

�1/8
¨

1+

�

β

α

�1/8

+

�

1− β
1−α

�1/8
«

,

(20)

17

m
=

�

α

β

�1/4

+

�

1−α
1− β

�1/4

+

�

α(1−α)
β(1− β)

�1/4

− 2

�

α(1−α)
β(1− β)

�1/8
¨

1+

�

α

β

�1/8

+

�

1−α
1− β

�1/8
«

.

(21)

3. Some New Modular Equations of Degree Four

In this section, we establish several modular equations of degree four using Ramanujan’s

modular equations.

Set

P :=
ϕ(q)

ϕ(q4)
and Q :=

ϕ(qn)

ϕ(q4n)
. (22)



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 928

Employing (6) and (7) with q = e−y , we find that

2

P
− 1= (1−α)1/4 and

2

Q
− 1= (1− β)1/4, (23)

where β is of degree n over α.

Theorem 1. If X =
ϕ(q)ϕ(q2)

ϕ(q4)ϕ(q8)
and Y =

ϕ(q)ϕ(q8)

ϕ(q4)ϕ(q2)
, then

X 2 − 2X 3Y + X 4Y 2 − 4X Y + 4Y 2 − 4X Y 3 + 4X 2Y 2− 2X 3Y 3 + 2X 2Y 4 = 0. (24)

Proof. Employing (23) with n= 2 in (8), we find that

(Q2 − 2Q2P +Q2P2 − 4Q+ 4+ 4QP − 4P − 2QP2 + 2P2)

(−Q2 + 4Q− 4− 4QP + 4P + 2QP2 − 2P2) = 0.
(25)

By examining the first factor near q = 0, it can be seen that there is a neighbourhood about the

origin, where the first factor vanish but the second factor does not. By the identity theorem

first factor vanishes identically. Hence, by using the fact that X = PQ and Y = P/Q, we obtain

(24).

Theorem 2. If X =
ϕ(q)ϕ(q3)

ϕ(q4)ϕ(q12)
and Y =

ϕ(q)ϕ(q12)

ϕ(q4)ϕ(q3)
, then

Y 2 +
1

Y 2
− 12

�

Y +
1

Y

�

+ 12

�p
X +

2p
X

��p
Y +

1p
Y

�

= 4

�

X +
4

X

�

+ 30. (26)

Proof. Employing (23) with n= 3 in (9), we find that

24PQ2+ 24QP2− 12PQ3− 30Q2P2 − 12QP3+ 12P2Q3

+ 12P3Q2 − 4P3Q3 − 16PQ+ P4 +Q4 = 0.
(27)

Using the fact that X = PQ and Y = P/Q in (27), we obtain (26).

Theorem 3. If X =
ϕ(q)ϕ(q4)

ϕ(q4)ϕ(q16)
and Y =

ϕ(q)ϕ(q16)

ϕ(q4)ϕ(q4)
, then

X 4Y 2+ 6X 3Y + X 2 + 16Y 2 + 24X Y − 4
p

X Y
�

X 3Y + X 2+ 2X + 8Y
�

= 0. (28)

Proof. Employing (23) with n= 4 in (10), we find that

Q4+ 16− 32Q+ 24Q2− 8Q3+Q4P4 − 4Q4P3 + 6Q4P2 − 4Q4P = 0. (29)

Using the fact that X = PQ and Y = P/Q in (29), we obtain (28).



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 929

Theorem 4. If X =
ϕ(q)ϕ(q5)

ϕ(q4)ϕ(q20)
and Q =

ϕ(q)ϕ(q20)

ϕ(q4)ϕ(q5)
, then

Y 3 +
1

Y 3
− 70

�

Y 2 +
1

Y 2

�

− 785

�

Y +
1

Y

�

+ 160





p

Y 3 +
1
p

Y 3





×
�p

X +
2p
X

�

+ 80

�p
Y +

1p
Y

�





p

X 3 +
8
p

X 3
+ 10

�p
X +

2p
X

�





= 80

�

X +
4

X

��

5+ 2

�

Y +
1

Y

��

+ 16

�

X 2 +
16

X 2

�

+ 1620.

(30)

Proof. Employing (23) with n= 5 in (11), we find that

16384+ 147456PQ+1024aP2Q2 − 128aP5Q6 − 128aP6Q5+ 32aP6Q6

− 2048aP2Q3 + 1536aP2Q4 − 2048aP3Q2+ 4096aP3Q3 − 3072aP3Q4

+ 1536aP4Q2 − 3072aP4Q3 + 2720aP4Q4+ 1024aP3Q5 − 96aP3Q6

− 1184aP4Q5 + 1024aP5Q3 − 1184aP5Q4+ 176aP4Q6 − 96aP6Q3

+ 176aP6Q4− 512aP2Q5 + 48aP2Q6− 512aP5Q2+ 228864P2Q2 + 64Q6

+ 48aP6Q2 + 61440P2+ 61440Q2− 184320PQ2− 184320P2Q− 960P5Q4

+ 122880PQ3− 43776Q4P − 150528Q3P2 + 122880P3Q− 150528P3Q2

+ 52416P2Q4+ 96256P3Q3− 43776P4Q+ 52416P4Q2 + 6912Q5P

− 7872P2Q5 − 31872P3Q4 − 31872P4Q3 − 192Q6P + 192Q6P2 + 4224Q5P3

+ 9600Q4P4 + 6912P5Q− 7872P5Q2+ 4224P5Q3 − 64P3Q6− 960P4Q5

− 192P6Q+ 192P6Q2 − 64P6Q3 + 672aP5Q5− 49152P − 49152Q+ 64P6

− 40960Q3+ 14592Q4− 40960P3+ 14592P4− 2304Q5− 2304P5 = 0,

(31)

where a = (αβ)1/2. Isolating the terms having a on one side of the equation and squaring

both sides, we deduce that

(−256PQ− 1600P2Q2 + 640PQ2+ 640P2Q− 640PQ3+ 320Q4P + 1600Q3P2

− 640P3Q+ 1600P3Q2− 785P2Q4− 1620P3Q3 + 320P4Q− 785P4Q2− 70Q5P

+ 160P2Q5+ 800P3Q4+ 800P4Q3− 160Q5P3 − 400Q4P4 − 70P5Q+ 160P5Q2

− 160P5Q3+ 80P4Q5 + 80P5Q4+Q6 + P6 − 16P5Q5)(256PQ+ 1600P2Q2

− 640PQ2− 640P2Q+ P6Q6 + 640PQ3− 320Q4P − 1600Q3P2 + 640P3Q

− 1600P3Q2 + 815P2Q4 + 1620P3Q3− 320P4Q+ 815P4Q2 + 70Q5P − 190P2Q5

− 860P3Q4− 860P4Q3− 6Q6P + 15Q6P2 + 220Q5P3 + 490Q4P4 + 70P5Q

− 190P5Q2+ 220P5Q3− 20P3Q6 − 140P4Q5 − 140P5Q4 − 6P6Q+ 15P6Q2

− 20P6Q3 +Q6+ P6 + 15P4Q6 + 46P5Q5− 6P5Q6+ 15P6Q4 − 6P6Q5) = 0.

(32)



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 930

By examining the first factor near q = 0, it can be seen that there is a neighbourhood about the

origin, where the first factor vanish but the second factor does not. By the identity theorem

first factor vanishes identically. Hence, by using the fact that X = PQ and Y = P/Q, we obtain

(30).

Theorem 5. If X =
ϕ(q)ϕ(q7)

ϕ(q4)ϕ(q28)
and Y =

ϕ(q)ϕ(q28)

ϕ(q4)ϕ(q7)
, then

Y 4 +
1

Y 4
− 280

�

Y 3+
1

Y 3

�

− 28

�

Y 2+
1

Y 2

��

349+ 78

�

X +
4

X

��

− 56

�

Y +
1

Y

��

1079+ 310

�

X +
4

X

�

+ 24

�

X 2 +
16

X 2

��

+ 112

�p
Y +

1p
Y

�



515

�p
X +

2p
X

�

+ 90





p

X 3+
8
p

X 3





+4





p

X 5+
32
p

X 5







+ 56





p

Y 3 +
1
p

Y 3







40





p

X 3 +
8
p

X 3





+313

�p
X +

2p
X

��

+ 1176





p

Y 5 +
1
p

Y 5





�p
X +

2p
X

�

= 16

�

4

�

X 3+
64

X 3

�

+ 203

�

X 2 +
16

X 2

�

+ 2023

�

X +
4

X

��

+ 106330.

(33)

Proof. Employing (23) with n= 7 in (12), we find that

−4+ 2Q+ 2P − 2PQb+ PQc = 0, (34)

where b =
�

αβ
�1/8

and c =
�

αβ
�1/4

. Isolating the terms having b on one side of the equation

and squaring both sides, we deduce that

4P2Q2c − 16+ 16Q+ 16P + 8PQc− 4Q2− 8PQ− 4PQ2c− 4P2− 4P2Qc− P2Q2a = 0. (35)

Isolating the terms having c on one side of the equation and squaring both sides, we deduce

that

8aP2Q4− 32aP2Q3 + 32P2Q2a+ 16aP4Q4+ 80aP3Q3 − 32aP3Q2

− 256+ 8aP4Q2− 32aP4Q3 − 32aP3Q4 + 512P + 512Q− 768PQ

− 96P2Q2 + 384P2Q+ 384PQ2− 64P3Q− 16P4 − 384P2− 384Q2

− 64Q3P + 128P3+ 128Q3− 16Q4− P4Q4a2 = 0.

(36)

Isolating the terms having a on one side of the equation and squaring both sides and then

using the fact that X = PQ and Y = P/Q, we obtain (33).



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 931

Theorem 6. If P =
ϕ(q)

ϕ(q4)
and Q =

ϕ(q8)

ϕ(q32)
, then

�

P4 +
1

P4

��

−Q4+ 16Q3− 112Q2+ 448Q− 1120+
1792

Q
− 1792

Q2
+

1024

Q3
− 256

Q4

�

+

�

P3 +
1

P3

��

8Q4+
2048

Q4
− 8192

Q3
+

14336

Q2
− 14336

Q

�

+

�

P2 +
1

P2

�

�

−28Q4

−7168

Q4
+

28672

Q3
− 50176

Q2
+

50176

Q

�

+

�

P +
1

P

��

56Q4+
14336

Q4
− 57344

Q3
+

100352

Q2

−100352

Q

�

+ P3
�

8960+ 896Q2− 3584Q− 128Q3
�

+ 30848Q+
125400

Q
− 78144

+
1

P3

�

7936− 1536Q− 640Q2+ 384Q3
�

+
1

P2

�

−320Q3− 832Q2+ 9472Q− 29824
�

+ P2
�

448Q3− 3136Q2+ 12544Q− 31360
�

+ P
�

62720+ 6272Q2− 896Q3− 25088Q
�

+
1

P

�

61696− 23040Q+4736Q2− 384Q3
�

− 70Q4− 17920

Q4
+ 992Q3+

71680

Q3

− 7456Q2− 125400

Q2
= 0.

(37)

Proof. Proof of the identity (37) is similar to the proof of the identity (28) given above

except that in place of result (10), result (13) is used.

Theorem 7. If X =
ϕ(q)ϕ(q9)

ϕ(q4)ϕ(q36)
and Y =

ϕ(q)ϕ(q36)

ϕ(q4)ϕ(q9)
, then

Y 6 +
1

Y 6
− 908

�

Y 5 +
1

Y 5

�

− 83582

�

Y 4 +
1

Y 4

�

− 1369692

�

Y 3 +
1

Y 3

�

− 3

�

Y 2 +
1

Y 2

��

2657883+ 96832

�

X 2+
16

X 2

��

− 24

�

Y +
1

Y

�

×
�

892353+ 289628

�

X +
4

X

��

+ 17323008

�p
Y +

1p
Y

��p
X +

2p
X

�

+ 1831776





p

Y 3 +
1
p

Y 3









p

X 3 +
8
p

X 3



+ 21504





p

Y 5+
1
p

Y 5





×




p

X 5+
32
p

X 5



− 29469924= 128

�

2

�

X 4+
256

X 4

�

+420

�

X 3 +
64

X 3

�

+ 9987

�

X 2 +
16

X 2

�

+ 75426

�

X +
4

X

��

.

(38)

Proof. Employing (23) with n= 9 in (14) and (15), we find that

2Q2P − 4Q2P2c − 2aQ2P + 4cP2Q2 b+ 8cQ2P − 4cPQ2 b+ 4bP2Q2− 4Q2



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 932

4QP + 8bQ2− 4cQ2 − 12bPQ2+ 8cP bQ+ 2QP2a+ 4aQP − 4bP2Q

−2QP2− 8cPQ− 12cP2 bQ+ 8bPQ+ 8QP2c − 4aP2 − 4cP2 + 8cP2 b = 0. (39)

Isolating the terms having b on one side of the equation and squaring both sides, we deduce

that

72aQ4P2 − 80Q3P4ca+ 192Q3P3ca+ 144Q2P4ca− 48Q4P3ca+ 96Q2P2ac

− 144Q3P2ac + 96acP3Q− 240acP3Q2 + 48acQ4P2 + 16Q4P4ac − 112QP4ac

− 16aQ4Pc + 32cQ3aP − 16Q2P2 + 32Q3P − 32Q3P2 + 16Q2P3 + 16Q4P4c

− 4Q4P2 + 16Q4P + 240aP3Q3 + 16a2Q3P2 + 8a2Q3P3 + 192Q3P3c + 72Q2P4a

− 64Q3P4a− 4a2Q4P2 + 32acP4 − 32a2Q2P3 − 16aQ4P − 16QP4a− 4Q2P4a2

+ 16QP4a2 + 32a2QP3 − 16a2Q2P2 + 96Q2P2c − 240Q3P2c − 80cQ4P3 − 4Q2P4

− 240Q3P2a+ 96Q3Pa+ 96Q2P2a+ 96Q3Pc − 16QP4c + 96aP3Q− 112cQ4P

− 144Q2P3c − 16a2P4 − 16aP4+ 32cQ4− 16aQ4+ 48Q2P4c + 144cQ4P2 − 16Q4

− 64aQ4P3 + 16Q4P4a+ 32QP3c − 48Q3P4c − 240aP3Q2 + 8Q3P3 = 0.

(40)

By eliminating c and a in the similar manner and then using the fact that X = PQ and

Y = P/Q, we obtain (37).

Theorem 8. If X =
ϕ(q)ϕ(q13)

ϕ(q4)ϕ(q52)
and Y =

ϕ(q)ϕ(q52)

ϕ(q4)ϕ(q13)
, then

Y 7 +
1

Y 7
− 6734

�

Y 6 +
1

Y 6

�

− 13

�

Y 5 +
1

Y 5

��

173721+ 49184

�

X +
4

X

��

− 52

�

Y 4 +
1

Y 4

��

1803735+ 627156

�

X +
4

X

�

+ 85300

�

X 2 +
16

X 2

��

− 13

�

Y 3 +
1

Y 3

��

93381075+ 35045664

�

X +
4

X

�

+ 6898048

�

X 2 +
16

X 2

�

+522240

�

X 3 +
64

X 3

��

− 26

�

Y 2+
1

Y 2

��

255338797+ 99428448

�

X +
4

X

�

+22686560

�

X 2 +
16

X 2

�

+ 2651904

�

X 3 +
64

X 3

�

+ 112384

�

X 4 +
256

X 4

��

− 13

�

Y +
1

Y

��

1355929177+ 537574080

�

X +
4

X

�

+ 131058304

�

X 2 +
16

X 2

�

+18059264

�

X 3 +
64

X 3

�

+ 1173504

�

X 4 +
256

X 4

�

+ 24576

�

X 5 +
1024

X 5

��

+ 416

�p
Y +

1p
Y

�



35973930

�p
X +

2p
X

�

+ 11349767





p

X 3+
8
p

X 3





+ 2161984





p

X 5 +
32
p

X 5



+ 222176





p

X 7+
128
p

X 7



+ 9984





p

X 9 +
512
p

X 9







M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 933

+128





p

X 11 +
2048
p

X 11







+ 832





p

Y 3+
1
p

Y 3





�

9431769

�p
X +

2p
X

�

+ 2904505





p

X 3 +
8
p

X 3



+ 519072





p

X 5 +
32
p

X 5



+ 45976





p

X 7 +
128
p

X 7





+1408





p

X 9+
512
p

X 9







+ 416





p

Y 5 +
1
p

Y 5





�

4977453

�p
X +

2p
X

�

+1447562





p

X 3 +
8
p

X 3



+ 219840





p

X 5+
32
p

X 5



+ 12528





p

X 7+
128
p

X 7









+ 208





p

Y 7 +
1
p

Y 7







1197554

�p
X +

2p
X

�

+ 308857





p

X 3 +
8
p

X 3





+31104





p

X 5 +
32
p

X 5







+ 208





p

Y 9 +
1
p

Y 9





�

53242

�p
X +

2p
X

�

+10113





p

X 3 +
8
p

X 3







+ 106912





p

Y 11 +
1

p

Y 11





�p
X +

2p
X

�

= 24251297512

+ 9667573344

�

X +
4

X

�

+ 2403244896

�

X 2 +
16

X 2

�

+ 346604544

�

X 3 +
64

X 3

�

+ 24986624

�

X 4 +
256

X 4

�

+ 692224

�

X 5 +
1024

X 5

�

+ 4096

�

X 6 +
4096

X 6

�

.

(41)

Theorem 9. If X =
ϕ(q)ϕ(q17)

ϕ(q4)ϕ(q68)
and Y =

ϕ(q)ϕ(q68)

ϕ(q4)ϕ(q17)
, then

Y 9 +
1

Y 9
− 36754

�

Y 8 +
1

Y 8

�

− 17

�

Y 7 +
1

Y 7

��

1096631+ 719552

�

X +
4

X

��

− 272

�

Y 6 +
1

Y 6

��

2211163+ 874270

�

X 2+
16

X 2

�

+ 2693790

�

X +
4

X

��

+ 68

�

Y 5 +
1

Y 5

��

79079901− 15277312

�

X 3+
64

X 3

�

− 114408800

�

X 2 +
16

X 2

�

−157553008

�

X +
4

X

��

+ 136

�

Y 4 +
1

Y 4

��

2016688545− 10983072

�

X 4+
256

X 4

�

−159079808

�

X 3+
64

X 3

�

− 636257912

�

X 2 +
16

X 2

�

− 237316728

�

X +
4

X

��

+ 68

�

Y 3 +
1

X 3

��

39729209249+ 4069258864

�

X +
4

X

�

− 6641286816

�

X 2 +
16

X 2

�

−2422235008

�

X 3 +
64

X 3

�

− 296408768

�

X 4+
256

X 4

�

− 11767808

�

X 5 +
1024

X 5

��



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 934

+ 272

�

Y 2 +
1

Y 2

��

43899000931+ 7993457534

�

X +
4

X

�

− 4759955170

�

X 2 +
16

X 2

�

−2288693056

�

X 3 +
64

X 3

�

− 369671072

�

X 4 +
256

X 4

�

− 24953344

�

X 5 +
1024

X 5

�

−572928

�

X 6 +
4096

X 6

��

+ 34

�

Y +
1

Y

��

821963657831+ 177636127584

�

X +
4

X

�

−67278544896

�

X 2 +
16

X 2

�

− 39038801664

�

X 3+
64

X 3

�

− 7258679936

�

X 4 +
256

X 4

�

−610099200

�

X 5 +
1024

X 5

�

− 22257664

�

X 6+
4096

X 6

�

− 262144

�

X 7 +
16384

X 7

��

− 1088

�p
Y +

1p
Y

�







18819607015

�p
X +

2p
X

�

+ 80443210





p

X 3+
8
p

X 3





−2329762760





p

X 5+
32
p

X 5



− 714027520





p

X 7 +
128
p

X 7



− 92048640





p

X 9+
512
p

X 9





−5523712





p

X 11 +
2048
p

X 11



− 139264





p

X 13 +
8192
p

X 13



− 1024





p

X 15 +
32768
p

X 15











− 544





p

Y 3 +
1
p

Y 3











21182793502

�p
X +

2p
X

�

− 724326781





p

X 3 +
8
p

X 3





− 2970898992





p

X 5 +
32
p

X 5



− 832722704





p

X 7+
128
p

X 7



− 96598528





p

X 9 +
512
p

X 9





−4848384





p

X 11 +
2048
p

X 11



− 81920





p

X 13 +
8192
p

X 13











− 544





p

Y 5 +
1
p

Y 5





×







6400381178

�p
X +

2p
X

�

− 866738129





p

X 3 +
8
p

X 3



− 1160442480

�p

X 5

+
32
p

X 5



− 270503280





p

X 7 +
128
p

X 7



− 24447488





p

X 9 +
512
p

X 9



− 742144

�p

X 11

+
2048
p

X 11











− 1088





p

Y 7 +
1
p

Y 7





�

450392085

�p
X +

2p
X

�

− 185484849

�p

X 3

+
8
p

X 3



− 128003816





p

X 5 +
32
p

X 5



− 21820492





p

X 7 +
128
p

X 7



− 1134848

�p

X 9



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 935

+
512
p

X 9











− 1088





p

Y 9+
1
p

Y 9











14667567

�p
X +

2p
X

�

− 34598985





p

X 3 +
8
p

X 3





−13824368





p

X 5+
32
p

X 5



− 1300340





p

X 7 +
128
p

X 7











+ 544





p

Y 11 +
1

p

Y 11





×







3837042

�p
X +

2p
X

�

+ 5300551





p

X 3 +
8
p

X 3



+ 1065760





p

X 5 +
32
p

X 5











+ 1632





p

Y 13 +
1

p

Y 13











78306

�p
X +

2p
X

�

+ 41561





p

X 3 +
8
p

X 3











+ 1160896





p

Y 15 +
1

p

Y 15





�p
X +

2p
X

�

+ 36888130319124= 128

�

512

�

X 8 +
65536

X 8

�

+ 147968

�

X 7+
16384

X 7

�

+ 9450368

�

X 6 +
4096

X 6

�

+ 230775680

�

X 5+
1024

X 5

�

+ 2576886724

�

X 4 +
256

X 4

�

+ 13262885376

�

X 3+
64

X 3

�

+21325126145

�

X 2 +
16

X 2

�

− 65099344519

�

X +
4

X

��

.

(42)

Proofs of the identities (41) and (42) are similar to the proof of the identity (38) given

above except that in place of result (12), result (19) is used for proving (41); result (20) and

(21) is used for proving (42).

4. General Formulas for Explicit Evaluations of hk,n

We shall employ modular equations in Section 3 to establish several general formulas for

explicit evaluations of hk,n.

Theorem 10. For any positive real number n, we have

h2
4,4n − 2

p
2h4,nh2

4,4n − 2
p

2h4,4n + 2+ 4h4,nh4,4n − 2
p

2h4,n

− 2
p

2h2
4,nh4,4n + 2h2

4,n+ 2h2
4,nh2

4,4n = 0.
(43)

Proof. Employing (5) in (24), we obtain (43).



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 936

Corollary 1. We have

h4,4 =

�

2−
p

2
p

2

�

(
p

2+ 1), (44)

h4,1/4 =
(
p

2+
4
p

2)

2
, (45)

h4,2 = 1+
p

2−
pp

2+ 1, (46)

h4,1/2 =
1+
pp

2− 1p
2

. (47)

Proof. Proofs of (44) and (45)

Putting n= 1 in (43), we find that

h2− 4(
p

2+ 1)h+ 4+ 2
p

2= 0, (48)

where h := h4,4.

Solving the above equation, we obtain (44) and (45).

Proof. Proofs of (46) and (47)

Putting n= 1/2 in (43), we find that

(h2− 2(1+
p

2)h+ 2+
p

2)(h2− 2(
p

2− 1)h+ 2−p2) = 0, (49)

where h := h4,2.

Since the roots of the second factor are imaginary, we deduce that

h2 − 2(1+
p

2)h+ 2+
p

2= 0. (50)

On solving the above equation, we obtain (46) and (47).

Theorem 11. For any positive real number n, we have

N2 +
1

N2
− 12

�

N +
1

N

�

+ 12
p

2

�p
M +

1p
M

��p
N +

1p
N

�

= 8

�

M +
1

M

�

+ 30,

(51)

where M = h4,nh4,9n and N =
h4,n

h4,9n

.

Proof. From the equations (5) and (26), we obtain (51).

Corollary 2. We have

h4,9 = 5− 3
p

2+ 3
p

3− 2
p

6−
q

93− 66
p

2+ 54
p

3− 38
p

6, (52)



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 937

h4,1/9 = 5− 3
p

2+ 3
p

3− 2
p

6+

q

93− 66
p

2+ 54
p

3− 38
p

6, (53)

h4,3 =
�p

2− 1
�

�p
3+ 1p

2

�

, (54)

h4,1/3 =
�p

2+ 1
�

�p
3− 1p

2

�

. (55)

Proof. [Proofs of (52) and (53)] Putting n= 1 in (51), we find that

x2− (20− 12
p

2)x − 32+ 24
p

2= 0, (56)

where x = h4,9 +
1

h4,9

.

On solving the above equation and x > 1, we deduce that

x = 10− 6
p

2+ 2

p

51− 36
p

2. (57)

Again solving the above equation, we obtain (52) and (53).

Proof. [Proofs of (54) and (55)] Putting n= 1/3 in (51), we find that

(x2+ 2
p

2x − 10)(x −p2)2 = 0, (58)

where x = h4,3 +
1

h4,3

.

Since x > 1, we find that

x2+ 2
p

2x − 10= 0. (59)

Solving the above equation, we obtain (54) and (55).

Theorem 12. For any positive real number n, we have

N3+
1

N3
− 70

�

N2+
1

N2

�

− 785

�

N +
1

N

�

+ 160
p

2











p

N3+
1

p

N3





×
�p

M +
1p
M

�

+

�p
N +

1p
N

�





p

M3 +
1

p

M3
+ 5

�p
M +

1p
M

�











= 160

�

M +
1

M

��

5+ 2

�

N +
1

N

��

+ 64

�

M2 +
1

M2

�

+ 1620,

(60)

where M = h4,nh4,25n and N =
h4,n

h4,25n

.



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 938

Proof. Employing (5) in (30), we obtain (60).

Corollary 3. We have

h4,25 = (
p

2− 1)4
�

18+ 8
p

2+ 3
p

5− 2

p

27
p

5+ 12
p

10− 30
p

2− 20

�

, (61)

h4,1/25 = (
p

2− 1)4
�

18+ 8
p

2+ 3
p

5+ 2

p

27
p

5+ 12
p

10− 30
p

2− 20

�

, (62)

h4,5 =

È

17+ 7
p

5− 4
p

29+ 13
p

5

2
−
È

15+ 7
p

5− 4
p

29+ 13
p

5

2
, (63)

h4,1/5 =

È

17+ 7
p

5− 4
p

29+ 13
p

5

2
+

È

15+ 7
p

5− 4
p

29+ 13
p

5

2
. (64)

Proof. [Proof of (61) and (62)] Putting n= 1 in (60), we find that

(h4− 456h3+ 320
p

2h3 − 674h2+ 480
p

2h2− 456h+ 320
p

2h+ 1)(h+ 1)2 = 0, (65)

where h := h4,25.

Since h+ 1 6= 0, we find that

h4 − 456h3+ 320
p

2h3 − 674h2+ 480
p

2h2 − 456h+ 320
p

2h+ 1= 0. (66)

The above equation reduces to

x2− 2−
�

456− 320
p

2
�

x − 674+ 480
p

2= 0, (67)

where x = h4,25 +
1

h4,25

.

On solving the above equation and x > 2, we deduce that

h4,25 +
1

h4,25

= 228− 160
p

2+ 102
p

5− 72
p

10. (68)

Solving the above equation, we obtain (61) and (62).

Proof. [Proof of (63) and (64)] Putting n= 1/5 in (60), we find that

(h4− 6h3+ 2h3
p

2− 10h2+ 10h2
p

2− 10h+ 6
p

2h− 3+ 2
p

2)

(h4+ 6h3+ 2h3
p

2− 10h2− 10h2
p

2+ 10h+ 6
p

2h− 3− 2
p

2)

(−h− 1+
p

2)2(−h+ 1+
p

2)2 = 0,

(69)

where h := h4,5.

The first factor of the equation (69) vanishes for the specific value of q = e−π
p

5/4 but the

other factors does not vanish. Hence,

h4 − 6h3+ 2h3
p

2− 10h2+ 10h2
p

2− 10h+ 6
p

2h− 3+ 2
p

2= 0. (70)



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 939

The above equation reduces to

x4− 12x3+ 12x2− 16x − 16= 0, (71)

where x = h4,5 −
1

h4,5

.

The above equation (71) can be written as

x2− (6+ 2
p

5)x − 2− 2
p

5= 0. (72)

Solving the above equation and x < 1, we find that

x = 3+
p

5− 2

p

4+ 2
p

5. (73)

Again solving the above equation, we obtain (63) and (64).

Theorem 13. For any positive real number n, we have

N4 +
1

N4
− 280

�

N3 +
1

N3

�

− 28

�

N2+
1

N2

��

349+ 156

�

M +
1

M

��

− 56

�

N +
1

N

��

1079+ 620

�

M +
1

M

�

+ 96

�

M2 +
1

M2

��

− 106330

+ 112

�p
N +

1p
N

�



515
p

2

�p
M +

1p
M

�

+ 180
p

2





p

M3 +
1

p

M3





+16
p

2





p

M5 +
1

p

M5







+ 56





p

N3+
1

p

N3







80
p

2





p

M3 +
1

p

M3





+313
p

2

�p
M +

1p
M

��

+ 1176
p

2





p

N5 +
1

p

N5





�p
M +

1p
M

�

= 32

�

16

�

M3 +
1

M3

�

+ 406

�

M2 +
16

M2

�

+ 2023

�

M +
4

M

��

,

(74)

where M = h4,nh4,49n and N =
h4,n

h4,49n

.

Proof. Employing (5) in (33), we obtain (74).

Corollary 4. We have

h4,7 =

�

3+
p

7p
2

�

�

2
p

2−p7
�

, (75)

h4,1/7 =

�

3−p7p
2

�

�

2
p

2+
p

7
�

. (76)



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 940

Proof. [Proof of (75) and (76)] Putting n= 1/7 in (74), we find that

(h2 − 12h+ 7
p

2h+ 1)(h2+ 12h+ 7
p

2h+ 1)(−h2+
p

2h− 1)2

(−h2+ 2h+ 3
p

2h− 1)2(−h2 − 2h+ 3
p

2h− 1)2 = 0,
(77)

where h := h4,7.

We observe that the first factor of the equation (77) vanishes for specific value of q = e−π
p

7/4,

but the other factors does not vanish. Hence, we obtain (75) and (76).

Theorem 14. For any positive real number n, we have

�

16M4 +
1

M4

�
�

−N4+ 16N3− 56N2+ 112
p

2N − 280+
224
p

2

N
− 224

N2

+
64
p

2

N3
− 16

N4

�

+ 16
p

2

�

M3 +
1

8M3

�
�

4N4+
64

N4
− 512

p
2

N3
+

1792

N2
− 1792

p
2

N

�

+ 8

�

M2 +
1

4M2

�
�

−28N4− 323

N4
+

1792
p

2

N3
− 6272

N2
+

6272
p

2

N

�

+ 1984
p

2N3

+ 32
p

2

�

M +
1

2M

�
�

7N4+
112

N4
− 448

p
2

N3
+

1568

N2
− 1568

p
2

N

�

− 280N4− 4480

N4

+ 512
p

2M3
�

35+ 7N2− 14
p

2N −p2N3
�

+
17920

p
2

N3
− 14912N2− 62700

N2

+
64
p

2

M3

�

31− 6
p

2N − 5N2+ 3
p

2N3
�

+ 1792M2
�p

2N3− 7N2+ 14
p

2N − 35
�

+
64

M2

�

−5
p

2N3 − 13N2+ 74
p

2N − 233
�

+ 30848
p

2N +
62700

p
2

N
− 78144

+ 1792
p

2M
�

35+ 7N2−p2N3− 14
p

2N
�

+
128
p

2

M

�

241− 90
p

2N + 37N2− 3
p

2N3
�

= 0,

(78)

where M = h4,n and N = h4,64n.

Proof. Employing (5) in (37), we obtain (78).

Corollary 5. We have

h4,8 =
p

2(
p

2+ 1)3/2





p
2+

pp
2+ 1−

q

4+

p

2+ 10
p

2



 (79)

h4,1/8 =
(
p

2− 1)3/2

p
2

�p
2+

pp
2+ 1−

Æ

4+
p

2+ 10
p

2

� (80)



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 941

Proof. [Proof of (79) and (80)] Putting n= 1/8 in (78), we find that

(−h4+ 16h3+ 12
p

2h3 − 36h2− 24
p

2h2+ 32h+ 24
p

2h− 12− 8
p

2)

(−h4− 16h3+ 12
p

2h3 − 36h2+ 24
p

2h2− 32h+ 24
p

2h− 12+ 8
p

2)

(h2 + 4h+ 2
p

2h− 2− 2
p

2)2(h2 − 4h+ 2
p

2h− 2+ 2
p

2)2 = 0,

(81)

where h := h4,8.

We observe that the first factor of the equation (81) vanishes for specific value of q = e−π
p

2,

but the other factors does not vanish. Hence, we obtain (79) and (80).

Theorem 15. For any positive real number n, we have

N6+
1

N6
− 908

�

N5 +
1

N5

�

− 83582

�

N4 +
1

N4

�

− 1369692

�

N3 +
1

N3

�

− 3

�

N2+
1

N2

��

2657883+ 387328

�

M2 +
1

M2

��

− 24

�

N +
1

N

�

×
�

892353+ 579256

�

M +
1

M

��

+ 17323008
p

2

�p
N +

1p
N

��p
M +

1p
M

�

+ 3663552
p

2





p

N3 +
1

p

N3









p

M3 +
1

p

M3



+ 86016
p

2





p

N5+
1

p

N5





×




p

M5 +
1

p

M5



− 29469924= 128

�

32

�

M4 +
1

M4

�

+3360

�

M3 +
1

M3

�

+ 39948

�

M2 +
1

M2

�

+ 150852

�

M +
1

M

��

,

(82)

where M = h4,nh4,81n and N =
h4,n

h4,81n

.

Proof. Employing (5) in (38), we obtain (82).

Corollary 6. We have

h4,9 = 5− 3
p

2+ 3
p

3− 2
p

6−
q

93− 66
p

2+ 54
p

3− 38
p

6, (83)

h4,1/9 = 5− 3
p

2+ 3
p

3− 2
p

6+

q

93− 66
p

2+ 54
p

3− 38
p

6. (84)

Proof. [Proof of (83) and (84)] Putting n= 1 in (82), we find that

(h4− 20h3+ 12
p

2h3 − 30h2+ 24h2
p

2− 20h+ 12
p

2h+ 1)

(h4+ 20h3+ 12
p

2h3 − 30h2− 24h2
p

2+ 20h+ 12
p

2h+ 1)

(−h4− 6h3+ 6
p

2h3− 14h2+ 8h2
p

2− 6h+ 6
p

2h− 1)2

(−h4+ 6h3+ 6
p

2h3− 14h2− 8h2
p

2+ 6h+ 6
p

2h− 1)2 = 0,

(85)



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 942

where h := h4,9.

We observe that the first factor of the equation (85) vanishes for specific value of q = e−π
p

9/4,

but the other factors does not vanish. Hence, we obtain (75) and (76).

Theorem 16. For any positive real number n, we have

N7 +
1

N7
− 6734

�

N6 +
1

N6

�

− 13

�

N5+
1

N5

��

173721+ 98368

�

M +
1

M

��

− 52

�

N4+
1

N4

��

1803735+ 1254312

�

M +
1

M

�

+ 341200

�

M2 +
1

M2

��

− 13

�

N3+
1

N3

��

93381075+ 70091328

�

M +
1

M

�

+ 27592192

�

M2 +
1

M2

�

+4177920

�

M3 +
1

M3

��

− 26

�

N2+
1

N2

��

255338797+ 198856896

�

M +
1

M

�

+90748240

�

M2 +
1

M2

�

+ 21215232

�

M3 +
1

M3

�

+ 1798144

�

M4 +
1

M4

��

− 13

�

N +
1

N

��

1355929177+ 1075148160

�

M +
1

M

�

+ 524233216

�

M2 +
1

M2

�

+144474112

�

M3 +
1

M3

�

+ 18776064

�

M4 +
1

M4

�

+ 786432

�

M5 +
1

M5

��

+ 416
p

2

�p
N +

1p
N

�



35973930

�p
M +

1p
M

�

+ 22699534





p

M3 +
1

p

M3





+ 8647936





p

M5 +
1

p

M5



+ 1777408





p

M7 +
1

p

M7



+ 159744





p

M9 +
1

p

M9





+4096





p

M11 +
1

p

M11







+ 832
p

2





p

N3+
1

p

N3





�

9431769

�p
M +

1p
M

�

+ 5809010





p

M3 +
1

p

M3



+ 2076288





p

M5 +
1

p

M5



+ 367808





p

M7 +
1

p

M7





+22528





p

M9 +
1

p

M9







+ 416
p

2





p

N5 +
1

p

N5





�

4977453

�p
M +

1p
M

�

+2895124





p

M3 +
1

p

M3



+ 879360





p

M5 +
1

p

M5



+ 100224





p

M7 +
1

p

M7









+ 208
p

2





p

N7 +
1

p

N7







1197554

�p
M +

1p
M

�

+ 617714





p

M3 +
1

p

M3







M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 943

+124416





p

M5 +
1

p

M5







+ 208
p

2





p

N9+
1

p

N9





�

53242

�p
M +

1p
M

�

+20226





p

M3 +
1

p

M3







+ 106912
p

2





p

N11 +
1

p

N11





�p
M +

1p
M

�

= 19335146688

�

M +
1

M

�

+ 9612979584

�

M2 +
1

M2

�

+ 2772836352

�

M3 +
1

M3

�

+ 399785984

�

M4 +
1

M4

�

+ 22151168

�

M5 +
1

M5

�

+ 262144

�

M6 +
1

M6

�

+ 24251297512,

(86)

where M = h4,nh4,169n and N =
h4,n

h4,169n

.

Proof. Employing (5) in (41), we obtain (86).

Corollary 7. We have

h4,13 = −
r

1279+ 355
p

13+ 12
p

a1

2
+

r

1281+ 355
p

13+ 12
p

a1

2
, (87)

h4,1/13 =

r

1279+ 355
p

13+ 12
p

a1

2
+

r

1281+ 355
p

13+ 12
p

a1

2
. (88)

where a1 = 22733+ 6305
p

13.

Proof. [Proof of (87) and (88)] Putting n= 1/13 in (86), we find that

(h4 − 50h3+ 34
p

2h3 − 154h2+ 110
p

2h2 − 190h+ 134
p

2h− 99+ 70
p

2)

(h4 + 50h3+ 34
p

2h3 − 154h2− 110
p

2h2 + 190h+ 134
p

2h− 99− 70
p

2)

(−h4− 20h3+ 16
p

2h3 − 54h2+ 36
p

2h2− 44h+ 32
p

2h− 17+ 12
p

2)2

(−h4+ 20h3+ 16
p

2h3 − 54h2− 36
p

2h2+ 44h+ 32
p

2h− 17− 12
p

2)2

(−h+ 1+
p

2)2(−h− 1+
p

2)2 = 0,

(89)

where h := h4,13.

We observe that the first factor of the equation (89) vanishes for specific value of q = e−π
p

13/4,

but the other factors does not vanish. Hence, we find that

x4− 116x2− 144− 100x3− 240x = 0, (90)

where x = h4,13 +
1

h4,13

.

The above equation can be written as

(x2− 50x − 14
p

13x − 10
p

13− 34)(x2− 50x + 14
p

13x + 10
p

13− 34) = 0. (91)



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 944

Since the roots of second factor are imaginary, we deduce that

x2− 50x − 14
p

13x − 10
p

13− 34= 0. (92)

On solving the above equation, we obtain (87) and (88).

Theorem 17. For any positive real number n, we have

N9+
1

N9
− 36754

�

N8+
1

N8

�

− 17

�

N7+
1

N7

��

1096631+ 1439104

�

M +
1

M

��

− 272

�

N6 +
1

N6

��

2211163+ 3497080

�

M2 +
1

M2

�

+ 5387580

�

M +
1

M

��

+ 68

�

N5+
1

N5

��

79079901− 122218496

�

M3 +
1

M3

�

− 457635200

�

M2 +
1

M2

�

−315106016

�

M +
1

M

��

+ 136

�

N4+
1

N4

��

2016688545− 175729152

�

M4 +
1

M4

�

−1272638464

�

M3 +
1

M3

�

− 2545031648

�

M2 +
1

M2

�

− 474633456

�

M +
1

M

��

+ 68

�

N3+
1

M3

��

39729209249− 376569856

�

M5 +
1

M5

�

+ 8138517728

�

M +
1

M

�

−26565147264

�

M2 +
1

M2

�

− 19377880064

�

M3 +
1

M3

�

− 4742540288

�

M4 +
1

M4

��

+ 272

�

N2 +
1

N2

��

43899000931− 36667392

�

M6 +
1

M6

�

+ 15986915068

�

M +
1

M

�

−19039820680

�

M2 +
1

M2

�

− 1839544448

�

M3 +
1

M3

�

− 5914737152

�

M4 +
1

M4

�

−798507008

�

M5 +
1

M5

��

+ 34

�

N +
1

N

��

821963657831+ 355272255168

�

M +
1

M

�

−269114179584

�

M2 +
1

M2

�

− 312310413312

�

M3 +
1

M3

�

− 1424490496

�

M6 +
1

M6

�

−1161388787976

�

M4 +
1

M4

�

− 19523174400

�

M5 +
1

M5

�

− 33554432

�

M7 +
1

M7

��

− 1088
p

2

�p
N +

1p
N

�







18819607015

�p
M +

1p
M

�

+ 160886420





p

M3 +
1

p

M3





−9319051040





p

M5 +
1

p

M5



− 5712220160





p

M7 +
1

p

M7



− 1472778240

�p

M9

+
1

p

M9



− 176758784





p

M11 +
1

p

M11



− 8912896





p

M13 +
1

p

M13







M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 945

−131072





p

M15 +
1

p

M15











− 544
p

2





p

N3+
1

p

N3





�

21182793502

�p
M +

1p
M

�

− 1448653562





p

M3 +
1

p

M3



− 11883595968





p

M5 +
1

p

M5



− 6661781632

�p

M7

+
1

p

M7



− 1545576448





p

M9 +
1

p

M9



− 155148288





p

M11 +
1

p

M11





−5242880





p

M13 +
1

p

M13











− 544
p

2





p

N5 +
1

p

N5





�

6400381178

�p
M +

1p
M

�

−1733476258





p

M3 +
1

p

M3



− 4641769920





p

M5 +
1

p

M5



− 2164026240

�p

M7

+
1

p

M7



− 391159808





p

M9 +
1

p

M9



− 23748608





p

M11 +
1

p

M11











− 1088
p

2





p

N7+
1

p

N7











450392085

�p
M +

1p
M

�

− 370969698





p

M3 +
1

p

M3





− 512015264





p

M5 +
1

p

M5



− 174563936





p

M7 +
1

p

M7



− 18157568

�p

M9

+
1

p

M9











− 1088
p

2





p

N9 +
1

p

N9





�

14667567

�p
M +

1p
M

�

− 69197970

�p

M3

+
1

p

M3



− 55297472





p

M5 +
1

p

M5



− 10402720





p

M7 +
1

p

M7











+ 544
p

2





p

N11 +
1

p

N11











3837042

�p
M +

1p
M

�

+ 10601102





p

M3 +
1

p

M3





+4263040





p

M5 +
1

p

M5











+ 1632
p

2





p

N13 +
1

p

N13





�

78306

�p
M +

1p
M

�



M. Naika, K. Bairy, M. Manjunatha / Eur. J. Pure Appl. Math, 3 (2010), 924-947 946

+83122





p

M3 +
1

p

M3











+ 1160896
p

2





p

N15 +
1

p

N15





�p
M +

1p
M

�

= 128

�

131072

�

M8 +
1

M8

�

+ 18939904

�

M7 +
1

M7

�

+ 106103083008

�

M3 +
1

M3

�

+ 604823552

�

M6 +
1

M6

�

+ 7384821760

�

M5 +
1

M5

�

+ 41230187584

�

M4 +
1

M4

�

+85300504580

�

M2 +
1

M2

�

− 130198689038

�

M +
1

M

��

− 36888130319124

(93)

where M = h4,nh4,289n and N =
h4,n

h4,289n

.

Proof. Employing (5) in (42), we obtain (93).

Corollary 8. We have

h4,17 =
v −
p

v2− 4

2
, (94)

h4,1/17 =
v +
p

v2− 4

2
, (95)

where

v =
�p

2− 1
�2

�

11+
p

2−
p

17
�p

2− 1
�

+ 2

q

p

17
�

9− 2
p

2
�

− 2
�

13
p

2− 6
�

�

.

Proof. [Proof of (94) and (95)] Putting n= 1/17 in (93), we find that

(h8− 116h7+ 76
p

2h7 − 608h6+ 456
p

2h6 − 1516h5+ 1012h5
p

2− 1858h4

+ 1392h4
p

2− 1516h3+ 1012
p

2h3− 608h2+ 456
p

2h2 − 116h+ 76
p

2h+ 1)

(h8+ 116h7+ 76
p

2h7 − 608h6− 456
p

2h6 + 1516h5+ 1012h5
p

2− 1858h4

− 1392h4
p

2+ 1516h3+ 1012
p

2h3− 608h2− 456
p

2h2 + 116h+ 76
p

2h+ 1)

(−h4− 50h3+ 38
p

2h3− 62h2+ 40
p

2h2 − 50h+ 38
p

2h− 1)2(h− 1)2

(−h4+ 50h3+ 38
p

2h3− 62h2− 40
p

2h2 + 50h+ 38
p

2h− 1)2(h+ 1)2 = 0,

(96)

where h := h4,17.

We observe that the first factor of the equation (96) vanishes for specific value of q = e−π
p

17/4,

but the other factors does not vanish hence, we find that

x4− 612x2− 640+ 76
p

2x3+ 784
p

2x − 116x3− 1168x + 456
p

2x2+ 480
p

2= 0, (97)

where x = h4,17 +
1

h4,17

.

Solving the above equation, we obtain (94) and (95).



REFERENCES 947

ACKNOWLEDGEMENTS The first and second authors are thankful to DST for their support

under the research project SR/S4/MS:509/07.

References

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2005.

[2] B. C. Berndt, Analytic Eisenstein series, theta-functions, and series relations in the spirit

of Ramanujan, J. Reine Angew. Math., 303/304:332–365, 1978.

[3] B. C. Berndt, Ramanujan’s Notebooks, Part III, Springer-Verlag, New York, 1991.

[4] B. C. Berndt, Ramanujan’s Notebooks, Part V, Springer–Verlag, New York, 1998.

[5] B. C. Berndt, L. A. Goldberg, Analytic properties of arithmetic sums arasing in the theory

of the classical theta-functions, SIAM J. Math. Anal., 15:208–220, 1984.

[6] M. S. Mahadeva Naika and S. Chandan Kumar, Some new modular equations and their

applications, (communicated).

[7] S. Ramanujan, Notebooks (2 volumes), Tata Institute of Fundamental Research, Bombay,

1957.

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