<A>Identities from some class invariant
We observe that
1
3
 17  3 21
9  3 21 
  5  21 
1. 



8
8
8


21  3
8
1
3
 11  6 3
96 3  1 1
  (3 4  4  3 )
2. 


 2
2
2


5  17
19  5 17  5  17
3.




2
2
8

17  3 

8 

6
1
1
3
 47  12 15
45  12 15 
4  15  15 4
 
4. 



2
2
2


5.

73  42 3  72  42 3

1
3
72 3
3 2 3

4
4
1

3
1
1
3
3
2
2
 191  72 7  6  7 4 (3  7)
189  72 7  6  7 4 (3  7) 
2
2

6.. 



2
2




 
7  6 7 
3 7  6 7
=
3
1
4
1
4
1
3
 93 6
73 6 
  3 6 

7. 


2
2
4


8.
 18  3 33 
17  3 33

1
3
6 1
4
9  33
1  33

8
8
In order to prove the such form of identity ,it is hard to compute directly. Therefore
we must to find other method to show them .The following are two methods to prove
those identities.
The first method is to use the class invariant of Ramanujan’s.
Before we prove them, we introduce one definition and two theorem.

Defintion: (a;q)   (1- a q n), q  1, and  (q) = (-q; q 2) . .If q=exp(-  n ),where n
n=0
is a positive rational number , the two class invariants Gn and gn are defined by
1
1
Gn = 2 4 q 24 (q) and gn= 2  4 q 24 (q) .
1

1

Theorem1:Let p= G 4n + G -4n .Then
G9 n
1
 2
2
2
p - 2 + ( p 1 ) p(
2
= G np + p - 1

2



1
6
4)

2
p
- 4 +(p
2
2
2 3
 1p ) (
.


4)
Theorem2:Let p= g 4n - g -4n .Then

g9 n = g np + p 2
1
 2
2
2
p + 4 + ( p 1 ) p(
+ 1

2


1
6
4)

2
p
+ 2 +(p
2
2
2 3
 1p ) (
.


Now, we prove the eight identities.
1. On the table of Gn from the Ramanujan’s notebook Part V(p.189), we get
1
 5  21  6  5  21
G63= 2 

 

2
8


21  3 
 . (＊11)
8 

1
4
By theorem 1,
 2
2
2
2
2
2
p - 2 + (p  1)(p  4)
p - 4 + (p  1)(p  4)
2
G 63 = G 7 p + p -1 


2
2



1
6
   
4
From table of Gn, G7= 2 4 .Then p= 2 4  2 4
1
1
1
-4
=
5
.Thus
2
1
1
6
3
1 5 +
21   17 + 3 21
9 + 3 21 
 .(＊12)
G 63 = 2 4 

 

8
8
 2  

Compare (＊11) and (＊12), we complete the proof.
2. On the table of Gn from the Ramanujan’s notebook Part V(p.189), we get
1
1 1
 3  13  4
3 4  4  3 
. (＊21)
G117= 
  2 3  13  6 


2
 2 


1
3



4)
1
 3  13  4
Since G13= 
 , p= 13 . By theorem 1,
 2 
1
4
1
3
 11  6 3
 3  13 
96 3 


G117 = 

. (＊22)
 2 3  13


2
2
2




Compare (＊21) and (＊22), we complete the proof.


1
6
3. On the table of Gn from the Ramanujan’s notebook Part V(p.189), we get
G153=
Since G17=

5  17

8
5  17

8
 5  17
G153 = 


8

17  3
8

1
3
 37  9 17
33  9 17 

 . (＊31)



4
4


2
17  3
5  17
, p=
. By theorem 1,
8
2

17  3  5  17
19  5 17



8
2
2


1
6
1

 1 7 5 17


(1 9
4
5 1 7 ) ( 1 3

5 17 )


 1 3 5 1 7
4

(1 9
2
,(＊32)where (19  5 17)(13  5 17)  20  4 17 .
Compare (＊31) and (＊32), we complete the proof.
4. On the table of Gn from the Ramanujan’s notebook Part V(p.189), we get
 1 5 
G225= 
 2  3
 2 

From table of Gn, G25=

1
3
 4  15  15 14 

 . (＊41)


2


1 5
.Then p=7.Thus
2
1
3
 47 +12 15
 1 5 
45 +12 15 


G 225 = 

.(＊42)
 7  4 3


2
2
2




Compare (＊41) and (＊42), we complete the proof.


1
6
5. On the table of Gn from the Ramanujan’s notebook Part V(p.189), we get
3
5 1 7 ) (1 3


5 17

G333=
6  37

Since G37= 6  37

G 333 = 6  37

 7
1
4
1
4
 7
1
4

3  2 37
1
6
 7  2 3  3 2 3 

 . (＊51)


2


, p= 2 37 . By theorem 1,
3  2 37

1
6
73  42 3  72  42 3
 .(＊ )
1
3
52
Compare (＊51) and (＊52), we complete the proof.
6. On the table of Gn from the Ramanujan’s notebook Part V(p.189), we get
G441=
3 7
2 3
2


1
6
2 7  74 7
2
 
7  6 7 
3 7  6 7
3

1
4
1
4
. (＊61)

1
4  7  74
2 7  74 7
9  4 7 9  3 7 74

Since G49=
, G 494 
.

2
2
2
2 2
1
Then p= 9  47 .
p + p -1  9  4 7  2 3 16  6 7
2
= 9  4 7  2 3(3  7)
= (2  3)(3 3  2 7)
 3 7 
= (2  3)

2


3
By theorem 1,
G 441 =
1
2 7  74 7
2
3
 191  72 7  A
3 7
189  72 7  A 

2 3 



2
2
2




1
6
.(＊62),where
A= (192  72 7)(189  72 7)  6 2016  762 7  6  7 4
1
3
2
(3  7) .
2
Compare (＊61) and (＊62), we complete the proof.
7. On the table of gn from the Ramanujan’s notebook Part V(p.189), we get
 (2 
g90=
Since g10=
5)( 5  6)

1
6
 3 6



4

6 1 
 . (＊71)
4 

1
1 5
 (2  5) 6 , p= 5 . By theorem 2,
2
1
3
 93 6
73 6 
 .(＊72)
g90 = 2  5
5 6 



2
2


Compare (＊71) and (＊72), we complete the proof.


1
6

1
6
8. On the table of gn from the Ramanujan’s notebook Part V(p.189), we get
g198=

1  2 4 2  33)
1
6
 9  33
1  33 
 . (＊81)

8
8 


 
Since g22= 1  2 , p= 4 2 . By theorem 2,

g198 = 1  2 4 2  33

1
6
18  3 33  17  3 33
 .(＊ )
1
3
82
Compare (＊81) and (＊82), we complete the proof.
The second method is to use the following theorem.
1
1
Theorem 3:Suppose that c= (a 2 - b) 3 .Then we can write (a + b) 3  x + y ,where
4x3-3cx=a and y=x2-c.
For example, consider the identity 2.
Let a =
11  6 3
11  6 3
96 3
,b=
,then c=1.Compute 4x3-3x=
.Square both
2
2
2
side and let t=x2. Thus, t(4t-3)2=
11  6 3
3
 t=1+
. Hence,
2
4
3
1
4  3 ;y=
.The proof is complete.
2
4
The other identities are similar to the above.
x=
We can use theorem 3 to construct more identities of this form . In order to get such
identity , we must give x and c firstly. Then a and y are got by the condition of
theorem 3. For example, let x=
3 2
,c=1.Since 4x3-3cx=a and
4
y=x2-c,a=

1
3 2
2 1
1 2
and y=
.And c= (a 2 - b) 3 , then b=
.Thus,
2
4
2
3 2
1 2

2
2

1
3
=
3 2
+
4
2 1
.
4
<B>Some identities from  n
2 4 3  ( 3  1)
1. 1  1  (2  3) 
24 2
4
8
 5  17
2. 8 1  1  


8

17  3 

8 

24
1
2


4  17  1 
 3  4  17 


4
4


4
5 1 5 1
8
3. 8 1  1  ( 5  2) 

2
2
The three special identities are from 9 , 17 ,  25 ,respectively, by using the theorem
for  n (PartV ,p.290)and the definition of  n .i.e.,
 7  17
3  17



4
4





1
1
1
1
8
12
24
8
 n = G -12
(1- 1- G -24
.
 (1 1G -24
n (G n - G n  1) =
n )  2 n
n )
2
2
Proof of identity 1:
From the theorem for  n on the Ramanujan’s notebook PartV(p.290),

1
9  
2
3
2
 
 2 9
1






4
8
3
3  1

4 

3

4
1
1
8
 3 1  2 3  3
 

 
4
2

 
1

 1 3 3
And G9  
 ,then
2


 2 
9
1
8
3 1 
1
  5 ( 2 3  ( 3  1)) .
4  24

4
 8 1  1  (2  3) .Thus the proof is
complete.
The other identities are similar to the above .
In order to construct the equality of the form, we must introduce the following
theorem.
Theorem 4:If r is any positive real number and t=

r- r 2  1  


4

1
1 
t + - t -  
2
2 


r 1
, then
8
8
1
t + 1
2
2

1
t + 1 
2
 .

2


12
If we let G 12
n =r, then we can also get the above identities. For instant, let r= G 9 ,then


1 -12 12
1 -12
24
t=7+4 3 and  9  G 9 (G 9 - G 9  1) = G 9

2
2


 3 3
1
 G 9- 12


2
4


3
4
1
t + 1
2
2

1
t + 1 
2


2


8
8
1
1
 ,by theorem 4. And  9 = (1- 1- G 9-24 ) ,then

2

 3 3

(1- 1- G9-24 ) = G 9-12

4

8
3 1 
 .Thus the identity is hold.
4 

To construct the equality of the form, we need to give an arbitrary r (r  1)and
consider the identity r-1(r- r 2  1 )=1- 1- r -2 . Then by using theorem 4 , we can get
such identities.
For instant, let r=17, then t=
1- 1- r -2 =(1-
1
17
3
.Thus,
2


1 
-1
2
288 );r (r- r  1 )=
17 


1
t + 1
2
2

1
t + 1 
2


2


8
1
 
17 

2 1
2
1
 8117
2 1 

2 

8
1

2 8 =817  8 


2 1
2
2 1 
.
2 

<C>More identities from class invariant
Moreover, if we compute class invariants via Kronecker’s limit formula or modular
equation and compare the result with the table of Gn (PartV ,p.189), then we can get
other identities.
[Kronecker’s limit formula]
 9  65
1
1  65
G 65 , ( 74  10 65  (5  65))  


4
8
8

3
 63 3
23 3 
1
 
G 69 , 



4
4
2






2
 187 108 3  9  6 3 
3
 6  11
2  11  1
  (2 22  7 2  182  56 11)
G 77 , 


4
4  2


3
 18  9 3
14  9 3 
1
 
G141 , 

( 7855  4536 3  9(7  4 3))


4
4
2


 17  145
9  145
G145 , 2 29  5 5  240  20 145  


8
8

G 213 ,
G 301 ,
G 445 ,




3
 21  12 3
1
3
19  12 3 

(135619  78300 3) 
(87  50 3)  



2
2
2
2


3
 46  7 43
1
42  7 43 

( 7(25941  3956 43)  (301  46 43))  



4
2
2


 13  89
5  89
71320  7560 89  (189  20 89)  


8
8





3
6
 113  5 505
105  5 505 

G 505 , (130 5  29 101)  169440  7540 505  



8
8


3
[modular equation]
 7  65
1
G 65 , (3  65  3 5  13)  


4
8


G 69 , 5  23
2

2
3 3  23
2



1
2
3


65  1 

8 

2
748  432 3  747  432 3



2
17  9 3  31419  18144 3  4 3  47




  7  247 
3
2
G141 ,
2 14  9 3
G145 ,
 17  145
9  145
241  20 145  240  20 145  


8
8





3
G 213 ,
3
 5 3  71  2  59  7 71 
2 19  12 3 (41  24 3)  542475  313200 3  
 

2
2

 



1
2
In Ramanujan’s notebook PartV(p.265), there is an another table of G n which are
computed via class field theory. The following are the identities attained from
comparing between the table of G n which are computed via class field theory and the
table of G n ( p.189).
 13  3 
1. G 65 , 
 2 


1
4
 5 1 


 2 
1
4
 7  65



8

 11  7 
2. G 77 , 8  3 7 


2




1
8
1
8
1
4
 23  8 11
19  8 11 





4
4


1
 7  47 12
4 3  47 
 
2


3. G141 ,


4. G145 ,
 2  5   5  2 29 
1
4
1
8




1
4
1
2
65  1 

8 


31420  18144 3  31419  18144 3
1
2
 12  145
8  145 





4
4


1
 1  5  3 5  41  4  7  41
5. G 205 , 

 
 2 

2
8


 
41  1 

8 


1
12
1
1
4
2
 2025  315 41
2017  315 41   25  3 41
17  3 41 
 





 

8
8
8
8

 

6. G 213 ,
1
1
 5 3  71  8  59  7 71 12

 
 
2
2



7. G 265 ,




 2  5   7  2 53 
1
4
1
4

542476  313200 3  542475  313200 3
 16  265
12  265



4
4

 23 43  57 7 
8. G 301 , 8  3 7 


2




1
8
1
8
1
2



1
4
 1199  84 43
1195  84 43 





4
4


1
 21  445  4  13  89
5  89
2  5 

 
2
8
8

 
9. G 445 ,

1
12




1
1
4
2
 71325  7560 89
71321  7560 89   85  9 89
77  9 89 
 





 

4
4
8
8

 

1
4
1
 1 5 
2  5 
 10  101
2


10. G 505 ,


1
4
2
 292  13 505
288  13 505 





4
4


11. G 553 ,
1
1
2
4
 96  11 79
100  11 79   19163  2156 79
19159  2156 79 

 




 

4
4
4
4

 

<D>Other identities
1.
A
B
P
Q
C
The triangle ABC is a right triangle. The two curves are arcs of circles centered at B
and C. Thus, c:=BA=BQ and b:=CA=CP. Let a=BC. Underneath the figure, two
equalities are given.
(b+c- b2  c 2 )2=2( b2  c 2 -b)( b2  c 2 -c) and
( 3 (a  b) - 3 a 2  ab  b2 )3=3( 3 a3  b3 -a)( 3 a3  b3 -b)
2
2.Let x2=y+a,y2=z+a,z2=x+a, then
x3+
1 2
1
1
x (1+ 4a  7 )- x(2a+1- 4a  7 )+ (a-2-a 4a  7 )=0
2
2
2
x3+
1 2
1
1
x (1- 4a  7 )- x(2a+1+ 4a  7 )+ (a-2+a 4a  7 )=0.
2
2
2
Let A= 4a  7 .The roots of (*1) are
1+ A 2
1
2A
x1  
4 a - A s it na (n1
6
3
3
3 3
1
x2  
1+ A 2

1
2A
4 a - A s i n ( t a n1
6
3
3 3
3 3
x3  
1+ A 2
 1
2A  1
4a - A sin(  tan 1
)
6
3
3 3
3 3
)
1
)
The roots of (*2) are
A -1 2
1
2A  1
x4 
+
4a + A sin( tan 1
)
6
3
3
3 3
x4 
A -1 2
 1
2A  1
+
4a + A sin(  tan 1
)
6
3
3 3
3 3
x4 
A -1 2
 1
2A  1
+
4a + A sin(  tan 1
)
6
3
3 3
3 3
3. m 3 4m - 8n + n 3 4m + n =
1
3

3
2
2
(4m + n) + 3 4(m - 2n)(4m + n) - 3 2(m - 2n)

pf: (u2-2uv-2v2)2= u4-4 u3v+8uv3+4v4
=u(u3+8v3)+v(4v3-4u3) (*)
let m=u3 +8v3 ,n=4v3-4u3
1

2
3
3  4 m - 8 n
4 m - 8 n =u3 6 u = 6



1
3
2
4 m + n = v3 6 
3
3  4 m + n
v
=
6

3
Use (*), then the proof is complete.
Moreover, (u2+auv-bv2)2 , where a2=2b and a,b are rational number, then
(u2+auv-bv2)2 =u(u3-2abv3)+v(b2v3+2au3)
let m= u3-2abv3 ,n= b2v3+2au3
Then, by the same procedure, we can get another identity .
For instant, let a=2,b=2.Thus,
m 3 4m + 8n + n 3 4m - n =
1
3

3
2
2
(4m - n) + 3 4(m + 2n)(4m - n) - 3 2(m + 2n)

4. 3 (m 2 + mn + n 2) 3 (m - n)(m + 2n)(2m + n) + 3mn 2 + n 3 - m3
(m - n) m + 2n  3 (2m + n) m - n  3 (m + 2n) 2m + n 
=
+
9
9
9
2
2
2
3
1
5.
1
3
1
3
 cos40    cos80    cos20 
1
3
 s e c 4 0  
1
3
s e c 80
3
3
  ( 3 9  2) 
2

1
3
1
3


1
3
s e c26 0( 9  1 )
3
1
1
1
Theorem 5:  ,  ,  are roots of x3–ax2+bx-1=0, then  3   3   3   a  6  3t  3 ,
1
1
1
1
1
  3     3    3  b  6  3t  3 where t3–3(a+b+3)t-[ab+6(a+b)+9]=0.
Pf:Since 2cos20o, -2cos40o,and -2cos80o are roots of x3–3x-1=0.Let
a=0,b=-3,then t=- 3 9 .Thus the proof is complete.