`Equal Sums of Like Powers`

Algebraic Identities

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

Identities for ( h = 1, 2, 4 )

We start with the following Ramanujan 6-10-8 Identity and Hirschhorn 3-7-5 Identity ,

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

and ad = bc, then

64 F₆F₁₀= 45 F₈²(Ramanujan )

25 F₃F₇ = 21 F₅²(Hirschhorn )

5 F₃F₈= 8 F₅F₆(Chamberland )

16 F₆ F₇= 7 F₃ F₁₀( Chamberland )

F_-2F₃²= -3 F_-1² F₆( Chamberland )

245 F₃ F₁₁+ 330 F₇²= 539 F₅F₉( Chamberland )

300 F₆ F₁₄+ 308 F₁₀²= 525 F₈F₁₂( Chamberland )

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

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

if P₁= P₂= P₄= 0 , then

4 P₆P₁₀= 3 P₈²

P₃P₇= P₅²

P₃P₈= 2 P₅P₆

3 P₆ P₇= P₃ P₁₀

P_-2P₃²= -P_-1² P₆

P₃ P₁₁+ 2 P₇² = 3 P₅ P₉

9 P₆ P₁₄+ 11 P₁₀²= 18 P₈P₁₂

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

Let P_n= (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 P_n= (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   P₃P₈= 2 P₅P₆
if P₀= P₁= P₃= 0 , then   P₂P₇= 2 P₄P₅

if P_-1= P₀= P₂= 0 , then   P₁P₆= 2 P₃P₄+ P₁³P₄/ 12

if P_-2= P_-1= P₁= 0 , then   P₀P₅= 2 P₂P₃

if P₁= P₂= P₄= 0 , then   P₃P₁₀= 3 P₆P₇
if P₀= P₁= P₃= 0 , then   P₂P₉= 3 P₅P₆

if P_-1= P₀= P₂= 0 , then   P₁P₈= 3 P₄P₅+ P₁²P₃P₄/ 4+ P₁⁵P₄/ 240

if P_-2= P_-1= P₁= 0 , then   P₀P₇= 3 P₃P₄+ P₀²P₃P₄/ 4

if P₁= P₂= P₄= 0 , then   P₃² = P₀P₆
if P₀= P₁= P₃= 0 , then   P₂² = -P_-1P₅

if P_-1= P₀= P₂= 0 , then   P₁² = -P_-2P₄

if P_-2= P_-1= P₁= 0 , then   P₀² = -P_-3P₃+ P₀²P₃P_-3/ 4

if P₁= P₂= P₄= 0 , then   P_-1²= -P₀P_-2
if P₀= P₁= P₃= 0 , then   P_-2²= P_-1P_-3 - P_-1⁴/ 12

if P_-1= P₀= P₂= 0 , then   P_-3²= P_-2P_-4

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

if P_-2= P_-1= P₁= 0 , then   P₂²= P₀P₄

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

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

if P₁= P₃= P₄= 0 , then   P₅P₇= ( P₆ - P₂³/ 12 )²
if P₀= P₂= P₃= 0 , then   P₄P₆= ( P₅ + P₁⁵ / 720 )² + P₁²P₄²/ 12

if P₁= P₃= P₄= 0 , then   P₂²= - P_-1P₅( 1 - P₀²/ 4 )
if P₀= P₂= P₃= 0 , then   P₁²= - P_-2P₄( 12/ ( P_-1P₁ - 12 ) )²

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

if P₁= P₃= P₄= 0 , then   P₂⁴P₇= P₀²P₅³

if P_-1= P₁= P₂= 0 , then   P_-2²= -P_-4P₀

if P_-1= P₁= P₂= 0 , then   P_-5P₀= -2 P_-2P_-3

if P_-1= P₁= P₂= 0 , then   P_-3P₄= -P_-2P₃

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

Chen Shuwen found and proved the following identities in 2017.

Let P_n= (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 P₀P₆= ( P₁³/ 60 - 12 P₅ / P₁²)²

if P₁= P₂= P₆= 0 , then P₃⁴= P₀P₄( 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 P_n= (a₁ⁿ+ a₂ⁿ+ a₃ⁿ + a₄ⁿ- b₁ⁿ- b₂ⁿ- b₃ⁿ- b₄ⁿ ) / n , and P₀= 2 ( a₁ a₂ a₃a₄- b₁ b₂ b₃ b₄ ) / (a₁ a₂ a₃a₄+ b₁ b₂ b₃ b₄ )

if P₁= P₂= P₃= P₅= 0 , then   P₄ P₉= 2 P₆P₇
if P₀= P₁= P₂= P₄= 0 , then   P₃P₈= 2 P₅P₆
if P_-1= P₀= P₁= P₃= 0 , then   P₂P₇= 2 P₄P₅
if P_-2= P_-1= P₀= P₂= 0 , then   P₁P₆= 2 P₃P₄ + P₁³P₄/ 12
if P_-3= P_-2= P_-1= P₁= 0 , then   P₀P₅= 2 P₂P₃

if P₁= P₂= P₃= P₅= 0 , then   P₄³ = P₀( P₄P₈- P₆² )
if P₀= P₁= P₂= P₄= 0 , then   P₃³= P_-1( P₃P₇- P₅² )
if P_-1= P₀= P₁= P₃= 0 , then   P₂³= P_-2( P₂P₆- P₄²- P₂⁴/ 12 )
if P_-2= P_-1= P₀= P₂= 0 , then   P₁³= P_-3( P₁P₅- P₃²- P₁³ P₃/ 12 + P₁⁶/ 720 )
if P_-3= P_-2= P_-1= P₁= 0 , then   P₀³= P_-4( P₀P₄- P₂² )( 1- P₀² / 4 )

if P₁= P₂= P₃= P₅= 0 , then   P₇ P₀²= -P₄²P_-1 ( 1- P₀² / 4 )
if P₀= P₁= P₂= P₄= 0 , then   P₆ P_-1²= -P₃²P_-2
if P_-1= P₀= P₁= P₃= 0 , then   P₅ P_-2²= -P₂²P_-3
if P_-2= P_-1= P₀= P₂= 0 , then   P₄ P_-3²= -P₁²P_-4
if P_-3= P_-2= P_-1= P₁= 0 , then   P₃ P_-4²= -P₀²P_-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 P_n= (a₁ⁿ+ a₂ⁿ+ a₃ⁿ + a₄ⁿ- b₁ⁿ- b₂ⁿ- b₃ⁿ- b₄ⁿ ) / n , and P₀= 2 ( a₁ a₂ a₃a₄- b₁ b₂ b₃ b₄ ) / (a₁ a₂ a₃a₄+ b₁ b₂ b₃ b₄ )

if P₁= P₂= P₄= P₆= 0 , then    P₃⁴ P₈= P₀( P₃P₇- P₅² )²

if P₀= P₁= P₃= P₅= 0 , then    P₂⁴ P₇= -P_-1( P₂P₆- P₄² - P₂⁴/ 12 )²

if P_-1= P₀= P₂= P₄= 0 , then    P₁⁴ P₆= -P_-2( P₁P₅- P₃² - P₁³ P₃/ 12 + P₁⁶/ 720 )²

if P_-2= P_-1= P₁= P₂= 0 , then

P_-3P_-5/ P_-4= P₃P₅/ 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 F_n= a₁ⁿ+ a₂ⁿ+ a₃ⁿ - b₁ⁿ- b₂ⁿ- b₃ⁿ
if F₂= F₆= 0 , then

F₄² F₁₀= 20 ∏_i³₌₁∏_j³₌₁( a_i²- b_j² )

By referring to Piezas's above results, Chen Shuwen found and proved the following simplified identities in 2017.

Let R_n= (a₁ⁿ+ a₂ⁿ+ a₃ⁿ - b₁ⁿ- b₂ⁿ- b₃ⁿ ) / n
if R₁= R₃= 0 , then   R₂² R₅= ∏_i³₌₁∏_j³₌₁( a_i- b_j )

Let R_n= (a₁ⁿ+ a₂ⁿ+ a₃ⁿ + a₄ⁿ- b₁ⁿ- b₂ⁿ- b₃ⁿ- b₄ⁿ ) / n
if R₁= R₂= R₄= 0 , then   R₃² ( R₅²- R₃ R₇ )= ∏_i⁴₌₁∏_j⁴₌₁( a_i- b_j )

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

if R₁= R₂= R₃= R₅= 0 , then   R₄³ ( R₄ R₉- 2 R₆ R₇ )= ∏_i⁵₌₁∏_j⁵₌₁( a_i- b_j )

Let R_n= (a₁ⁿ+ a₂ⁿ+ a₃ⁿ + a₄ⁿ+ a₅ⁿ+ a₆ⁿ- b₁ⁿ- b₂ⁿ- b₃ⁿ- b₄ⁿ - b₅ⁿ - b₆ⁿ ) / n

if R₁= R₂= R₃= R₄= R₆= 0 , then   R₅³ ( R₅ R₈²+ 2 R₅ R₇ R₉ - R₁₁ R₅²- R₇³)= ∏_i⁶₌₁∏_j⁶₌₁( a_i- b_j )

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

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

R₁ S+ R₃³ - 2R₂ R₃ R₄ + R₂² R₅= ∏_i³₌₁∏_j³₌₁( a_i- b_j )

where S is a polynomial containing 9 items.

This identity can be simplified as below,
    if R₁= R₂= 0 , then   R₃³= ∏_i³₌₁∏_j³₌₁( a_i- b_j )

    if R₁= R₃= 0 , then   R₂² R₅= ∏_i³₌₁∏_j³₌₁( a_i- b_j )

    if R₁= R₄= 0 , then   R₃³+ R₂² R₅= ∏_i³₌₁∏_j³₌₁( a_i- b_j )

    if R₁= R₅= 0 , then   R₃³ - 2R₂ R₃ R₄= ∏_i³₌₁∏_j³₌₁( a_i- b_j )

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 F_n= 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 F_n= a₁ⁿ+ a₂ⁿ+ a₃ⁿ + a₄ⁿ- b₁ⁿ- b₂ⁿ- b₃ⁿ- b₄ⁿ ,and F₀= a₁ a₂ a₃a₄- b₁ b₂ b₃ b₄
if F₀= F₂= F₄= 0 , then

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

Define F_n= a₁ⁿ+ a₂ⁿ+ a₃ⁿ + a₄ⁿ- b₁ⁿ- b₂ⁿ- b₃ⁿ- b₄ⁿ
if F₂= F₄= 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 F_n= a₁ⁿ+ a₂ⁿ+ a₃ⁿ + a₄ⁿ + a₅ⁿ- b₁ⁿ- b₂ⁿ- b₃ⁿ- b₄ⁿ - b₅ⁿ
if F₂= F₄= 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 R_n= (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 a₂a₃

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 R_n= (a₁ⁿ+ a₂ⁿ+ a₃ⁿ + a₄ⁿ- b₁ⁿ- b₂ⁿ- b₃ⁿ- b₄ⁿ ) / n , and R₀= 2 ( a₁ a₂ a₃a₄- b₁ b₂ b₃ b₄ ) / (a₁ a₂ a₃a₄+ b₁ b₂ b₃ b₄ )
if R₀= R₁= 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 a₂a₃a₄

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

if R₁= R₂= 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 a₂a₃a₄                              (Chen)

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

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

if R₁= R₂= 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 R_n= (a₁ⁿ+ a₂ⁿ+ a₃ⁿ + a₄ⁿ+ a₅ⁿ+ a₆ⁿ- b₁ⁿ- b₂ⁿ- b₃ⁿ- b₄ⁿ - b₅ⁿ - b₆ⁿ ) / n

if R₁= R₂= 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

Reference for this page

[R1] Ramanujan 6-10-8 Identity
[R2] Hirschhorn 3-7-5 Identity
[R3] Tito Piezas III, Generalizing Ramanujan's 6-10-8 Identity
[R4] Tito Piezas III, How to prove this nice 10th power identity for ......
[R5] Tito Piezas III, How did Ramanujan discover this identity?
[R6] Chamberland,M. Ramanujan's 6-8-10 Equation and Beyond , Mathematics Magazine , 2009 , 82(2):135-140
[R7] Berndt,B., Bhargava,S. A Remarkable Identity found in Ramanujan's Third Notebook. Glasgow Mathematical Journal, 34:341-345, 1992.
[R8] Chamberland,M. Using Integer Relations Algorithms for Finding Relationships among Functions, Tapas in Experimental Mathematics,
Tewodros Amdeberhan and Victor Moll eds., Contemporary Mathematics, 257:127-133, 2008.
[R9] Titus Piezas III, A Collection of Algebraic Identities, 2009
[R10] Tito Piezas III, Theorem for Equal Sums of Like Powers x1^8+x2^8+x3^8+…