# Equal Sums of Like Powers

## Algebraic Identities

( The Generalization of Ramanujan's 6-10-8 Identity )

Identities for ( h = 1, 2, 4 )

Let   Fn = ( a + b + c )n + ( b + c + d )n -( c + d + a )n -( d + a + b )n +( a - d )n -( b - c )n

64 F6 F10 = 45 F82                                              ( Ramanujan )

25 F3 F7 = 21 F52                                                ( Hirschhorn )

5 F3 F8 = 8 F5 F6                                                 ( Chamberland )

16 F6 F7 = 7 F3 F10                                            ( Chamberland )

F-2 F32 = -3 F-12 F6                                            ( Chamberland )

245 F3 F11 + 330 F72 = 539 F5 F9                   ( Chamberland )

300 F6 F14 + 308 F102 = 525 F8 F12               ( Chamberland )

• Simplified equivalent expressions of the above identities were found by Chen Shuwen as below.

Let   Pn = (a1n + a2n + a3n - b1n - b2n - b3n ) / n

if  P1 = P2 = P4 = 0 , then

4 P6 P10 = 3 P82

P3 P7 = P52

P3 P8 = 2 P5 P6

3 P6 P7 = P3 P10

P-2 P32 = -P-12 P6

P3 P11 + 2 P72 = 3 P5 P9

9 P6 P14 + 11 P102 = 18 P8 P12

• More interesting results have been obtained by Chen Shuwen in 2017.

Let   Pn = (a1n + a2n + a3n - b1n - b2n - b3n ) / n , and P0 = 2 ( a1 a2 a3 - b1 b2 b3 ) / (a1 a2 a3 + b1 b2 b3 )

if  P1 = P2 = P4 = 0 , then

P5 / P3 = P7 / P5 = P8 / 2P6 = 2P10 / 3P8 = ( P7 / P3 )1/2 = ( P10 / 3P6 )1/2 = (a12 + a22 + a32 ) / 2

P8 / 2P5 = P3 / P0 = P6 / P3 = P10 / 3P7 = (a1 a2 a3 + b1 b2 b3 ) / 2

3P52 / P10 = P32 / P6 = -P-12 / P-2 = P0 = 2 * (a1 a2 a3 - b1 b2 b3 ) / (a1 a2 a3 + b1 b2 b3 )

and

P3 = a1 a2 a3 - b1 b2 b3

P5 = ( a1 a2 a3 - b1 b2 b3 ) (a12 + a22 + a32 ) / 2

P6 = ( a12 a22 a32 - b12 b22 b32 ) / 2

P7 = ( a1 a2 a3 - b1 b2 b3 ) (a12 + a22 + a32 )2 / 4

P8 = ( a12 a22 a32 - b12 b22 b32 ) (a12 + a22 + a32 ) / 2

P10 = 3( a12 a22 a32 - b12 b22 b32 ) (a12 + a22 + a32 )2 / 8

Identities for ( h = 1, 2, 4 ) , ( h = 0, 1, 3 ) , ( h = -1, 0, 2 ) , ( h = -2, -1, 1 )

• Chen Shuwen found and proved the following identities in 2017.

Let   Pn = (a1n + a2n + a3n - b1n - b2n - b3n ) / n , and P0 = 2 ( a1 a2 a3 - b1 b2 b3 ) / (a1 a2 a3 + b1 b2 b3 )

• if  P= P= P4 = 0 , then   P3 P8 = 2 P5 P6

if  P= P= P3 = 0 , then   P2 P7 = 2 P4 P5

if  P-1 = P= P2 = 0 , then   P1 P6 = 2 P3 P4 + P13 P4 / 12

if  P-2 = P-1 = P1 = 0 , then   P0 P5 = 2 P2 P

• if  P= P= P4 = 0 , then   P3 P10 = 3 P6 P7

if  P= P= P3 = 0 , then   P2 P= 3 P5 P6

if  P-1 = P= P2 = 0 , then   P1 P= 3 P4 P5 + P12 P3 P4 / 4 + P15 P4 / 240

if  P-2 = P-1 = P1 = 0 , then   P0 P= 3 P3 P4 + P02 P3 P4 / 4

• if  P= P= P4 = 0 , then   P32  =  P0 P6

if  P= P= P3 = 0 , then   P22  = -P-1 P5

if  P-1 = P= P2 = 0 , then   P12  = -P-2 P4

if  P-2 = P-1 = P1 = 0 , then   P02  = -P-3 P3 + P02 P3 P-3 / 4

• if  P= P= P4 = 0 , then   P-12  = -P0 P-2

if  P= P= P3 = 0 , then   P-22  = P-1 P-3 - P-14 / 12

if  P-1 = P= P2 = 0 , then   P-32  = P-2 P-4

if  P-2 = P-1 = P1 = 0 , then   P-42  = P-3 P-5

• if  P-2 = P-1 = P1 = 0 , then   P22  = P0 P4

• if  P-2 = P-1 = P1 = 0 , then   P-3 P2 = -P-4 P3

Identities for ( h = 1, 3, 4 ) , ( h = 0, 2, 3 ) , ( h = -1, 1, 2 ) , ( h = -2, 0, 1 )

• Chen Shuwen found and proved the following identities in 2017.

Let   Pn = (a1n + a2n + a3n - b1n - b2n - b3n ) / n , and P0 = 2 ( a1 a2 a3 - b1 b2 b3 ) / (a1 a2 a3 + b1 b2 b3 )

• if  P= P3 = P4 = 0 , then   P5 P7 = ( P6 - P23 / 12 )2

if  P= P2 = P3 = 0 , then   P4 P6 = ( P5 + P15 / 720 )2 + P12 P42 / 12

• if  P= P3 = P4 = 0 , then   P22 = - P-1 P5 ( 1 - P02 / 4 )

if  P= P2 = P3 = 0 , then   P12 = - P-2 P4 ( 12 / ( P-1 P1 - 12 ) )2

if  P-1 = P1 = P2 = 0 , then   P02 = - P-3 P3 ( 1 - P02 / 4 )

• if  P1 = P3 = P4 = 0 , then   P24 P7 = P02 P53
• if  P-1 = P1 = P2 = 0 , then   P-22  = -P-4 P0
• if  P-1 = P1 = P2 = 0 , then   P-5 P= -2 P-2 P-3
• if  P-1 = P1 = P2 = 0 , then   P-3 P= -P-2 P3

Identities for ( h = 2, 3, 4 ) , ( h = 1, 2, 6 )

• Chen Shuwen found and proved the following identities in 2017.

Let   Pn = (a1n + a2n + a3n - b1n - b2n - b3n ) / n , and P0 = 2 ( a1 a2 a3 - b1 b2 b3 ) / (a1 a2 a3 + b1 b2 b3 )

• if  P2 = P3 = P4 = 0 , then    P0 P6 = ( P13 / 60 - 12 P5 / P12 )2

• if  P1 = P2 = P6 = 0 , then    P34 = P0 P4 ( P42 - 2 P3 P5 )

Identities for ( h = 1, 2, 3, 5 ) , ( h = 0, 1, 2, 4 ) , ( h = -1, 0, 1, 3 ) , ( h = -2, -1, 0, 2 ) ,( h = -3, -2, -1, 1 )

• Chen Shuwen found and proved the following identities in 2017.

Let   Pn = (a1n + a2n + a3n + a4n - b1n - b2n - b3n - b4n ) / n , and P0 = 2 ( a1 a2 a3 a4 - b1 b2 b3 b4 ) / (a1 a2 a3 a4 + b1 b2 b3 b4 )

• if  P= P= P3  = P5 = 0 , then   P4 P9 = 2 P6 P7

if  P= P= P2  = P4 = 0 , then   P3 P8 = 2 P5 P6

if  P-1 = P= P1  = P3 = 0 , then   P2 P7 = 2 P4 P5

if  P-2 = P-1 = P0  = P2 = 0 , then   P1 P6 = 2 P3 P4 + P13 P4 / 12

if  P-3 = P-2 = P-1 = P1 = 0 , then   P0 P5 = 2 P2 P3

• if  P= P= P3  = P5 = 0 , then   P43 = P0 ( P4 P8 - P62 )

if  P= P= P2  = P4 = 0 , then   P33 = P-1 ( P3 P7 - P52 )

if  P-1 = P= P1  = P3 = 0 , then   P23 = P-2 ( P2 P6 - P42 - P24 / 12 )

if  P-2 = P-1 = P0  = P2 = 0 , then   P13 = P-3 ( P1 P5 - P32 - P13 P3 / 12 + P16 / 720 )

if  P-3 = P-2 = P-1 = P1 = 0 , then   P03 = P-4 ( P0 P4 - P22 ) ( 1 - P02 / 4 )

• if  P= P= P3  = P5 = 0 , then   P7 P02  = -P42 P-1 ( 1 - P02 / 4 )

if  P= P= P2  = P4 = 0 , then   P6 P-12 = -P32 P-2

if  P-1 = P= P1  = P3 = 0 , then   P5 P-22 = -P22 P-3

if  P-2 = P-1 = P0  = P2 = 0 , then   P4 P-32 = -P12 P-4

if  P-3 = P-2 = P-1 = P1 = 0 , then   P3 P-42 = -P02 P-5  / ( 1 - P02 / 4 )

Identities for ( h = 1, 2, 4, 6 ) , ( h = 0, 1, 3, 5 ) , ( h = -1, 0, 2, 4 ) and ( h = -2, -1, 1, 2 )

• Chen Shuwen found and proved the following identities in 2017 - 2019.

Let   Pn = (a1n + a2n + a3n + a4n - b1n - b2n - b3n - b4n ) / n , and P0 = 2 ( a1 a2 a3 a4 - b1 b2 b3 b4 ) / (a1 a2 a3 a4 + b1 b2 b3 b4 )

• if  P= P= P4  = P6 = 0 , then    P34 P8  P0 ( P3 P7 - P52 )2
• if  P= P= P3  = P5 = 0 , then    P24 P7 = -P-1 ( P2 P6 - P42 - P24 / 12 )2
• if  P-1 = P= P2  = P4 = 0 , then    P14 P6 = -P-2 ( P1 P5 - P32 - P13 P3/ 12 + P16/ 720 )2

• if  P-2 = P-1  = P1  = P2 = 0 , then

P-3 P-5 / P-4 = P3 P5 / P4        ( Perfect !!! )

P6 = P52 / P4  + P32 / P0

P7 = P53 / P42 + 2 P3 P4 / P0

P8 = P54 / P43 + 2 P3 P5 / P0 + P42 / P0

Identities for ( k = 1, 3 ) , ( k = 1, 2, 4 ) , ( k = 1, 2, 3, 5 ) and ( k = 1, 2, 3, 4, 6 )

• Tito Piezas III gave the below results.
• Define   Fn = a1n + a2n + a3n - b1n - b2n - b3n

if  F2 = F6 = 0 , then

F42 F10 = 20 ∏i3=1j3=1( ai2 - bj2 )

• By referring to Piezas's above results, Chen Shuwen found and proved the following simplified identities in 2017.
• Let   Rn = (a1n + a2n + a3n - b1n - b2n - b3n ) / n

if  R1 = R3 = 0 , then   R22 R5 = ∏i3=1j3=1( ai - bj )

• Let   Rn = (a1n + a2n + a3n + a4n - b1n - b2n - b3n - b4n ) / n

if  R1 = R= R4 = 0 , then   R32 ( R52 - R3 R7 ) = ∏i4=1j4=1( ai - bj )

• Let   Rn = (a1n + a2n + a3n + a4n + a5n - b1n - b2n - b3n - b4n - b5n ) / n

if  R1 = R= R3 = R5 = 0 , then   R43 ( R4 R9 - 2 R6 R7 ) = ∏i5=1j5=1( ai - bj )

• Let   Rn = (a1n + a2n + a3n + a4n + a5+ a6n - b1n - b2n - b3n - b4n - b5n - b6n ) / n

if  R1 = R= R3 = R4 = R6 = 0 , then   R53 ( R5 R82 + 2 R5 R7 R9 - R11 R52 - R73 ) = ∏i6=1j6=1( ai - bj )

Identities for ( k = 1, 2 ) , ( k = 1, 3 ) , ( k = 1, 4 ) and ( k = 1, 5 )

• Let   Rn = (a1n + a2n + a3n - b1n - b2n - b3n ) / n ，Chen Shuwen found and proved that,

R1 S + R33 - 2R2 R3 R4 + R22 R5 = ∏i3=1j3=1( ai - bj )

where S is a polynomial containing 9 items.

• This identity can be simplified as below,

if  R1 = R= 0 , then   R33 = ∏i3=1j3=1( ai - bj )

if  R1 = R= 0 , then   R22 R5 = ∏i3=1j3=1( ai - bj )

if  R1 = R= 0 , then   R33 + R22 R5 = ∏i3=1j3=1( ai - bj )

if  R1 = R= 0 , then   R33 - 2R2 R3 R4 = ∏i3=1j3=1( ai - bj )

Identities for ( k = 1, 2 ) and ( k = 0, 1 ) , ( k = 1, 2, 3 ) and ( k = 0, 1, 2 ) , ( k = 1, 2, 3, 4 ) and ( k = 1, 2, 3, 4, 5 )

• Tito Piezas III gave the below results. ( Reference [R5], [R9], [R10] )
• Define   Fn = a1n + a2n + a3n - b1n - b2n - b3n

if  F2 = F4 = 0 , then

32 F6 F10 = 15 ( m + 1 ) F82 , where m = ( a14 + a24 + a34 ) / ( a12 + a22 + a32 )2

3 F8 = 4 F6  ( a12 + a22 + a32 )

• Define   Fn = a1n + a2n + a3n + a4n - b1n - b2n - b3n - b4n ,and F0 = a1 a2 a3 a4 - b1 b2 b3 b4

if  F0 = F= F4 = 0 , then

32 F6 F10 = 15 ( m + 1 ) F82 , where m = ( a14 + a24 + a34 + a44 ) / ( a12 + a22 + a32 + a42 )2

• Define   Fn = a1n + a2n + a3n + a4n - b1n - b2n - b3n - b4n

if  F2 = F= F6 = 0 , then

25 F8 F12 = 12 ( m + 1 ) F102 , where m = ( a14 + a24 + a34 + a44 ) / ( a12 + a22 + a32 + a42 )2

4 F10 = 5 F8  ( a12 + a22 + a32 + a42 )

• Define   Fn = a1n + a2n + a3n + a4n  + a5n - b1n - b2n - b3n - b4n - b5n

if  F2 = F= F6 = F8 = 0 , then

72 F10 F14 = 35 ( m + 1 ) F122 , where m = ( a14 + a24 + a34 + a44 + a54 ) / ( a12 + a22 + a32 + a42 + a52 )2

5 F12 = 6 F10  ( a12 + a22 + a32 + a42 + a52 )

• By referring to Piezas's above results, Chen Shuwen found and proved the following simplified identities in 2017.
• Let   Rn = (a1n + a2n + a3n - b1n - b2n - b3n ) / n , and R0 = 2 ( a1 a2 a3 - b1 b2 b3 ) / (a1 a2 a3 + b1 b2 b3 )

if  R0 = R1 = 0 , then

R2 R4 / R32 + R23 / 2R32 = ( m + 1 ) / 2 , where m = ( a12 + a22 + a32 ) / ( a1 + a2 + a3 )2

R3 / R2 = a1 + a2 + a3

R2 / R-1 = - a1 a2 a3

if  R1 = R2 = 0 , then (Piezas)

R3 R5 / R42 = ( m + 1 ) / 2 , where m = ( a12 + a22 + a32 ) / ( a1 + a2 + a3 )2

R4 / R3 = a1 + a2 + a3

R3 / R0 = ( a1 a2 a3 + b1 b2 b3 ) / 2                               (Chen)

• Let   Rn = (a1n + a2n + a3n + a4n - b1n - b2n - b3n - b4n ) / n , and R0 = 2 ( a1 a2 a3 a4 - b1 b2 b3 b4 ) / (a1 a2 a3 a4 + b1 b2 b3 b4 )

if  R0 = R= R2 = 0 , then

R3 R5 / R42 = ( m + 1 ) / 2 , where m = ( a12 + a22 + a32 + a42 ) / ( a1 + a2 + a3 + a4 )2

R4 / R3 = a1 + a2 + a3 + a4

R3 / R-1 = - a1 a2 a3 a4

R-2 / R-1 - R-1 / 2 = 1 / a1 + 1 / a2 + 1 / a3 + 1 / a4

if  R1 = R= R3 = 0 , then   (Piezas)

R4 R6 / R52 = ( m + 1 ) / 2 , where m = ( a12 + a22 + a32 + a42 ) / (a1 + a2 + a3 + a4 )2

R5 / R4 = a1 + a2 + a3 + a4

R4 ( 1 + R0 / 2 ) / R0 = - a1 a2 a3 a4                              (Chen)

R-1 / R0 - R-1 / 2 = 1 / a1 + 1 / a2 + 1 / a3 + 1 / a4        (Chen)

• Let   Rn = (a1n + a2n + a3n + a4n + a5n - b1n - b2n - b3n - b4n - b5n ) / n

if  R1 = R= R3 = R4 = 0 , then     (Piezas)

R5 R7 / R62 = ( m + 1 ) / 2 , where m = ( a12 + a22 + a32 + a42 + a52 ) / ( a1 + a2 + a3 + a4 + a5 )2

• Let   Rn = (a1n + a2n + a3n + a4n + a5+ a6n - b1n - b2n - b3n - b4n - b5n - b6n ) / n

if  R1 = R= R3 = R4 = R5 = 0 , then     (Piezas)

R6 R8 / R72 = ( m + 1 ) / 2 , where m = ( a12 + a22 + a32 + a42 + a52 + a62 ) / ( a1 + a2 + a3 + a4 + a5 + a6 )2