Annales Mathematicae et Informaticae
41 (2013) pp. 41–55
Proceedings of the
15th International Conference on Fibonacci Numbers and Their Applications
Institute of Mathematics and Informatics, Eszterházy Károly College
Eger, Hungary, June 25–30, 2012
Identities in the spirit of Ramanujan’s
amazing identity
Curtis Cooper
Department of Mathematics and Computer Science
University of Central Missouri
Warrensburg, MO 64093
U.S.A.
[email protected]
Abstract
Motivated by an amazing identity by Ramanujan in his “lost notebook”,
a proof of Ramanujan’s identity suggested by Hirschhorn using an algebraic
identity, and an algorithm by Chen to find such an algebraic identity, we
will establish several identities similar to Ramanujan’s amazing identity. For
example, if
X
9 + 3609x − 135x2
,
an xn =
1 − 6888x + 6888x2 − x3
n≥0
X
bn xn =
10 − 1478x + 172x2
,
1 − 6888x + 6888x2 − x3
cn xn =
12 + 1146x + 138x2
,
1 − 6888x + 6888x2 − x3
n≥0
X
n≥0
then
a3n + b3n = c3n + 1.
Keywords: Ramanujan, identity
MSC: 11A55
1. Introduction
In his “lost notebook”, Ramanujan [4] stated the following amazing identity. If
X
n≥0
an xn =
1 + 53x + 9x2
,
1 − 82x − 82x2 + x3
41
42
C. Cooper
X
bn xn =
2 − 26x − 12x2
,
1 − 82x − 82x2 + x3
cn xn =
2 + 8x − 10x2
,
1 − 82x − 82x2 + x3
n≥0
X
n≥0
then
a3n + b3n = c3n + (−1)n .
Hirschhorn [2] demonstrated that using the algebraic identity from the “lost notebook”,
(x2 + 7xy − 9y 2 )3 + (2x2 − 4xy + 12y 2 )3 = (2x2 + 10y 2 )3 + (x2 − 9xy − y 2 )3 , (1.1)
Ramanujan could have proved his identity. Chen [1] gave an algorithm to produce
similar algebraic identities and Ramanujan-like identities. Our goal is to use this
procedure to find explicit algebraic identities and Ramanujan-like identities.
2. Third power algebraic identity to Ramanujan-like
identity
The following algebraic identity was suggested by Chen [1] and the theorem and
proof were suggested by Hirschhorn [2].
Theorem 2.1. Let
(r1 x2 + s1 xy + t1 y 2 )3 + (r2 x2 + s2 xy + t2 y 2 )3
2
2 3
2
(2.1)
2 3
=(r3 x + s3 xy + t3 y ) + (x − s4 xy − t4 y ) ,
be an algebraic identity in variables x and y and integer constants r1 , r2 , r3 , s1 ,
s2 , s3 , s4 , t1 , t2 , t3 , and t4 . Then if
X
an xn =
r1 + (s1 s4 + t1 − r1 t4 )x − t1 t4 x2
,
1 − (s24 + t4 )x − (s24 t4 + t24 )x2 + t34 x3
bn xn =
r2 + (s2 s4 + t2 − r2 t4 )x − t2 t4 x2
,
1 − (s24 + t4 )x − (s24 t4 + t24 )x2 + t34 x3
cn xn =
r3 + (s3 s4 + t3 − r3 t4 )x − t3 t4 x2
1 − (s24 + t4 )x − (s24 t4 + t24 )x2 + t34 x3
n≥0
X
n≥0
X
n≥0
then
a3n + b3n = c3n + (−t4 )3n .
Proof. Let w0 = 0, w1 = 1, and
wn+2 = s4 wn+1 + t4 wn .
Identities in the spirit of Ramanujan’s amazing identity
The generating function for the sequence {wn } is given by
w(x) =
X
wn xn =
n≥0
x
.
1 − s4 x − t4 x2
Now, if x = wn+1 and y = wn , then
2
x2 − s4 xy − t4 y 2 = wn+1
− s4 wn+1 wn − t4 wn2
2
= wn+1
− wn (s4 wn+1 + t4 wn )
2
= wn+1
− wn wn+2 = (−t4 )n .
The last equality can be proved by induction on n.
Now, let
2
an = r1 x2 + s1 xy + t1 y 2 = r1 wn+1
+ s1 wn+1 wn + t1 wn2 ,
2
bn = r2 x2 + s2 xy + t2 y 2 = r2 wn+1
+ s2 wn+1 wn + t2 wn2 ,
2
cn = r3 x2 + s3 xy + t3 y 2 = r3 wn+1
+ s3 wn+1 wn + t3 wn2 .
We can show that
a3n + b3n = c3n + (−t4 )3n .
But, using generating function techniques, we can show that
X
wn2 xn =
n≥0
X
2
wn+1
xn =
n≥0
X
wn wn+1 xn =
n≥0
1−
(s24
x − t 4 x2
,
+ t4 )x − (s24 t4 + t24 )x2 + t34 x3
1 − t4 x
,
1 − (s24 + t4 )x − (s24 t4 + t24 )x2 + t34 x3
s4 x
.
1 − (s24 + t4 )x − (s24 t4 + t24 )x2 + t34 x3
Hence,
X
an xn =
r1 + (s1 s4 + t1 − r1 t4 )x − t1 t4 x2
,
1 − (s24 + t4 )x − (s24 t4 + t24 )x2 + t4 x3
bn xn =
r2 + (s2 s4 + t2 − r2 t4 )x − t2 t4 x2
,
1 − (s24 + t4 )x − (s24 t4 + t24 )x2 + t34 x3
cn xn =
r3 + (s3 s4 + t3 − r3 t4 )x − t3 t4 x2
,
1 − (s24 + t4 )x − (s24 t4 + t24 )x2 + t34 x3
n≥0
X
n≥0
X
n≥0
and the proof is complete.
43
44
C. Cooper
3. Search for third power algebraic identities
We will attempt to find particular integer constants involving all the r’s, s’s, and
t’s which satisfy equation (2.1) with the following procedure.
Procedure to search for third power algebraic identities
1. Pick one particular set of integers r1 , r2 , and r3 such that
r13 + r23 = r33 + 1.
(3.1)
2. Select a collection of sets of integers t1 , t2 , t3 , and t4 such that
t31 + t32 = t33 − t34 .
(3.2)
Also, select a range of integer values for s1 and s2 to search.
a. For each t1 , t2 , t3 , t4 , s1 , and s2 , compute s3 and s4 using the equations
s3 =
s1 t21 + s2 t22 + r12 s1 t24 + r22 s2 t24
,
r32 t24 + t23
s4 = r32 s3 − r12 s1 − r22 s2 .
Make sure these constants can be computed and that they are integers.
b. Check the following conditions.
3r1 t21 + 3s21 t1 + 3r2 t22 + 3s22 t2 = 3r3 t23 + 3s23 t3 + 3t24 − 3s24 t4 ,
6r1 s1 t1 + s31 + 6r2 s2 t2 + s32 = 6r3 s3 t3 + s33 + 6s4 t4 − s34 ,
3r12 t1 + 3r1 s21 + 3r22 t2 + 3r2 s22 = 3r32 t3 + 3r3 s23 − 3t4 + 3s24 .
c. If all the above conditions are satisfied (every equation is true), the
resulting collection of r’s, s’s, and t’s form an algebraic identity satisfying
equation (2.1).
To prove that the procedure above will produce an algebraic identity, cube the
trinomials in (2.1) to obtain
t31 y 6 + 3s1 t21 xy 5 + (3r1 t21 + 3s21 t1 )x2 y 4 + (6r1 s1 t1 + s31 )x3 y 3
+
+
t32 y 6
=
t33 y 6
+
+
(3r12 t1 + 3r1 s21 )x4 y 2 + 3r12 s1 x5 y + r13 x6
3s2 t22 xy 5 + (3r2 t22 + 3s22 t2 )x2 y 4 + (6r2 s2 t2
(3r22 t2 + 3r2 s22 )x4 y 2 + 3r22 s2 x5 y + r23 x6
+ s32 )x3 y 3
+ 3s3 t23 xy 5 + (3r3 t23 + 3s23 t3 )x2 y 4 + (6r3 s3 t3 + s33 )x3 y 3
+ (3r32 t3 + 3r3 s23 )x4 y 2 + 3r32 s3 x5 y + r33 x6
− t34 y 6 − 3s4 t24 xy 5 + (3t24 − 3s24 t4 )x2 y 4 + (6s4 t4 − s34 )x3 y 3
(3.3)
Identities in the spirit of Ramanujan’s amazing identity
45
+ (−3t4 + 3s24 )x4 y 2 − 3s4 x5 y + x6 .
Collecting like terms in (3.3), we obtain the following equation.
(t31 + t32 )y 6 + (3s1 t21 + 3s2 t22 )xy 5 + (3r1 t21 + 3s21 t1 + 3r2 t22 + 3s22 t2 )x2 y 4
+
+
=
(t33
−
+
+
(3.4)
(6r1 s1 t1 + s31 + 6r2 s2 t2 + s32 )x3 y 3 + (3r12 t1 + 3r1 s21 + 3r22 t2 + 3r2 s22 )x4 y 2
(3r12 s1 + 3r22 s2 )x5 y + (r13 + r23 )x6
t34 )y 6 + (3s3 t23 − 3s4 t24 )xy 5 + (3r3 t23 + 3s23 t3 + 3t24 − 3s24 t4 )x2 y 4
(6r3 s3 t3 + s33 + 6s4 t4 − s34 )x3 y 3 + (3r32 t3 + 3r3 s23 − 3t4 + 3s24 )x4 y 2
(3r32 s3 − 3s4 )x5 y + (r33 + 1)x6 .
Step 1 in the procedure insures that the coefficients of x6 in the algebraic
identity are equal. In addition, we would like r1 , r2 , and r3 to be positive integers.
For Ramanujan’s algebraic identity this condition is trivially true since
13 + 23 = 23 + 1.
Other trivial values of r1 , r2 , and r3 which satisfy (3.1) are r1 = 1 and r2 = r3 = r,
where r is a positive integer.
Appendix I gives positive integer values of r1 , r2 , and r3 (r1 < r2 and r2 6= r3 )
which satisfy (3.1). These values were determined by a C++ program.
In step 2, we select a collection of t’s satisfying (3.4) to try. This guarantees
that the coefficients of y 6 in the algebraic identity are equal. In the spirit of
Ramanujan, we assume t4 = ±1. To obtain nontrivial results, we also require that
t1 6= t2 , t1 6= −t2 , t1 6= −1, and t2 6= −1. Otherwise, some of the t’s could be
positive or negative integers and cancel each other. Appendix II contains some of
the t’s which satisfy (3.2). Again, this appendix was constructed with the help of
a C++ program.
Also, in step 2 we search a range of integers s1 and s2 (via a C++ program).
Some typical ranges for s1 and s2 were from −1500 to 1500. With the r’s, t’s, s1 ,
and s2 fixed, the constants left are s3 and s4 . For step 2a, we compute integers
s3 and s4 . The formulas in step 2a are equivalent to the equations equating the
coefficients in the xy 5 and x5 y terms in (3.4). These equations are
3s1 t21 + 3s2 t22 = 3s3 t23 − 3s4 t24 ,
3r12 s1 + 3r22 s2 = 3r32 s3 − 3s4 .
Step 2a merely solves them for s3 and s4 since they are linear equations in those
two variables. We also require that s4 > 0.
For step 2b, the conditions we check are the equations resulting from equating
the coefficients of the terms x2 y 4 , x3 y 3 and x4 y 2 on each side of equation (3.4). In
step 2c, if all of these conditions are satisfied, the constants determine an algebraic
identity.
46
C. Cooper
4. Third power results
We found the following results. The constants in each row of the following table
satisfy (2.1). We include the leading coefficient of 1 in the last trinomial. Recall
that the form of the last trinomial is x2 − s4 xy − t4 y 2 .
r1 , s1 , t1
1,556,-65601
1,61,-791
1,7,-9
1,-25,135
1,-227,11161
9,412,-11161
9,-126,3753
9,45,-135
9,-169,791
9,-1539,65601
3753,-126,9
11161,3481,-791
11161,412,-9
r2 , s2 , t2
2,-364,83802
2,-40,1010
2,-4,12
2,-32,138
2,-292,11468
10,-180,14258
10,236,-3230
10,-20,172
10,-180,812
10,-1640,67402
4528,200,-8
11468,-1300,1010
11468,-112,12
r3 , s3 , t3
2,-36,67402
2,-4,812
2,0,10
2,-36,172
2,-328,14258
12,112,11468
12,96,2676
12,12,138
12,-220,1010
12,-2004,83802
5262,84,6
14258,1292,812
14258,180,10
1, s4 , t4
1,756,1
1,83,-1
1,9,1
1,9,1
1,83,-1
1,756,1
1,430,-1
1,83,-1
1,9,1
1,83,-1
1,430,-1
1,6887,-1
1,756,1
The bounds on s1 and s2 varied depending on the speed of the search. Note
that the third row is the algebraic identity discovered by Ramanujan. This gives
Ramanujan’s amazing identity. The eighth row gives the algebraic identity
(9x2 + 45xy − 135y 2 )3 + (10x2 − 20xy + 172y 2 )3
=(12x2 + 12xy + 138y 2 )3 + (x2 − 83xy + y 2 )3 .
This produces the Ramanujan-like identity result found in the abstract. The seventh row gives the algebraic identity
(9x2 − 126xy + 3753y 2 )3 + (10x2 + 236xy − 3230y 2 )3
=(12x2 + 96xy + 2676y 2 )3 + (x2 − 430y + y 2 )3 .
This produces the following Ramanujan-like identity. If
X
an xn =
9 − 54172x + 3753x2
,
1 − 184899x + 184899x2 − x3
bn x n =
10 + 98260x − 3230x2
,
1 − 184899x + 184899x2 − x3
cn xn =
12 + 43968x + 2676x2
,
1 − 184899x + 184899x2 − x3
n≥0
X
n≥0
X
n≥0
then
a3n + b3n = c3n + 1.
47
Identities in the spirit of Ramanujan’s amazing identity
5. Fourth power algebraic identities to Ramanujanlike identities
McLaughlin [3] found ten sequences whose sums of their first through fifth powers
are equal. We will not be so ambitious. The following identity was suggested by
Chen [1] and the theorem and proof were suggested by Hirschhorn [2].
Theorem 5.1. Let
(x2 + s1 xy + t1 y 2 )4 + (mx2 + s2 xy + t2 y 2 )4 + (nx2 + s3 xy + t3 y 2 )4
2
2 4
2
2 4
2
(5.1)
2 4
=(mx + s4 xy + t4 y ) + (nx + s5 xy + t5 y ) + (x − s6 xy − t6 y ) ,
be an algebraic identity in variables x and y and integer constants m, n, s1 , s2 , s3 ,
s4 , s5 , s6 , t1 , t2 , t3 , t4 , t5 , and t6 . Then if
X
an xn =
1 + (s1 s6 + t1 − t6 )x − t1 t6 x2
,
1 − (s26 + t6 )x − (s26 t6 + t26 )x2 + t36 x3
bn xn =
m + (s2 s6 + t2 − mt6 )x − t2 t6 x2
,
1 − (s26 + t6 )x − (s26 t6 + t26 )x2 + t36 x3
cn xn =
n + (s3 s6 + t3 − nt6 )x − t3 t6 x2
,
1 − (s26 + t6 )x − (s26 t6 + t26 )x2 + t36 x3
dn xn =
m + (s4 s6 + t4 − mt6 )x − t4 t6 x2
,
1 − (s26 + t6 )x − (s26 t6 + t26 )x2 + t36 x3
en xn =
n + (s5 s6 + t5 − nt6 )x − t5 t6 x2
1 − (s26 + t6 )x − (s26 t6 + t26 )x2 + t36 x3
n≥0
X
n≥0
X
n≥0
X
n≥0
X
n≥0
then
a4n + b4n + c4n = d4n + e4n + (−t6 )4n .
Proof. The proof of this theorem is similar to the proof of Theorem 2.1.
6. Search for fourth power algebraic identities
We will attempt to find particular integer constants involving m, n, and all the s’s
and t’s which satisfy equation (5.1) with the following procedure.
Procedure to search for fourth power algebraic identities
1. Pick one particular set of integers m and n.
2. Select a collection of sets of integers t1 , t2 , t3 , t4 , t5 , and t6 = ±1 such that
t41 + t42 + t43 = t44 + t45 + 1. Also, select a range of integer values for s1 , s2 , s3 ,
and s4 to search.
48
C. Cooper
a. For each t1 , t2 , t3 , t4 , t5 , t6 , s1 , s2 , s3 , and s4 , compute s5 and s6 using
the equations
s5 =
s1 t31 + s2 t32 + s3 t33 − s4 t34 − s1 t36 − m3 s2 t36 − n3 s3 t36 + m3 s4 t36
,
n3 t36 + t35
s6 = −s1 − m3 s2 − n3 s3 + m3 s4 + n3 s5 .
Make sure these constants can be computed and that they are integers.
b. Check the following conditions.
4t31 + 6s21 t21 + 4mt32 + 6s22 t22 + 4nt33 + 6s23 t23
= 4mt34 + 6s24 t24 + 4nt35 + 6s25 t25 − 4t36 + 6s26 t26 ,
12s1 t21 + 4s31 t1 + 12ms2 t22 + 4s32 t2 + 12ns3 t23 + 4s33 t3
= 12ms4 t24 + 4s34 t4 + 12ns5 t25 + 4s35 t5 − 12s6 t26 + 4s36 t6 ,
6t21 + 12s21 t1 + s41 + 6m2 t22 + 12ms22 t2 + s42 + 6n2 t23 + 12ns23 t3 + s43
= 6m2 t24 + 12ms24 t4 + s44 + 6n2 t25 + 12ns25 t5 + s45 + 6t26 − 12s26 t6 + s46 ,
12s1 t1 + 4s31 + 12m2 s2 t2 + 4ms32 + 12n2 s3 t3 + 4ns33
= 12m2 s4 t4 + 4ms34 + 12n2 s5 t5 + 4ns35 + 12s6 t6 − 4s36 ,
4t1 + 6s21 + 4m3 t2 + 6m2 s22 + 4n3 t3 + 6n2 s23
= 4m3 t4 + 6m2 s24 + 4n3 t5 + 6n2 s25 − 4t6 + 6s26 .
c. If all the above conditions are satisfied (every equation is true), the
resulting collection of m, n, s’s, and t’s form an algebraic identity satisfying equation (5.1).
The proof that this procedure yields an algebraic identity is similar to the previous
procedure.
We need to make a couple of remarks. First of all, we pick positive integers
m and n with m < n. Again, in the spirit of Ramanujan, we assume t6 = ±1.
We first note that once a solution is found, we have many other similar solutions
since every one of the t’s could be positive or negative. We list out the nontrivial
values of the t’s (1 < t1 < t2 < t3 and t1 ≤ t4 ) in Appendix III. This appendix was
constructed with the help of a C++ program. Some typical ranges for s1 , s2 , s3 ,
and s4 were from −20 to 20. Finally, we require that s6 > 0.
7. Fourth power results
We found the following results. The constants in each row of the following table
satisfy (5.1). Again, we include the leading coefficient of 1 in the last trinomial.
Recall that the form of the last trinomial is x2 − s6 xy − t6 y 2 .
49
Identities in the spirit of Ramanujan’s amazing identity
m = 1 and n = 2
1, s1 , t1
1,-4,4
1,-3,4
1,-1,4
1,-4,5
1,0,5
1,-4,5
1,-5,6
1,-4,6
1,0,6
1,-6,7
1,-4,7
1,-4,7
1,-7,8
1,-5,8
1,-3,8
1,-6,8
1, s2 , t2
1,-6,9
1.-8,9
1,-2,9
1,-6,6
1,-2,6
1,-5,6
1,-10,23
1,-12,23
1,-4,23
1,-7,14
1,-12,14
1,0,14
1,-6,11
1,-10,11
1,0,11
1,-6,11
2, s3 , t3
2,-10,13
2,-11,13
2,-3,13
2,-10,11
2,-2,11
2,-9,11
2,-15,29
2,-16,29
2,-4,29
2,-13,21
2,-16,21
2,-4,21
2,-13,19
2,-15,19
2,-3,19
2,-12,19
1, s4 , t4
1,-7,11
1,-9,12
1,7,-12
1,-7,9
1,7,-9
1,-7,10
1,-11,26
1,-13,27
1,11,-27
1,-9,18
1,-13,19
1,9,-19
1,-9,16
1,-11,16
1,9,-16
1,-9,17
2, s5 , t5
2,-10,12
2,-11,11
2,-3,-11
2,-10,10
2,-2,-10
2,-9,9
2,-15,27
2,-16,26
2,-4,-26
2,-13,19
2,-16,18
2,-4,-18
2,-13,17
2,-15,17
2,-3,-17
2,-12,16
1, s6 , t6
1,3,-1
1,2,1
1,10,-1
1,3,-1
1,9,1
1,2,1
1,4,-1
1,3,1
1,15,-1
1,4,-1
1,3,1
1,13,-1
1,4,-1
1,4,-1
1,12,1
1,3,1
m = 2 and n = 3
1, s1 , t1
1,-1,7
1,-8,8
1,0,8
1,-7,8
1,-8,10
1,-4,10
1,-3,11
1,0,11
1,-8,13
1,4,-13
1,-1,14
1,-8,15
1,-5,16
1,-6,19
1,-2,21
1,14,-116
2, s2 , t2
2,-2,14
2,-10,11
2,-2,11
2,-9,11
2,-12,19
2,0,19
2,0,16
2,-4,39
2,-10,13
2,0,-13
2,-3,41
2,-12,19
2,-1,55
2,0,57
2,-6,64
2,0,-155
3, s3 , t3
3,-3,21
3,-18,19
3,-2,19
3,-16,19
3,-20,29
3,-4,29
3,-3,27
3,-4,50
3,-18,26
3,4,-26
3,-4,55
3,-20,34
3,-6,71
3,-6,76
3,10,-113
3,14,-271
2, s4 , t4
2,10,-19
2,-14,16
2,10,-16
2,-13,17
2,-16,26
2,12,-26
2,12,-23
2,16,-46
2,-14,22
2,4,-22
2,17,-49
2,-16,30
2,19,-65
2,20,-68
2,10,-112
2,12,-236
3, s5 , t5
3,-6,-18
3,-17,17
3,-5,-17
3,-15,16
3,-19,25
3,-7,-25
3,-6,-24
3,-9,-45
3,-17,23
3,3,-23
3,-9,-50
3,-19,29
3,-11,-64
3,-11,-69
3,6,-69
3,11,-235
1, s6 , t6
1,16,-1
1,3,-1
1,15,1
1,2,1
1,3,1
1,19,-1
1,18,1
1,25,-1
1,3,-1
1,1,1
1,26,1
1,3,1
1,30,-1
1,31,1
1,22,-1
1,1,-1
m = 3 and n = 5
1, s1 , t1
1,-2,21
3, s2 , t2
3,-4,41
5, s3 , t3
5,6,-71
3, s4 , t4
3,6,-69
5, s5 , t5
5,4,-49
1, s6 , t6
1,22,-1
50
C. Cooper
The bounds on s1 , s2 , s3 , and s4 varied depending on the speed of the search. The
first row of the table for m = 1 and n = 2 gives the algebraic identity
(x2 − 4xy + 4y 2 )4 + (x2 − 6xy + 9y 2 )4 + (2x2 − 10xy + 13y 2 )4
= (x2 − 7xy + 11y 2 )4 + (2x2 − 10xy + 12y 2 )4 + (x2 − 3xy + y 2 )4 .
This produces the following Ramanujan-like identity. If
X
an xn =
1 − 7x + 4x2
,
1 − 8x + 8x2 − x3
bn x n =
1 − 8x + 9x2
,
1 − 8x + 8x2 − x3
cn x n =
2 − 15x + 13x2
,
1 − 8x + 8x2 − x3
dn xn =
1 − 9x + 11x2
,
1 − 8x + 8x2 − x3
en xn =
2 − 16x + 12x2
,
1 − 8x + 8x2 − x3
n≥0
X
n≥0
X
n≥0
X
n≥0
X
n≥0
then
a4n + b4n + c4n = d4n + e4n + 1.
The row in the table for m = 3 and n = 5 gives the algebraic identity
(x2 − 2xy + 21y 2 )4 + (3x2 − 4xy + 41y 2 )4 + (5x2 + 6xy − 71y 2 )4
= (3x2 + 6xy − 69y 2 )4 + (5x2 + 4xy − 49y 2 )4 + (x2 − 22xy + y 2 )4 .
This produces the following Ramanujan-like identity. If
X
an xn =
1 − 22x + 21x2
,
1 − 483x + 483x2 − x3
bn x n =
3 − 44x + 41x2
,
1 − 483x + 483x2 − x3
cn xn =
5 + 66x + 71x2
,
1 − 483x + 483x2 − x3
dn xn =
3 + 66x + 69x2
,
1 − 483x + 483x2 − x3
en xn =
5 + 44x + 49x2
,
1 − 483x + 483x2 − x3
n≥0
X
n≥0
X
n≥0
X
n≥0
X
n≥0
Identities in the spirit of Ramanujan’s amazing identity
then
51
a4n + b4n + c4n = d4n + e4n + 1.
8. Questions
The previous data suggests several questions.
1. In the third power case, we were unable to find any nontrivial algebraic
identities like (2.1) with r1 = 1 and r2 = r3 = r where r ≥ 3. We would like
to know if any exist and if so, what are they?
2. We were unable to find any fourth power algebraic identities of the form
(r1 x2 + s1 xy + t1 y 2 )4 + (r2 x2 + s2 xy + t2 y 2 )4 + (r3 x2 + s3 xy + t3 y 2 )4
= (r4 x2 + s4 xy + t4 y 2 )4 + (x2 − s5 xy − t5 y 2 )4 ,
where the r’s are positive integers and the s’s and t’s are nontrivial. Do such
identities exist?
3. In the fourth power case, we found algebraic identities for every pair we tried
where m if a positive integer and n = m + 1. Is this always true? In addition,
is there any other algebraic identity where n 6= m + 1 other than the one we
found where m = 3 and n = 5?
52
C. Cooper
Appendix I: r13 + r23 = r33 + 1
r1
9
64
73
135
244
334
368
577
1010
1033
1126
1945
3088
3097
3753
3987
4083
4609
5700
5856
6562
7364
9001
10876
11161
11767
11980
13294
15553
16617
r2
10
94
144
235
729
438
1537
2304
1897
1738
5625
11664
21609
3518
4528
9735
8343
36864
38782
9036
59049
83692
90000
31180
11468
41167
131769
19386
186624
35442
r3
12
103
150
249
738
495
1544
2316
1988
1852
5640
11682
21630
4184
5262
9953
8657
36888
38823
9791
59076
83711
90030
31615
14258
41485
131802
21279
186660
36620
r1
19774
20848
24697
26914
27238
27784
27835
30376
35131
36865
38305
39892
44218
49193
50313
59728
65601
99457
107258
135097
158967
190243
191709
198550
243876
294121
336820
372106
434905
590896
r2
257049
152953
345744
44521
33412
35385
72629
455625
76903
589824
51762
151118
751689
50920
80020
182458
67402
222574
278722
439312
312915
219589
579621
713337
547705
325842
583918
444297
780232
734217
r3
257088
153082
345786
47584
38599
40362
73967
455670
79273
589872
57978
152039
751740
63086
86166
184567
83802
229006
283919
443530
326033
259495
586529
718428
563370
391572
619111
518292
822898
844422
53
Identities in the spirit of Ramanujan’s amazing identity
Appendix II: t31 + t32 = t33 − t34
t1
-9
6
-9
8
-8
-6
-8
9
-6
9
6
8
-12
9
-12
10
-10
-9
-10
12
-9
12
9
10
-103
64
-103
94
-94
-64
-94
103
-64
103
64
94
t2
6
-9
8
-9
-6
-8
9
-8
9
-6
8
6
9
-12
10
-12
-9
-10
12
-10
12
-9
10
9
64
-103
94
-103
-64
-94
103
-94
103
-64
94
64
t3
-8
-8
-6
-6
-9
-9
6
6
8
8
9
9
-10
-10
-9
-9
-12
-12
9
9
10
10
12
12
-94
-94
-64
-64
-103
-103
64
64
94
94
103
103
t4
1
1
1
1
-1
-1
-1
-1
-1
-1
1
1
-1
-1
-1
-1
1
1
1
1
1
1
-1
-1
-1
-1
-1
-1
1
1
1
1
1
1
-1
-1
t1
-144
71
-144
138
-138
-71
-138
144
-71
144
71
138
-150
73
-150
144
-144
-73
-144
150
-73
150
73
144
-172
135
-172
138
-138
-135
-138
172
-135
172
135
138
t2
71
-144
138
-144
-71
-138
144
-138
144
-71
138
71
73
-150
144
-150
-73
-144
150
-144
150
-73
144
73
135
-172
138
-172
-135
-138
172
-138
172
-135
138
135
t3
-138
-138
-71
-71
-144
-144
71
71
138
138
144
144
-144
-144
-73
-73
-150
-150
73
73
144
144
150
150
-138
-138
-135
-135
-172
-172
135
135
138
138
172
172
t4
1
1
1
1
-1
-1
-1
-1
-1
-1
1
1
-1
-1
-1
-1
1
1
1
1
1
1
-1
-1
1
1
1
1
-1
-1
-1
-1
-1
-1
1
1
54
C. Cooper
Appendix III: t41 + t42 + t43 = t44 + t45 + 1
t1
2
2
2
2
2
2
2
2
3
3
3
3
3
3
3
3
3
3
4
4
4
4
4
4
4
4
4
4
4
4
5
5
6
6
6
6
6
6
t2
31
31
35
35
47
47
148
148
6
6
7
7
7
7
21
21
24
24
9
9
18
18
41
41
49
49
76
76
83
83
6
6
14
14
19
19
23
23
t3
47
47
47
47
173
173
191
191
21
21
8
8
44
44
36
36
111
111
13
13
19
19
103
103
75
75
105
105
100
100
11
11
37
37
31
31
29
29
t4
14
49
19
50
71
172
56
206
16
19
2
9
24
43
2
37
77
104
11
12
6
22
58
101
25
78
54
110
32
110
9
10
22
36
9
32
26
27
t5
49
14
50
19
172
71
206
56
19
16
9
2
43
24
37
2
104
77
12
11
22
6
101
58
78
25
110
54
110
32
10
9
36
22
32
9
27
26
t1
6
6
6
6
6
6
6
6
6
6
7
7
7
7
7
7
7
7
7
7
8
8
8
8
8
8
9
9
9
9
9
9
10
10
10
10
10
10
t2
25
25
29
29
31
31
47
47
138
138
14
14
27
27
57
57
76
76
109
109
11
11
43
43
109
109
25
25
34
34
197
197
14
14
19
19
39
39
t3
29
29
47
47
41
41
71
71
165
165
21
21
157
157
73
73
107
107
148
148
19
19
51
51
132
132
34
34
193
193
200
200
103
103
29
29
41
41
t4
15
32
23
48
24
43
43
72
100
178
18
19
109
147
9
79
83
104
121
142
16
17
47
48
62
144
30
31
152
171
45
236
80
92
25
26
32
45
t5
32
15
48
23
43
24
72
43
178
100
19
18
147
109
79
9
104
83
142
121
17
16
48
47
144
62
31
30
171
152
236
45
92
80
26
25
45
32
Identities in the spirit of Ramanujan’s amazing identity
55
References
[1] Chen, K.-W., Extensions of an amazing identity of Ramanujan, The Fibonacci Quarterly, Vol. 50 (2012), 227–230.
[2] Hirschhorn, M. D., An amazing identity of Ramanujan, Mathematics Magazine,
Vol. 68 (1995), 199–201.
[3] McLaughlin, J., An identity motivated by an amazing identity of Ramanujan, The
Fibonacci Quarterly, Vol. 48 (2010), 34–38.
[4] Ramanujan, S., The Lost Notebook and Other Unpublished Papers, New Delhi,
Narosa, 1988, p. 341.