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On certain functional equation related to derivations

  • Benjamin Marcen EMAIL logo and Joso Vukman
Published/Copyright: February 6, 2024

Abstract

In this article, we prove the following result. Let n 3 be some fixed integer and let R be a prime ring with char ( R ) ( n + 1 ) ! 2 n 2 . Suppose there exists an additive mapping D : R R satisfying the relation

2 n 2 D ( x n ) = i = 0 n 2 n 2 i x i D ( x 2 ) x n 2 i + ( 2 n 2 1 ) ( D ( x ) x n 1 + x n 1 D ( x ) ) + i = 1 n 2 k = 2 i ( 2 k 1 1 ) n k 2 i k + k = 2 n 1 i ( 2 k 1 1 ) n k 2 n i k 1 x i D ( x ) x n 1 i

for all x R . In this case, D is a derivation. This result is related to a classical result of Herstein, which states that any Jordan derivation on a prime ring with char ( R ) 2 is a derivation.

MSC 2010: 16W25; 16R60; 39B05

1 Introduction

Let R be an associative ring. Given an integer n > 1 , a ring R is said to be n -torsion free if, for x R , the condition n x = 0 implies x = 0 . As usual, the commutator x y y x will be denoted by [ x , y ] . A ring R is said to be prime if, for a , b R , the condition a R b = 0 implies that either a = 0 or b = 0 , and is said to be semiprime in case a R a = ( 0 ) implies a = 0 . We denote by char ( R ) , the characteristic of a prime ring R . An additive mapping D : R R is called a derivation if the equation D ( x y ) = D ( x ) y + x D ( y ) holds for all x , y R . A derivation D is inner in case there exists a R such that D ( x ) = [ a , x ] holds for all x R . An additive mapping D : R R is called a Jordan derivation in case D ( x 2 ) = D ( x ) x + x D ( x ) is fulfilled for all x R . Clearly, every derivation is a Jordan derivation. The converse is in general not true. A classical result of Herstein [1] asserts that any Jordan derivation on a prime ring with char ( R ) 2 is a derivation (see [2] for a brief proof). Cusack [3] generalized Herstein theorem to 2-torsion free semiprime rings (see [4] for an alternative proof, and [5] for a proof in the setting of algebras). Beidar et al. [6] have fairly generalized Herstein theorem. For results related to Herstein theorem, the reader is referred to [79]. Brešar [10] has proved the following result (see [11] for a generalization).

Theorem 1

Let R be a 2-torsion free semiprime ring and let D : R R be an additive mapping satisfying the relation

(1) D ( x y x ) = D ( x ) y x + x D ( y ) x + x y D ( x )

for all pairs x , y R . In this case, D is a derivation.

An additive mapping D , which maps an arbitrary ring R into itself and satisfies relation (1) for all pairs x , y R , is called a Jordan triple derivation. One can easily prove that any Jordan derivation on an arbitrary 2-torsion free ring is a Jordan triple derivation, which means that Theorem 1 generalizes Cusack’s generalization of Herstein theorem.

The above result represents a motivation for many other results (see [1214]). Vukman [15] conjectured that in case there exists an additive mapping D : R R , where R is a 2-torsion free semiprime ring, satisfying the relation

(2) 2 D ( x y x ) = D ( x y ) x + x y D ( x ) + D ( x ) y x + x D ( y x )

for all pairs x , y R , then D is a derivation. By our knowledge this conjecture is in general still an open problem. Putting x for y in relations (1) and (2) we obtain

(3) D ( x 3 ) = D ( x ) x 2 + x D ( x ) x + x 2 D ( x )

and

(4) 2 D ( x 3 ) = D ( x 2 ) x + x 2 D ( x ) + D ( x ) x 2 + x D ( x 2 ) .

Recently, Fošner et al. [8] proved the following result regarding relation (4), which is related to Vukman’s conjecture mentioned above.

Theorem 2

Let R be a prime ring with char ( R ) 2 and let D : R R be an additive mapping satisfying the relation

2 D ( x 3 ) = D ( x 2 ) x + x 2 D ( x ) + D ( x ) x 2 + x D ( x 2 )

for all x R . In this case, D is a derivation.

Any Jordan derivation can be written in the form D ( x y + y x ) = D ( x ) y + x D ( y ) + D ( y ) x + y D ( x ) , which gives, putting y = x 2 , relation (4). Therefore, one can conclude that Theorem 2 generalizes Herstein theorem. Relation (3) leads to the following functional equation:

D ( x n ) = i = 1 n x i 1 D ( x ) x n i ,

which was studied on prime rings by Beidar et al. [6].

Putting x 2 , x 3 , , x n 2 for y in (2) we obtain after some calculations (see [16] for the details) the following functional equation:

(5) 2 n 2 D ( x n ) = i = 0 n 2 n 2 i x i D ( x 2 ) x n 2 i + ( 2 n 2 1 ) ( D ( x ) x n 1 + x n 1 D ( x ) ) + i = 1 n 2 k = 2 i ( 2 k 1 1 ) n k 2 i k + k = 2 n 1 i ( 2 k 1 1 ) n k 2 n i k 1 x i D ( x ) x n 1 i ,

where n 3 is some fixed integer. It is our aim in this article to prove the following result.

Theorem 3

Let n 3 be some fixed integer and let R be a prime ring with char ( R ) ( n + 1 ) ! 2 n 2 , and let D : R R be an additive mapping satisfying the relation

(6) 2 n 2 D ( x n ) = i = 0 n 2 n 2 i x i D ( x 2 ) x n 2 i + ( 2 n 2 1 ) ( D ( x ) x n 1 + x n 1 D ( x ) ) + i = 1 n 2 k = 2 i ( 2 k 1 1 ) n k 2 i k + k = 2 n 1 i ( 2 k 1 1 ) n k 2 n i k 1 x i D ( x ) x n 1 i

for all x R . In this case, D is a derivation.

In case n = 3 , Theorem 3 reduces to Theorem 2. For functional equations related to derivations we refer to [1720] where further references can be found.

2 Main results

In the proof of Theorem 3, we shall use as the main tool the theory of functional identities (Beidar-Brešar-Chebotar theory). The theory of functional identities considers set-theoretic mappings on rings that satisfy some identical relations. When treating such relations one usually concludes that the form of the maps involved can be described, unless the ring is very special. We refer the reader to [21] for an introductory account on functional identities and to [22] for full treatment of this theory.

We would also like to mention the latest relevant article by Brešar introducing the theory of functional identities and their applications [23].

Let R be a ring and let X be a subset of R . By C ( X ) we denote the set { r R [ r , X ] = 0 } . Let m N and let E : X m 1 R , p : X m 2 R be arbitrary mappings. In the case when m = 1 , this should be understood as that E is an element in R and p = 0 . Let 1 i < j m and define E i , p i j , p j i : X m R by

E i ( x ¯ m ) = E ( x 1 , , x i 1 , x i + 1 , , x m ) , p i j ( x ¯ m ) = p j i ( x ¯ m ) = ( x 1 , , x i 1 , x i + 1 , , x j 1 , x j + 1 , , x m ) ,

where x ¯ m = ( x 1 , , x m ) X m .

Let I , J { 1 , , m } , and for each i I , j J let E i , F j : X m 1 R be arbitrary mappings. Consider functional identities

(7) i I E i i ( x ¯ m ) x i + j J x j F j j ( x ¯ m ) = 0 ( x ¯ m X m ) ,

(8) i I E i i ( x ¯ m ) x i + j J x j F j j ( x ¯ m ) C ( X ) ( x ¯ m X m ) .

A natural possibility when (7) and (8) are fulfilled is when there exist mappings p i j : X m 2 R , i I , j J , i j , and λ k : X m 1 C ( X ) , k I J , such that

(9) E i i ( x ¯ m ) = j J , j i x j p i j i j ( x ¯ m ) + λ i i ( x ¯ m ) , F j j ( x ¯ m ) = i I , j i p i j i j ( x ¯ m ) x i λ j j ( x ¯ m ) , λ k = 0 if k I J

for all x ¯ m X m , i I , j J . We shall say that every solution of the form (9) is a standard solution of (7) and (8).

The case when one of the sets I or J is empty is not excluded. The sum over the empty set of indexes should be simply read as zero. So, when J = 0 (resp. I = 0 ), (7) and (8) reduce to

(10) i I E i i ( x ¯ m ) x i = 0 resp. j J x j F j j ( x ¯ m ) = 0 ( x ¯ m X m ) ,

(11) i I E i i ( x ¯ m ) x i C ( X ) resp. j J x j F j j ( x ¯ m ) C ( X ) ( x ¯ m X m ) .

In that case the (only) standard solution is

(12) E i = 0 , i I ( resp. F j = 0 , j J ) .

A d -freeness of X will play an important role in this article. For a definition of d -freeness we refer the reader to [24]. Under some natural assumptions one can establish that various subsets (such as ideals, Lie ideals, the sets of symmetric or skew symmetric elements in a ring with involution) of certain types of rings are d -free. We refer the reader to [25] and [26] for results of this kind. Let us mention that a prime ring R is a d -free subset of its maximal right ring of quotients, unless R satisfies the standard polynomial identity of degree less than 2 d (see [26, Theorem 2.4]).

Let R be an algebra over a commutative ring ϕ and let

(13) p ( x 1 , x 2 , , x n ) = π S n x π ( 1 ) x π ( 2 ) x π ( n )

be a fixed multilinear polynomial in noncommuting indeterminates x i over ϕ . Here, S n stands for the symmetric group of order n . Let be a subset of R closed under p , i.e., p ( x ¯ n ) for all x 1 , x 2 , , x n , where x ¯ n = ( x 1 , x 2 , , x n ) . We shall consider a mapping D : R satisfying

(14) 2 n 2 D ( p ( x ¯ n ) ) = = π S n D ( x π ( 1 ) x π ( 2 ) ) x π ( 3 ) x π ( n ) + π S n x π ( 1 ) x π ( n 2 ) D ( x π ( n 1 ) x π ( n ) ) + π S n i = 1 n 3 n 2 i x π ( 1 ) x π ( i ) D ( x π ( i + 1 ) x π ( i + 2 ) ) x π ( i + 3 ) x π ( n ) + π S n ( ( 2 n 2 1 ) ( D ( x π ( 1 ) ) x π ( 2 ) x π ( n ) + x π ( 1 ) x π ( n 1 ) D ( x π ( n ) ) ) ) + π S n i = 1 n 2 k = 2 i ( 2 k 1 1 ) n k 2 i k + k = 2 n 1 i ( 2 k 1 1 ) n k 2 n i k 1 × x π ( 1 ) x π ( i ) D ( x π ( i + 1 ) ) x π ( i + 2 ) x π ( n ) )

for all x 1 , x 2 , , x n . Let us mention that the idea of considering the expression [ p ( x ¯ n ) , p ( y ¯ n ) ] in its proof is taken from [27]. For the proof of Theorem 3, we need Theorem 4, which might be of independent interest.

Theorem 4

Let be a 2 n -free Lie subring of R closed under p. If D : R is an additive mapping satisfying (14), then D is a derivation.

Proof

For any a R and x ¯ n n we have

(15) [ p ( x ¯ n ) , a ] = p ( [ x 1 , a ] , x 2 , , x n ) + p ( x 1 , [ x 2 , a ] , x 3 , , x n ) + + p ( x 1 , x 2 , , [ x n 1 , a ] , x n ) + p ( x 1 , x 2 , , x n 1 , [ x n , a ] )

and therefore

2 n 2 D ( [ p ( x ¯ n ) , a ] ) = π S n ( D ( [ x π ( 1 ) x π ( 2 ) , a ] ) x π ( 3 ) x π ( n ) + D ( x π ( 1 ) x π ( 2 ) ) [ x π ( 3 ) x π ( n ) , a ] ) + π S n ( [ x π ( 1 ) x π ( n 2 ) , a ] D ( x π ( n 1 ) x π ( n ) ) + x π ( 1 ) x π ( n 2 ) D ( [ x π ( n 1 ) x π ( n ) , a ] ) ) + π S n i = 1 n 3 n 2 i ( [ x π ( 1 ) x π ( i ) , a ] D ( x π ( i + 1 ) x π ( i + 2 ) ) x π ( i + 3 ) x π ( n ) + x π ( 1 ) x π ( i ) D ( [ x π ( i + 1 ) x π ( i + 2 ) , a ] ) x π ( i + 3 ) x π ( n ) + x π ( 1 ) x π ( i ) D ( x π ( i + 1 ) x π ( i + 2 ) ) [ x π ( i + 3 ) x π ( n ) , a ] ) ) + π S n ( ( 2 n 2 1 ) ( D ( [ x π ( 1 ) , a ] ) x π ( 2 ) x π ( n ) + x π ( 1 ) x π ( n 1 ) D ( [ x π ( n ) , a ] ) + D ( x π ( 1 ) ) [ x π ( 2 ) x π ( n ) , a ] + [ x π ( 1 ) x π ( n 1 ) , a ] D ( x π ( n ) ) ) ) + π S n i = 1 n 2 ( τ i , n ) ( [ x π ( 1 ) x π ( i ) , a ] D ( x π ( i + 1 ) ) x π ( i + 2 ) x π ( n ) + x π ( 1 ) x π ( i ) D ( [ x π ( i + 1 ) , a ] ) x π ( i + 2 ) x π ( n ) + x π ( 1 ) x π ( i ) D ( x π ( i + 1 ) ) [ x π ( i + 2 ) x π ( n ) , a ] ) ) ,

where

τ i , n = k = 2 i ( 2 k 1 1 ) n k 2 i k + k = 2 n 1 i ( 2 k 1 1 ) n k 2 n i k 1 .

In particular,

(16) 2 n 2 D ( [ p ( x ¯ n ) , p ( y ¯ n ) ] ) = π S n ( D ( [ x π ( 1 ) x π ( 2 ) , p ( y ¯ n ) ] ) x π ( 3 ) x π ( n ) + D ( x π ( 1 ) x π ( 2 ) ) [ x π ( 3 ) x π ( n ) , p ( y ¯ n ) ] ) + π S n ( [ x π ( 1 ) x π ( n 2 ) , p ( y ¯ n ) ] D ( x π ( n 1 ) x π ( n ) ) + x π ( 1 ) x π ( n 2 ) D ( [ x π ( n 1 ) x π ( n ) , p ( y ¯ n ) ] ) ) + π S n i = 1 n 3 n 2 i ( [ x π ( 1 ) x π ( i ) , p ( y ¯ n ) ] D ( x π ( i + 1 ) x π ( i + 2 ) ) x π ( i + 3 ) x π ( n ) + x π ( 1 ) x π ( i ) D ( [ x π ( i + 1 ) x π ( i + 2 ) , p ( y ¯ n ) ] ) x π ( i + 3 ) x π ( n ) + x π ( 1 ) x π ( i ) D ( x π ( i + 1 ) x π ( i + 2 ) ) [ x π ( i + 3 ) x π ( n ) , p ( y ¯ n ) ] ) ) + π S n ( ( 2 n 2 1 ) ( D ( [ x π ( 1 ) , p ( y ¯ n ) ] ) x π ( 2 ) x π ( n ) + x π ( 1 ) x π ( n 1 ) D ( [ x π ( n ) , p ( y ¯ n ) ] ) + D ( x π ( 1 ) ) [ x π ( 2 ) x π ( n ) , p ( y ¯ n ) ] + [ x π ( 1 ) x π ( n 1 ) , p ( y ¯ n ) ] D ( x π ( n ) ) ) ) + π S n i = 1 n 2 ( τ i , n ) ( [ x π ( 1 ) x π ( i ) , p ( y ¯ n ) ] D ( x π ( i + 1 ) ) x π ( i + 2 ) x π ( n ) + x π ( 1 ) x π ( i ) D ( [ x π ( i + 1 ) , p ( y ¯ n ) ] ) x π ( i + 2 ) x π ( n ) + x π ( 1 ) x π ( i ) D ( x π ( i + 1 ) ) [ x π ( i + 2 ) x π ( n ) , p ( y ¯ n ) ] ) )

for all x ¯ n , y ¯ n n .

Let us introduce the mapping f ( x π ( l ) , p ( y ¯ n ) ) = 2 n 2 D ( [ x π ( l ) , p ( y ¯ n ) ] ) = 2 n 2 D ( [ p ( y ¯ n ) , x π ( l ) ] ) for the ease of writing and presenting the results.

It is easy to verify that for l = 1 , 2 , , i 1 , i , i + 1 , , n 1 , n , we have (by (16))

(17) f ( x π ( l ) , p ( y ¯ n ) ) = 2 n 2 D ( [ x π ( l ) , p ( y ¯ n ) ] ) = 2 n 2 D ( [ p ( y ¯ n ) , x π ( l ) ] ) = σ S n ( D ( [ x π ( l ) , y σ ( 1 ) y σ ( 2 ) ] ) y σ ( 3 ) y σ ( n ) + D ( y σ ( 1 ) y σ ( 2 ) ) [ x π ( l ) , y σ ( 3 ) y σ ( n ) ] ) + σ S n ( [ x π ( l ) , y σ ( 1 ) y σ ( n 2 ) ] D ( y σ ( n 1 ) y σ ( n ) ) + y σ ( 1 ) y σ ( n 2 ) D ( [ x π ( l ) , y σ ( n 1 ) y σ ( n ) ] ) ) + σ S n i = 1 n 3 n 2 i ( [ x π ( l ) , y σ ( 1 ) y σ ( i ) ] D ( y σ ( i + 1 ) y σ ( i + 2 ) ) y σ ( i + 3 ) y σ ( n ) + y σ ( 1 ) y σ ( i ) D ( [ x π ( l ) , y σ ( i + 1 ) y σ ( i + 2 ) ] ) y σ ( i + 3 ) y σ ( n ) + y σ ( 1 ) y σ ( i ) D ( y σ ( i + 1 ) y σ ( i + 2 ) ) [ x π ( l ) , y σ ( i + 3 ) y σ ( n ) ] ) ) + σ S n ( ( 2 n 2 1 ) ( D ( [ x π ( l ) , y σ ( 1 ) ] ) y σ ( 2 ) y σ ( n ) + y σ ( 1 ) y σ ( n 1 ) D ( [ x π ( l ) , y σ ( n ) ] ) + D ( y σ ( 1 ) ) [ x π ( l ) , y σ ( 2 ) y σ ( n ) ] + [ x π ( l ) , y σ ( 1 ) y σ ( n 1 ) ] D ( y σ ( n ) ) ) ) + σ S n i = 1 n 2 ( τ i , n ) ( [ x π ( l ) , y σ ( 1 ) y σ ( i ) ] D ( y σ ( i + 1 ) ) y σ ( i + 2 ) y σ ( n ) + y σ ( 1 ) y σ ( i ) D ( [ x π ( l ) , y σ ( i + 1 ) ] ) y σ ( i + 2 ) y σ ( n ) + y σ ( 1 ) y σ ( i ) D ( y σ ( i + 1 ) ) [ x π ( l ) , y σ ( i + 2 ) y σ ( n ) ] ) )

for all x ¯ n , y ¯ n n .

In exactly the same way for l = 1 , 2 , , i 1 , i , i + 1 , , n 1 , we express f ( x π ( l ) x π ( l + 1 ) , p ( y ¯ n ) ) .

Therefore, we have

(18) f ( x π ( l ) x π ( l + 1 ) , p ( y ¯ n ) ) = 2 n 2 D ( [ x π ( l ) x π ( l + 1 ) , p ( y ¯ n ) ] ) = 2 n 2 D ( [ p ( y ¯ n ) , x π ( l ) x π ( l + 1 ) ] ) = σ S n ( D ( [ x π ( l ) x π ( l + 1 ) , y σ ( 1 ) y σ ( 2 ) ] ) y σ ( 3 ) y σ ( n ) + D ( y σ ( 1 ) y σ ( 2 ) ) [ x π ( l ) x π ( l + 1 ) , y σ ( 3 ) y σ ( n ) ] ) + σ S n ( [ x π ( l ) x π ( l + 1 ) , y σ ( 1 ) y σ ( n 2 ) ] D ( y σ ( n 1 ) y σ ( n ) ) + y σ ( 1 ) y σ ( n 2 ) D ( [ x π ( l ) x π ( l + 1 ) , y σ ( n 1 ) y σ ( n ) ] ) ) + σ S n i = 1 n 3 n 2 i ( [ x π ( l ) x π ( l + 1 ) , y σ ( 1 ) y σ ( i ) ] D ( y σ ( i + 1 ) y σ ( i + 2 ) ) y σ ( i + 3 ) y σ ( n ) + y σ ( 1 ) y σ ( i ) D ( [ x π ( l ) x π ( l + 1 ) , y σ ( i + 1 ) y σ ( i + 2 ) ] ) y σ ( i + 3 ) y σ ( n ) + y σ ( 1 ) y σ ( i ) D ( y σ ( i + 1 ) y σ ( i + 2 ) ) [ x π ( l ) x π ( l + 1 ) , y σ ( i + 3 ) y σ ( n ) ] ) ) + σ S n ( ( 2 n 2 1 ) ( D ( [ x π ( l ) x π ( l + 1 ) , y σ ( 1 ) ] ) y σ ( 2 ) y σ ( n ) + y σ ( 1 ) y σ ( n 1 ) D ( [ x π ( l ) x π ( l + 1 ) , y σ ( n ) ] ) + D ( y σ ( 1 ) ) [ x π ( l ) x π ( l + 1 ) , y σ ( 2 ) y σ ( n ) ] + [ x π ( l ) x π ( l + 1 ) , y σ ( 1 ) y σ ( n 1 ) ] D ( y σ ( n ) ) ) ) + σ S n i = 1 n 2 ( τ i , n ) ( [ x π ( l ) x π ( l + 1 ) , y σ ( 1 ) y σ ( i ) ] D ( y σ ( i + 1 ) ) y σ ( i + 2 ) y σ ( n ) + y σ ( 1 ) y σ ( i ) D ( [ x π ( l ) x π ( l + 1 ) , y σ ( i + 1 ) ] ) y σ ( i + 2 ) y σ ( n ) + y σ ( 1 ) y σ ( i ) D ( y σ ( i + 1 ) ) [ x π ( l ) x π ( l + 1 ) , y σ ( i + 2 ) y σ ( n ) ] ) )

for all x ¯ n , y ¯ n n .

Equation (16) can now be rewritten as

(19) 2 2 n 4 D ( [ p ( x ¯ n ) , p ( y ¯ n ) ] ) = π S n σ S n ( f ( x π ( 1 ) x π ( 2 ) , y σ ( 1 ) y σ ( n ) ) x π ( 3 ) x π ( n ) + 2 n 2 D ( x π ( 1 ) x π ( 2 ) ) [ x π ( 3 ) x π ( n ) , y σ ( 1 ) y σ ( n ) ] ) + π S n σ S n ( 2 n 2 [ x π ( 1 ) x π ( n 2 ) , y σ ( 1 ) y σ ( n ) ] D ( x π ( n 1 ) x π ( n ) ) + x π ( 1 ) x π ( n 2 ) f ( x π ( n 1 ) x π ( n ) , y σ ( 1 ) y σ ( n ) ) ) + π S n σ S n i = 1 n 3 n 2 i ( 2 n 2 [ x π ( 1 ) x π ( i ) , y σ ( 1 ) y σ ( n ) ] × D ( x π ( i + 1 ) x π ( i + 2 ) ) x π ( i + 3 ) x π ( n ) + x π ( 1 ) x π ( i ) f ( x π ( i + 1 ) x π ( i + 2 ) , y σ ( 1 ) y σ ( n ) ) x π ( i + 3 ) x π ( n ) + 2 n 2 x π ( 1 ) x π ( i ) D ( x π ( i + 1 ) x π ( i + 2 ) ) [ x π ( i + 3 ) x π ( n ) , y σ ( 1 ) y σ ( n ) ] ) ) + π S n σ S n ( ( 2 n 2 1 ) ( f ( x π ( 1 ) , y σ ( 1 ) y σ ( n ) ) x π ( 2 ) x π ( n ) + x π ( 1 ) x π ( n 1 ) f ( x π ( n ) , y σ ( 1 ) y σ ( n ) ) + 2 n 2 D ( x π ( 1 ) ) [ x π ( 2 ) x π ( n ) , y σ ( 1 ) y σ ( n ) ] + 2 n 2 [ x π ( 1 ) x π ( n 1 ) , y σ ( 1 ) y σ ( n ) ] D ( x π ( n ) ) ) ) + π S n σ S n i = 1 n 2 ( τ i , n ) ( 2 n 2 [ x π ( 1 ) x π ( i ) , y σ ( 1 ) y σ ( n ) ] D ( x π ( i + 1 ) ) x π ( i + 2 ) x π ( n ) + x π ( 1 ) x π ( i ) f ( x π ( i + 1 ) , y σ ( 1 ) y σ ( n ) ) x π ( i + 2 ) x π ( n ) + 2 n 2 x π ( 1 ) x π ( i ) D ( x π ( i + 1 ) ) [ x π ( i + 2 ) x π ( n ) , y σ ( 1 ) y σ ( n ) ] ) )

for all x ¯ n , y ¯ n n .

On other way, the last equation can now be rewritten as

(20) 2 2 n 4 D ( [ p ( x ¯ n ) , p ( y ¯ n ) ] ) = π S n σ S n ( f ( y σ ( 1 ) y σ ( 2 ) , x π ( 1 ) x π ( n ) ) y σ ( 3 ) y σ ( n ) + 2 n 2 D ( y σ ( 1 ) y σ ( 2 ) ) [ x π ( 1 ) x π ( n ) , y σ ( 3 ) y σ ( n ) ] ) + π S n σ S n ( 2 n 2 [ x π ( 1 ) x π ( n ) , y σ ( 1 ) y σ ( n 2 ) ] D ( y σ ( n 1 ) y σ ( n ) ) y σ ( 1 ) y σ ( n 2 ) f ( y σ ( n 1 ) y σ ( n ) , x π ( 1 ) x π ( n ) ) ) + π S n σ S n i = 1 n 3 n 2 i ( 2 n 2 [ x π ( 1 ) x π ( n ) , y σ ( 1 ) y σ ( i ) ] D ( y σ ( i + 1 ) y σ ( i + 2 ) ) y σ ( i + 3 ) y σ ( n ) y σ ( 1 ) y σ ( i ) f ( y σ ( i + 1 ) y σ ( i + 2 ) , x π ( 1 ) x π ( n ) ) y σ ( i + 3 ) y σ ( n ) + 2 n 2 y σ ( 1 ) y σ ( i ) D ( y σ ( i + 1 ) y σ ( i + 2 ) ) [ x π ( 1 ) x π ( n ) , y σ ( i + 3 ) y σ ( n ) ] ) ) + π S n σ S n ( ( 2 n 2 1 ) ( f ( y σ ( 1 ) , x π ( 1 ) x π ( n ) ) y σ ( 2 ) y σ ( n ) y σ ( 1 ) y σ ( n 1 ) f ( y σ ( n ) , x π ( 1 ) x π ( n ) ) + 2 n 2 D ( y σ ( 1 ) ) [ x π ( 1 ) x π ( n ) , y σ ( 2 ) y σ ( n ) ] + 2 n 2 [ x π ( 1 ) x π ( n ) , y σ ( 1 ) y σ ( n 1 ) ] D ( y σ ( n ) ) ) ) + π S n σ S n i = 1 n 2 ( τ i , n ) ( 2 n 2 [ x π ( 1 ) x π ( n ) , y σ ( 1 ) y σ ( i ) ] D ( y σ ( i + 1 ) ) y σ ( i + 2 ) y σ ( n ) y σ ( 1 ) y σ ( i ) f ( y σ ( i + 1 ) , x π ( 1 ) x π ( n ) ) y σ ( i + 2 ) y σ ( n ) + 2 n 2 y σ ( 1 ) y σ ( i ) D ( y σ ( i + 1 ) ) [ x π ( 1 ) x π ( n ) , y σ ( i + 2 ) y σ ( n ) ] ) )

for all x ¯ n , y ¯ n n .

Comparing relations (19) and (20) we arrive at

(21) 0 = π S n σ S n ( f ( x π ( 1 ) x π ( 2 ) , y σ ( 1 ) y σ ( n ) ) x π ( 3 ) x π ( n ) + 2 n 2 D ( x π ( 1 ) x π ( 2 ) ) [ x π ( 3 ) x π ( n ) , y σ ( 1 ) y σ ( n ) ] ) + π S n σ S n ( f ( y σ ( 1 ) y σ ( 2 ) , x π ( 1 ) x π ( n ) ) y σ ( 3 ) y σ ( n ) 2 n 2 D ( y σ ( 1 ) y σ ( 2 ) ) [ x π ( 1 ) x π ( n ) , y σ ( 3 ) y σ ( n ) ] ) + π S n σ S n ( 2 n 2 [ x π ( 1 ) x π ( n 2 ) , y σ ( 1 ) y σ ( n ) ] D ( x π ( n 1 ) x π ( n ) ) + x π ( 1 ) x π ( n 2 ) f ( x π ( n 1 ) x π ( n ) , y σ ( 1 ) y σ ( n ) ) ) + π S n σ S n ( 2 n 2 [ x π ( 1 ) x π ( n ) , y σ ( 1 ) y σ ( n 2 ) ] D ( y σ ( n 1 ) y σ ( n ) ) + y σ ( 1 ) y σ ( n 2 ) f ( y σ ( n 1 ) y σ ( n ) , x π ( 1 ) x π ( n ) ) ) + π S n σ S n i = 1 n 3 n 2 i ( 2 n 2 [ x π ( 1 ) x π ( i ) , y σ ( 1 ) y σ ( n ) ] × D ( x π ( i + 1 ) x π ( i + 2 ) ) x π ( i + 3 ) x π ( n ) + x π ( 1 ) x π ( i ) f ( x π ( i + 1 ) x π ( i + 2 ) , y σ ( 1 ) y σ ( n ) ) x π ( i + 3 ) x π ( n ) + 2 n 2 x π ( 1 ) x π ( i ) D ( x π ( i + 1 ) x π ( i + 2 ) ) [ x π ( i + 3 ) x π ( n ) , y σ ( 1 ) y σ ( n ) ] ) ) + π S n σ S n i = 1 n 3 n 2 i ( 2 n 2 [ x π ( 1 ) x π ( n ) , y σ ( 1 ) y σ ( i ) ] × D ( y σ ( i + 1 ) y σ ( i + 2 ) ) y σ ( i + 3 ) y σ ( n ) + y σ ( 1 ) y σ ( i ) f ( y σ ( i + 1 ) y σ ( i + 2 ) , x π ( 1 ) x π ( n ) ) y σ ( i + 3 ) y σ ( n ) 2 n 2 y σ ( 1 ) y σ ( i ) D ( y σ ( i + 1 ) y σ ( i + 2 ) ) [ x π ( 1 ) x π ( n ) , y σ ( i + 3 ) y σ ( n ) ] ) ) + π S n σ S n ( ( 2 n 2 1 ) ( f ( x π ( 1 ) , y σ ( 1 ) y σ ( n ) ) x π ( 2 ) x π ( n ) + x π ( 1 ) x π ( n 1 ) f ( x π ( n ) , y σ ( 1 ) y σ ( n ) ) + 2 n 2 D ( x π ( 1 ) ) [ x π ( 2 ) x π ( n ) , y σ ( 1 ) y σ ( n ) ] + 2 n 2 [ x π ( 1 ) x π ( n 1 ) , y σ ( 1 ) y σ ( n ) ] D ( x π ( n ) ) ) ) + π S n σ S n ( ( 2 n 2 1 ) ( f ( y σ ( 1 ) , x π ( 1 ) x π ( n ) ) y σ ( 2 ) y σ ( n ) + y σ ( 1 ) y σ ( n 1 ) f ( y σ ( n ) , x π ( 1 ) x π ( n ) ) 2 n 2 D ( y σ ( 1 ) ) [ x π ( 1 ) x π ( n ) , y σ ( 2 ) y σ ( n ) ] 2 n 2 [ x π ( 1 ) x π ( n ) , y σ ( 1 ) y σ ( n 1 ) ] D ( y σ ( n ) ) ) ) + π S n σ S n i = 1 n 2 ( τ i , n ) ( 2 n 2 [ x π ( 1 ) x π ( i ) , y σ ( 1 ) y σ ( n ) ] × D ( x π ( i + 1 ) ) x π ( i + 2 ) x π ( n ) + x π ( 1 ) x π ( i ) f ( x π ( i + 1 ) , y σ ( 1 ) y σ ( n ) ) x π ( i + 2 ) x π ( n ) + 2 n 2 x π ( 1 ) x π ( i ) D ( x π ( i + 1 ) ) [ x π ( i + 2 ) x π ( n ) , y σ ( 1 ) y σ ( n ) ] ) ) + π S n σ S n i = 1 n 2 ( τ i , n ) ( 2 n 2 [ x π ( 1 ) x π ( n ) , y σ ( 1 ) y σ ( i ) ] D ( y σ ( i + 1 ) ) y σ ( i + 2 ) y σ ( n ) + y σ ( 1 ) y σ ( i ) f ( y σ ( i + 1 ) , x π ( 1 ) x π ( n ) ) y σ ( i + 2 ) y σ ( n ) 2 n 2 y σ ( 1 ) y σ ( i ) D ( y σ ( i + 1 ) ) [ x π ( 1 ) x π ( n ) , y σ ( i + 2 ) y σ ( n ) ] ) ) .

Let s : Z Z be a mapping defined by s ( i ) = i n . For each σ S n the mapping s 1 σ s : { n + 1 , , 2 n } { n + 1 , , 2 n } will be denoted by σ ¯ . In the continuation of the article we will write x σ ¯ ( n + i ) instead of y σ ( i ) for i = 1 , 2 , , n .

Now using the theory of functional identities and exposing x π ( n ) in (21) from the right side we obtain

π S n σ S n ( f ( x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + 1 ) x σ ¯ ( 2 n ) ) x π ( 3 ) x π ( n 1 ) 2 n 2 D ( x π ( 1 ) x π ( 2 ) ) x σ ¯ ( n + 1 ) x σ ¯ ( 2 n ) x π ( 3 ) x π ( n 1 ) ) + π S n σ S n ( 2 n 2 D ( x σ ¯ ( n + 1 ) x σ ¯ ( n + 2 ) ) x σ ¯ ( n + 3 ) x σ ¯ ( 2 n ) x π ( 1 ) x π ( n 1 ) ) + π S n σ S n ( ( 2 n 2 1 ) ( f ( x π ( 1 ) , x σ ¯ ( n + 1 ) x σ ¯ ( 2 n ) ) x π ( 2 ) x π ( n 1 ) 2 n 2 D ( x π ( 1 ) ) x σ ¯ ( n + 1 ) x σ ¯ ( 2 n ) x π ( 2 ) x π ( n 1 ) ) ) + π S n σ S n ( ( 2 n 2 1 ) ( 2 n 2 D ( x σ ¯ ( n + 1 ) ) x σ ¯ ( n + 2 ) x σ ¯ ( 2 n ) x π ( 1 ) x π ( n 1 ) ) ) = i = 1 2 n 1 x i p 2 n , i ( x ¯ 2 n 1 i ) + λ 2 n ( x ¯ 2 n 1 )

for all x ¯ 2 n 1 2 n 1 , where x ¯ 2 n 1 i = ( x 1 , , x i 1 , x i + 1 , , x 2 n 1 ) , p 2 n , i : 2 n 2 R , and λ 2 n : 2 n 1 C ( ) . Again using the theory of functional identities a few more times we can conclude after ( n 3 ) steps that

π S n σ S n ( f ( x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + 1 ) x σ ¯ ( 2 n ) ) 2 n 2 D ( x π ( 1 ) x π ( 2 ) ) x σ ¯ ( n + 1 ) x σ ¯ ( 2 n ) ) + π S n σ S n ( 2 n 2 D ( x σ ¯ ( n + 1 ) x σ ¯ ( n + 2 ) ) x σ ¯ ( n + 3 ) x σ ¯ ( 2 n ) x π ( 1 ) x π ( 2 ) ) + π S n σ S n ( ( 2 n 2 1 ) ( f ( x π ( 1 ) , x σ ¯ ( n + 1 ) x σ ¯ ( 2 n ) ) x π ( 2 ) 2 n 2 D ( x π ( 1 ) ) x σ ¯ ( n + 1 ) x σ ¯ ( 2 n ) x π ( 2 ) ) ) + π S n σ S n ( ( 2 n 2 1 ) ( 2 n 2 D ( x σ ¯ ( n + 1 ) ) x σ ¯ ( n + 2 ) x σ ¯ ( 2 n ) x π ( 1 ) x π ( 2 ) ) ) = i = 1 n + 2 x i p 2 n , i ( x ¯ n + 2 i ) + λ 2 n ( x ¯ n + 2 )

for all x ¯ n + 2 n + 2 , p 2 n , i : n + 1 R , and λ 2 n : n + 2 C ( ) .

Last equation can now be rewritten as

(22) π S n σ S n ( D ( [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + 1 ) x σ ¯ ( n + 2 ) ] ) x σ ¯ ( n + 3 ) x σ ¯ ( 2 n ) + D ( x σ ¯ ( n + 1 ) x σ ¯ ( n + 2 ) ) [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + 3 ) x σ ¯ ( 2 n ) ] + [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + 1 ) x σ ¯ ( 2 n 2 ) ] D ( x σ ¯ ( 2 n 1 ) x σ ¯ ( 2 n ) ) + x σ ¯ ( n + 1 ) x σ ¯ ( 2 n 2 ) D ( [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( 2 n 1 ) x σ ¯ ( 2 n ) ] ) + i = 1 n 3 n 2 i ( [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + 1 ) x σ ¯ ( n + i ) ] D ( x σ ¯ ( n + i + 1 ) x σ ¯ ( n + i + 2 ) ) x σ ¯ ( n + i + 3 ) x σ ¯ ( 2 n ) + x σ ¯ ( n + 1 ) x σ ¯ ( n + i ) D ( [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + i + 1 ) x σ ¯ ( n + i + 2 ) ] ) x σ ¯ ( n + i + 3 ) x σ ¯ ( 2 n ) + x σ ¯ ( n + 1 ) x σ ¯ ( n + i ) D ( x σ ¯ ( n + i + 1 ) x σ ¯ ( n + i + 2 ) ) [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + i + 3 ) x σ ¯ ( 2 n ) ] ) + ( 2 n 2 1 ) ( D ( [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + 1 ) ] ) x σ ¯ ( n + 2 ) x σ ¯ ( 2 n ) + x σ ¯ ( n + 1 ) x σ ¯ ( 2 n 1 ) D ( [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( 2 n ) ] ) + D ( x σ ¯ ( n + 1 ) ) [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + 2 ) x σ ¯ ( 2 n ) ] + [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + 1 ) x σ ¯ ( 2 n 1 ) ] D ( x σ ¯ ( 2 n ) ) )

+ i = 1 n 2 ( τ i , n ) ( [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + 1 ) x σ ¯ ( n + i ) ] D ( x σ ¯ ( n + i + 1 ) ) x σ ¯ ( n + i + 2 ) x σ ¯ ( 2 n ) + x σ ¯ ( n + 1 ) x σ ¯ ( n + i ) D ( [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + i + 1 ) ] ) x σ ¯ ( n + i + 2 ) x σ ¯ ( 2 n ) + x σ ¯ ( n + 1 ) x σ ¯ ( n + i ) D ( x σ ¯ ( n + i + 1 ) ) [ x π ( 1 ) x π ( 2 ) , x σ ¯ ( n + i + 2 ) x σ ¯ ( 2 n ) ] ) ) + π S n σ S n ( 2 n 2 D ( x π ( 1 ) x π ( 2 ) ) x σ ¯ ( n + 1 ) x σ ¯ ( 2 n ) ) + π S n σ S n ( 2 n 2 D ( x σ ¯ ( n + 1 ) x σ ¯ ( n + 2 ) ) x σ ¯ ( n + 3 ) x σ ¯ ( 2 n ) x π ( 1 ) x π ( 2 ) ) + π S n σ S n ( ( 2 n 2 1 ) ( D ( [ x π ( 1 ) , x σ ¯ ( n + 1 ) x σ ¯ ( n + 2 ) ] ) x σ ¯ ( n + 3 ) x σ ¯ ( 2 n ) x π ( 2 ) + D ( x σ ¯ ( n + 1 ) x σ ¯ ( n + 2 ) ) [ x π ( 1 ) , x σ ¯ ( n + 3 ) x σ ¯ ( 2 n ) ] x π ( 2 ) + [ x π ( 1 ) , x σ ¯ ( n + 1 ) x σ ¯ ( 2 n 2 ) ] D ( x σ ¯ ( 2 n 1 ) x σ ¯ ( 2 n ) ) x π ( 2 ) + x σ ¯ ( n + 1 ) x σ ¯ ( 2 n 2 ) D ( [ x π ( 1 ) , x σ ¯ ( 2 n 1 ) x σ ¯ ( 2 n ) ] ) x π ( 2 ) + i = 1 n 3 n 2 i ( [ x π ( 1 ) , x σ ¯ ( n + 1 ) x σ ¯ ( n + i ) ] D ( x σ ¯ ( n + i + 1 ) x σ ¯ ( n + i + 2 ) ) x σ ¯ ( n + i + 3 ) x σ ¯ ( 2 n ) x π ( 2 ) + x σ ¯ ( n + 1 ) x σ ¯ ( n + i ) D ( [ x π ( 1 ) , x σ ¯ ( n + i + 1 ) x σ ¯ ( n + i + 2 ) ] ) x σ ¯ ( n + i + 3 ) x σ ¯ ( 2 n ) x π ( 2 ) + x σ ¯ ( n + 1 ) x σ ¯ ( n + i ) D ( x σ ¯ ( n + i + 1 ) x σ ¯ ( n + i + 2 ) ) [ x π ( 1 ) , x σ ¯ ( n + i + 3 ) x σ ¯ ( 2 n ) ] x π ( 2 ) ) + ( 2 n 2 1 ) ( D ( [ x π ( 1 ) , x σ ¯ ( n + 1 ) ] ) x σ ¯ ( n + 2 ) x σ ¯ ( 2 n ) x π ( 2 ) + x σ ¯ ( n + 1 ) x σ ¯ ( 2 n 1 ) D ( [ x π ( 1 ) , x σ ¯ ( 2 n ) ] ) x π ( 2 ) + D ( x σ ¯ ( n + 1 ) ) [ x π ( 1 ) , x σ ¯ ( n + 2 ) x σ ¯ ( 2 n ) ] x π ( 2 ) + [ x π ( 1 ) , x σ ¯ ( n + 1 ) x σ ¯ ( 2 n 1 ) ] D ( x σ ¯ ( 2 n ) ) x π ( 2 ) ) + i = 1 n 2 ( τ i , n ) ( [ x π ( 1 ) , x σ ¯ ( n + 1 ) x σ ¯ ( n + i ) ] D ( x σ ¯ ( n + i + 1 ) ) x σ ¯ ( n + i + 2 ) x σ ¯ ( 2 n ) x π ( 2 ) + x σ ¯ ( n + 1 ) x σ ¯ ( n + i ) D ( [ x π ( 1 ) , x σ ¯ ( n + i + 1 ) ] ) x σ ¯ ( n + i + 2 ) x σ ¯ ( 2 n ) x π ( 2 ) + x σ ¯ ( n + 1 ) x σ ¯ ( n + i ) D ( x σ ¯ ( n + i + 1 ) ) [ x π ( 1 ) , x σ ¯ ( n + i + 2 ) x σ ¯ ( 2 n ) ] ) x π ( 2 ) ) + π S n σ S n ( ( 2 n 2 1 ) ( 2 n 2 D ( x π ( 1 ) ) x σ ¯ ( n + 1 ) x σ ¯ ( 2 n ) x π ( 2 ) ) ) + π S n σ S n ( ( 2 n 2 1 ) ( 2 n 2 D ( x σ ¯ ( n + 1 ) ) x σ ¯ ( n + 2 ) x σ ¯ ( 2 n ) x π ( 1 ) x π ( 2 ) ) ) = i = 1 n + 2 x i p 2 n , i ( x ¯ n + 2 i ) + λ 2 n ( x ¯ n + 2 )

for all x ¯ n + 2 n + 2 , p 2 n , i : n + 1 R , and λ 2 n : n + 2 C ( ) .

Using the theory of functional identities and first exposing x π ( 2 ) from the right side and then x σ ¯ ( 2 n ) in (22) we obtain

π S n σ S n ( ( 2 n 2 1 ) D ( [ x π ( 1 ) , x σ ¯ ( n + 1 ) x σ ¯ ( n + 2 ) ] ) x σ ¯ ( n + 3 ) x σ ¯ ( 2 n 1 ) + ( 2 n 2 1 ) D ( x σ ¯ ( n + 1 ) x σ ¯ ( n + 2 ) ) x π ( 1 ) x σ ¯ ( n + 3 ) x σ ¯ ( 2 n 1 ) + ( 2 n 2 1 ) 2 D ( [ x π ( 1 ) , x σ ¯ ( n + 1 ) ] ) x σ ¯ ( n + 2 ) x σ ¯ ( 2 n 1 ) + ( 2 n 2 1 ) 2 D ( x σ ¯ ( n + 1 ) ) x π ( 1 ) x σ ¯ ( n + 2 ) x σ ¯ ( 2 n 1 ) + ( 2 n 2 1 ) ( 2 n 2 D ( x π ( 1 ) ) x σ ¯ ( n + 1 ) x σ ¯ ( 2 n 1 ) ) ) = i = 1 n x i p 2 n , i ( x ¯ n i ) + λ 2 n ( x ¯ n )

for all x ¯ n n , p 2 n , i : n 1 R , and λ 2 n : n C ( ) .

Again using the theory of functional identities and exposing x σ ¯ ( n + 3 ) x σ ¯ ( 2 n 1 ) we obtain

π S n σ S n ( ( 2 n 2 1 ) D ( [ x π ( 1 ) , x σ ¯ ( n + 1 ) x σ ¯ ( n + 2 ) ] ) + ( 2 n 2 1 ) D ( x σ ¯ ( n + 1 ) x σ ¯ ( n + 2 ) ) x π ( 1 ) + ( 2 n 2 1 ) 2 D ( [ x π ( 1 ) , x σ ¯ ( n + 1 ) ] ) x σ ¯ ( n + 2 ) + ( 2 n 2 1 ) 2 D ( x σ ¯ ( n + 1 ) ) x π ( 1 ) x σ ¯ ( n + 2 ) + ( 2 n 2 1 ) ( 2 n 2 D ( x π ( 1 ) ) x σ ¯ ( n + 1 ) x σ ¯ ( n + 2 ) ) ) = x 1 t ( x n + 1 , x n + 2 ) + x n + 1 u ( x 1 , x n + 2 ) + x n + 2 v ( x 1 , x n + 1 ) + κ ( x 1 , x n + 1 , x n + 2 ) ,

where t , u , v : 2 R and κ : 3 C ( ) .

Putting x π ( 1 ) = x σ ¯ ( n + 1 ) = x σ ¯ ( n + 2 ) = x in the above relation gives

(23) 2 ( 2 n 2 1 ) D ( x 2 ) x 2 ( 2 n 2 1 ) D ( x ) x 2 = x t ( x , x ) + x u ( x , x ) + x v ( x , x ) + κ ( x , x , x ) ,

where t , u , v : 2 R , and κ : 3 C ( ) .

After the complete linearization of the above identity and considering that is a 6-free subset of R , we obtain

(24) 2 δ n D ( x y ) + 2 δ n D ( y x ) 2 δ n D ( x ) y 2 δ n D ( y ) x = x f ( y ) + y g ( x ) + λ ( x , y ) ,

where δ n = ( 2 n 2 1 ) , f , g : R and λ : 2 C ( ) . The next equation is clearly proved in the same way as the last one. Using the theory of functional identities and first exposing x π ( 1 ) x π ( n 1 ) x σ ¯ ( n + 1 ) x σ ¯ ( 2 n 2 ) from the left side we obtain

(25) 2 δ n D ( x y ) + 2 δ n D ( y x ) 2 δ n x D ( y ) 2 δ n y D ( x ) = f ( x ) y + g ( y ) x + λ ( x , y ) ,

where δ n = ( 2 n 2 1 ) , f , g : R , and λ : 2 C ( ) . Replacing the roles of denotations x and y in (24) and comparing the so-obtained identities leads to 0 = x f ( y ) + y g ( x ) y f ( x ) x g ( y ) + λ ( x , y ) λ ( y , x ) , which yields f ( x ) = g ( x ) and λ ( x , y ) = λ ( y , x ) for all x , y . Putting x for y in (24) leads to

(26) 4 δ n D ( x 2 ) = 4 δ n D ( x ) x + 2 x f ( x ) + λ ( x , x ) .

Using the same arguments, it follows from (25) that f ( x ) = g ( x ) and λ ( x , y ) = λ ( y , x ) for all x , y . Therefore,

4 δ n D ( x 2 ) = 4 δ n x D ( x ) + 2 f ( x ) x + λ ( x , x ) .

Comparing the above relations gives

x ( 4 δ n D ( x ) 2 f ( x ) ) ( 4 δ n D ( x ) 2 f ( x ) ) x C ( ) .

Hence, there exist r R and μ : C ( ) such that

4 δ n D ( x ) 2 f ( x ) = r x + μ ( x ) .

Considering 2 f ( x ) = 4 δ n D ( x ) r x μ ( x ) in (26) gives

(27) 4 δ n D ( x 2 ) = 4 δ n D ( x ) x + 4 δ n x D ( x ) x r x x μ ( x ) + λ ( x , x ) .

Now setting x n instead of y in (24) and multiplying this equation with 2 n 3 we obtain

2 n 3 4 δ n D ( x n + 1 ) = 2 n 3 2 δ n D ( x ) x n + 2 n 3 2 δ n D ( x n ) x + 2 n 3 x f ( x n ) + 2 n 3 x n f ( x ) + 2 n 3 λ ( x , x n )

and so

(28) 2 n 1 δ n D ( x n + 1 ) = 2 n 2 δ n D ( x ) x n + 2 n 2 δ n D ( x n ) x + 2 n 3 x f ( x n ) + 2 n 3 x n f ( x ) + 2 n 3 λ ( x , x n ) .

On the other hand, setting n + 1 instead of n into (6) and multiplying this equation with δ n we obtain

(29) 2 n 1 δ n D ( x n + 1 ) = δ n i = 0 n 1 n 1 i x i D ( x 2 ) x n 1 i + ( 2 n 1 1 ) δ n ( D ( x ) x n + x n D ( x ) ) + δ n i = 1 n 1 ( τ i , n + 1 ) x i D ( x ) x n i .

Now comparing equations (28) and (29) we obtain

(30) 0 = δ n i = 0 n 1 n 1 i x i D ( x 2 ) x n 1 i + ( 2 n 1 1 ) δ n ( D ( x ) x n + x n D ( x ) ) + δ n i = 1 n 1 ( τ i , n + 1 ) x i D ( x ) x n i 2 n 2 δ n D ( x ) x n 2 n 2 δ n D ( x n ) x 2 n 3 x f ( x n ) 2 n 3 x n f ( x ) 2 n 3 λ ( x , x n ) .

Multiplying last equation with 4 and considering 2 f ( x ) = 4 δ n D ( x ) r x μ ( x ) and 2 f ( x n ) = 4 δ n D ( x n ) r x n μ ( x n ) we obtain

0 = 4 δ n i = 0 n 1 n 1 i x i D ( x 2 ) x n 1 i + ( 2 n 1 1 ) 4 δ n ( D ( x ) x n + x n D ( x ) ) + 4 δ n i = 1 n 1 ( τ i , n + 1 ) x i D ( x ) x n i 2 n 2 4 δ n D ( x ) x n 2 n 2 4 δ n D ( x n ) x 2 n 2 x ( 4 δ n D ( x n ) r x n μ ( x n ) ) 2 n 2 x n ( 4 δ n D ( x ) r x μ ( x ) ) 2 n 1 λ ( x , x n ) .

And by using (27)

(31) 0 = i = 0 n 1 n 1 i x i ( 4 δ n D ( x ) x + 4 δ n x D ( x ) x r x x μ ( x ) + λ ( x , x ) ) x n 1 i + ( 2 n 1 1 ) 4 δ n ( D ( x ) x n + x n D ( x ) ) + 4 δ n i = 1 n 1 ( τ i , n + 1 ) x i D ( x ) x n i 2 n δ n D ( x ) x n 2 n δ n D ( x n ) x 2 n δ n x D ( x n ) + 2 n 2 x r x n + 2 n 2 x μ ( x n ) 2 n δ n x n D ( x ) + 2 n 2 x n r x + 2 n 2 x n μ ( x ) 2 n 1 λ ( x , x n ) .

Now considering equation (27) into (6) we obtain

(32) 2 n 2 4 δ n D ( x n ) = i = 0 n 2 n 2 i ( 4 δ n x i D ( x ) x n 1 i + 4 δ n x i + 1 D ( x ) x n 2 i x i + 1 r x n 1 i x n 1 μ ( x ) + x n 2 λ ( x , x ) ) + ( 2 n 2 1 ) 4 δ n ( D ( x ) x n 1 + x n 1 D ( x ) ) + 4 δ n i = 1 n 2 ( τ i , n ) x i D ( x ) x n 1 i .

Using equation (32) into equation (31) we obtain

(33) 0 = i = 0 n 1 n 1 i ( 4 δ n x i D ( x ) x n i + 4 δ n x i + 1 D ( x ) x n 1 i x i + 1 r x n i x n μ ( x ) + x n 1 λ ( x , x ) ) + ( 2 n 1 1 ) 4 δ n ( D ( x ) x n + x n D ( x ) ) + 4 δ n i = 1 n 1 ( τ i , n + 1 ) x i D ( x ) x n i 2 n δ n D ( x ) x n i = 0 n 2 n 2 i ( 4 δ n x i D ( x ) x n i + 4 δ n x i + 1 D ( x ) x n 1 i x i + 1 r x n i x n μ ( x ) + x n 1 λ ( x , x ) ) ( 2 n 2 1 ) 4 δ n ( D ( x ) x n + x n 1 D ( x ) x ) 4 δ n i = 1 n 2 ( τ i , n ) x i D ( x ) x n i i = 0 n 2 n 2 i ( 4 δ n x i + 1 D ( x ) x n 1 i + 4 δ n x i + 2 D ( x ) x n 2 i x i + 2 r x n 1 i x n μ ( x ) + x n 1 λ ( x , x ) ) ( 2 n 2 1 ) 4 δ n ( x D ( x ) x n 1 + x n D ( x ) ) 4 δ n i = 1 n 2 ( τ i , n ) x i + 1 D ( x ) x n 1 i + 2 n 2 x r x n + 2 n 2 x μ ( x n ) 2 n δ n x n D ( x ) + 2 n 2 x n r x + 2 n 2 x n μ ( x ) 2 n 1 λ ( x , x n ) .

After a precise calculation we obtain

0 = 2 n 2 x r x n + 2 n 2 x μ ( x n ) + 2 n 2 x n r x + 2 n 2 x n μ ( x ) 2 n 1 λ ( x , x n )

and so

0 = x r x n + x μ ( x n ) + x n r x + x n μ ( x ) 2 λ ( x , x n ) .

The complete linearization of this relation and using the theory of functional identities leads to 0 = r x n + μ ( x n ) + x n 1 r x + x n 1 μ ( x ) and λ ( x , x n ) = 0 . Again, after the complete linearization of the last relation and using the theory of functional identities leads to

(34) 0 = r x n 1 + x n 1 r + x n 2 μ ( x )

and μ ( x n ) = 0 .

Left multiplication of the last relation by y n 1 gives

0 = y n 1 r x n 1 + y n 1 x n 1 r + y n 1 x n 2 μ ( x ) .

Putting y x for x in (34) leads to

0 = r y n 1 x n 1 + y n 1 x n 1 r + y n 2 x n 2 μ ( y x ) .

Comparing the last two relations gives

0 = y n 1 r x n 1 r y n 1 x n 1 + μ ( x ) y n 1 x n 2 μ ( y x ) y n 2 x n 2 .

Complete linearization of the last equation and using the theory of functional identities (exposing everything from the right side) we obtain

0 = y n 1 r x r y n 1 x + μ ( x ) y n 1 μ ( y x ) y n 2

and so

0 = [ y n 1 , r ] x + ( μ ( x ) y n 2 μ ( y x ) y n 3 ) y .

Whence it follows that [ y n 1 , r ] = 0 and μ ( x ) y n 2 μ ( y x ) y n 3 = 0 for all x R . Last equation after using the theory of functional identities implies that μ ( y x ) = 0 and μ ( x ) = 0 . From (34) we now obtain

0 = 2 r x n 1 .

Since R is prime, the last relation implies r = 0 . Using this in (27) gives us

(35) 4 δ n D ( x 2 ) = 4 δ n D ( x ) x + 4 δ n x D ( x ) + λ ( x , x ) .

Complete linearization of the last equation implies

(36) 4 δ n D ( x y ) + 4 δ n D ( y x ) = 4 δ n D ( x ) y + 4 δ n D ( y ) x + 4 δ n x D ( y ) + 4 δ n y D ( x ) + λ ( x , y ) + λ ( y , x ) .

Now setting y = x 3 in the last equation we obtain

(37) 8 δ n D ( x 4 ) = 4 δ n D ( x ) x 3 + 4 δ n D ( x 3 ) x + 4 δ n x D ( x 3 ) + 4 δ n x 3 D ( x ) + 2 λ ( x , x 3 ) .

On the other hand, setting x 2 instead of x in (35) we obtain

(38) 4 δ n D ( x 4 ) = 4 δ n D ( x 2 ) x 2 + 4 δ n x 2 D ( x 2 ) + λ ( x 2 , x 2 ) .

Comparing equations (37) and (38) we obtain

(39) 0 = 4 δ n D ( x ) x 3 + 4 δ n D ( x 3 ) x + 4 δ n x D ( x 3 ) + 4 δ n x 3 D ( x ) + 2 λ ( x , x 3 ) 8 δ n D ( x 2 ) x 2 8 δ n x 2 D ( x 2 ) 2 λ ( x 2 , x 2 ) .

Now setting y = x 2 in (36) we obtain

(40) 8 δ n D ( x 3 ) = 4 δ n D ( x ) x 2 + 4 δ n D ( x 2 ) x + 4 δ n x D ( x 2 ) + 4 δ n x 2 D ( x ) + 2 λ ( x , x 2 ) .

Using (40) and (35) in (39) we obtain

0 = 8 δ n D ( x ) x 3 + 4 δ n D ( x ) x 3 + 4 δ n D ( x 2 ) x 2 + 4 δ n x D ( x 2 ) x + 4 δ n x 2 D ( x ) x + 2 λ ( x , x 2 ) x + 4 δ n x D ( x ) x 2 + 4 δ n x D ( x 2 ) x + 4 δ n x 2 D ( x 2 ) + 4 δ n x 3 D ( x ) + 2 x λ ( x , x 2 ) + 8 δ n x 3 D ( x ) + 4 λ ( x , x 3 ) 16 δ n D ( x 2 ) x 2 16 δ n x 2 D ( x 2 ) 4 λ ( x 2 , x 2 )

and so

0 = 12 δ n D ( x ) x 3 12 δ n D ( x 2 ) x 2 + 8 δ n x D ( x 2 ) x + 4 δ n x 2 D ( x ) x + 4 δ n x D ( x ) x 2 12 δ n x 2 D ( x 2 ) + 12 δ n x 3 D ( x ) + 4 x λ ( x , x 2 ) + 4 λ ( x , x 3 ) 4 λ ( x 2 , x 2 ) .

Using 4 δ n D ( x 2 ) = 4 δ n D ( x ) x + 4 δ n x D ( x ) + λ ( x , x ) in the last equation we obtain

0 = 12 δ n D ( x ) x 3 12 δ n D ( x ) x 3 12 δ n x D ( x ) x 2 3 x 2 λ ( x , x ) + 8 δ n x D ( x ) x 2 + 8 δ n x 2 D ( x ) x + 2 x 2 λ ( x , x ) + 4 δ n x 2 D ( x ) x + 4 δ n x D ( x ) x 2 12 δ n x 2 D ( x ) x 12 δ n x 3 D ( x ) 3 x 2 λ ( x , x ) + 12 δ n x 3 D ( x ) + 4 x λ ( x , x 2 ) + 4 λ ( x , x 3 ) 4 λ ( x 2 , x 2 )

and so

0 = 4 x 2 λ ( x , x ) + 4 x λ ( x , x 2 ) + 4 λ ( x , x 3 ) 4 λ ( x 2 , x 2 ) .

Complete linearization of the last equation and using the theory of functional identities it follows that 4 λ ( x 2 , x 2 ) = 4 λ ( x , x 3 ) and 0 = 4 x λ ( x , x ) + 4 λ ( x , x 2 ) . Again using the theory of functional identities it follows that 4 λ ( x , x 2 ) = 0 and 4 λ ( x , x ) = 0 . Consequently, it follows from (35) that

4 δ n D ( x 2 ) = 4 δ n D ( x ) x + 4 δ n x D ( x ) .

And so that D is a Jordan derivation. By Herstein theorem, D is a derivation, which completes the proof of the theorem.□

We are now in the position to prove Theorem 3.

Proof of Theorem 3

The complete linearization of (6) gives us (14). First suppose that R is not a PI ring (satisfying the standard polynomial identity of degree less than 6). According to Theorem 4 the mapping D is a derivation.

Assume now that R is a PI ring. It is well known that in this case R has a nonzero center (see [1]). Let c be a nonzero central element.

Picking any x R and setting x 1 = x 2 = x and x 3 = = x n = c in (14) we obtain

(41) n ( n 1 ) 2 n 3 D ( c n 2 x 2 ) = 2 + i = 1 n 3 n 2 i c n 2 D ( x 2 ) + 2 ( n 2 ) + i = 1 n 3 n 2 i 2 ( n 2 i ) c n 3 D ( c x ) x + 2 ( n 2 ) + i = 1 n 3 n 2 i 2 i c n 3 x D ( c x ) + n 2 2 + i = 1 n 3 n 2 i n 2 i 2 c n 4 D ( c 2 ) x 2 + n 2 2 + i = 1 n 3 n 2 i i 2 c n 4 x 2 D ( c 2 ) + i = 1 n 3 n 2 i i ( n 2 i ) c n 4 x D ( c 2 ) x + ( n 1 ) ( 2 n 2 1 ) + i = 1 n 2 ( n 1 i ) ( τ i , n ) c n 2 D ( x ) x + ( n 1 ) ( 2 n 2 1 ) + i = 1 n 2 ( i ) ( τ i , n ) c n 2 x D ( x ) + n 1 2 ( 2 n 2 1 ) + i = 1 n 2 n 1 i 2 ( τ i , n ) c n 3 D ( c ) x 2 + n 1 2 ( 2 n 2 1 ) + i = 1 n 2 i 2 ( τ i , n ) c n 3 x 2 D ( c ) + i = 1 n 2 i ( n 1 i ) ( τ i , n ) c n 3 x D ( c ) x .

Complete linearization of equation (41) and setting x 1 = c 2 , x 2 = x we obtain

(42) 2 n ( n 1 ) 2 n 3 D ( c n x ) = 2 2 + i = 1 n 3 n 2 i c n 2 D ( c 2 x ) + 2 ( n 2 ) + 2 i = 1 n 3 n 2 i ( n 2 i ) c n 3 D ( c 3 ) x + 4 ( n 2 ) + 2 i = 1 n 3 n 2 i ( n 2 i ) + 2 i = 1 n 3 n 2 i i c n 1 D ( c x ) + 2 ( n 2 ) + 2 i = 1 n 3 n 2 i i c n 3 x D ( c 3 ) + 2 ( ( n 1 ) ( 2 n 2 1 ) ) + i = 1 n 2 ( n 1 i ) ( τ i , n ) + i = 1 n 2 i ( τ i , n ) c n D ( x ) + 2 n 1 2 ( 2 n 2 1 ) + i = 1 n 2 n 1 i 2 ( τ i , n ) + i = 1 n 2 i ( n 1 i ) ( τ i , n ) c n 1 D ( c ) x + 2 n 1 2 ( 2 n 2 1 ) + i = 1 n 2 i 2 ( τ i , n ) + i = 1 n 2 i ( n 1 i ) ( τ i , n ) c n 1 x D ( c ) + 2 n 2 2 + i = 1 n 3 n 2 i n 2 i 2 + i = 1 n 3 n 2 i i ( n 2 i ) + ( n 1 ) ( 2 n 2 1 ) + i = 1 n 2 ( n 1 i ) ( τ i , n ) c n 2 D ( c 2 ) x + 2 n 2 2 + i = 1 n 3 n 2 i i 2 + i = 1 n 3 n 2 i i ( n 2 i ) + ( n 1 ) ( 2 n 2 1 ) + i = 1 n 2 ( i ) ( τ i , n ) c n 2 x D ( c 2 ) .

On the other hand, complete linearization of equation (41) and setting x 1 = c , x 2 = c x we obtain

(43) 2 n ( n 1 ) 2 n 3 D ( c n x ) = 2 2 + i = 1 n 3 n 2 i + 4 ( n 2 ) + 2 i = 1 n 3 n 2 i ( n 2 i ) + 2 i = 1 n 3 n 2 i ( i ) c n 2 D ( c 2 x ) + 2 ( ( n 1 ) ( 2 n 2 1 ) ) + i = 1 n 2 ( n 1 i ) ( τ i , n ) + i = 1 n 2 i ( τ i , n ) c n 1 D ( c x ) + 2 ( n 2 ) + 2 i = 1 n 3 n 2 i ( n 2 i ) + 2 n 2 2 + i = 1 n 3 n 2 i n 2 i 2 + i = 1 n 3 n 2 i i ( n 2 i ) c n 2 D ( c 2 ) x + 2 ( n 2 ) + 2 i = 1 n 3 n 2 i ( i ) + 2 n 2 2 + i = 1 n 3 n 2 i i 2 + i = 1 n 3 n 2 i i ( n 2 i ) c n 2 x D ( c 2 ) + ( n 1 ) ( 2 n 2 1 ) + i = 1 n 2 ( n 1 i ) ( τ i , n ) + 2 n 1 2 ( 2 n 2 1 ) + i = 1 n 2 n 1 i 2 ( τ i , n ) + i = 1 n 2 i ( n 1 i ) ( τ i , n ) c n 1 D ( c ) x + ( n 1 ) ( 2 n 2 1 ) + i = 1 n 2 ( i ) ( τ i , n ) + 2 n 1 2 ( 2 n 2 1 ) + i = 1 n 2 i 2 ( τ i , n ) + i = 1 n 2 i ( n 1 i ) ( τ i , n ) c n 1 x D ( c ) .

Comparing equations (42) and (43) we obtain

(44) 0 = 4 ( n 2 ) 2 i = 1 n 3 n 2 i ( n 2 i ) 2 i = 1 n 3 n 2 i ( i ) c n 2 D ( c 2 x ) + 2 ( n 2 ) + 2 i = 1 n 3 n 2 i ( i ) c n 3 D ( c 3 ) x + 4 ( n 2 ) + 4 i = 1 n 3 n 2 i ( i ) 2 ( n 1 ) ( 2 n 2 1 ) i = 1 n 2 ( n 1 i ) ( τ i , n ) i = 1 n 2 i ( τ i , n ) c n 1 D ( c x ) + 2 ( n 2 ) + 2 i = 1 n 3 n 2 i ( i ) c n 3 x D ( c 3 ) + 2 ( n 1 ) ( 2 n 2 1 ) + i = 1 n 2 ( n 1 i ) ( τ i , n ) + i = 1 n 2 i ( τ i , n ) c n D ( x ) + ( n 1 ) ( 2 n 2 1 ) i = 1 n 2 ( n 1 i ) ( τ i , n ) c n 1 D ( c ) x + ( n 1 ) ( 2 n 2 1 ) i = 1 n 2 ( i ) ( τ i , n ) c n 1 x D ( c ) + ( n 1 ) ( 2 n 2 1 ) + i = 1 n 2 ( n 1 i ) ( τ i , n ) 2 ( n 2 ) 2 i = 1 n 3 n 2 i ( i ) c n 2 D ( c 2 ) x + ( n 1 ) ( 2 n 2 1 ) + i = 1 n 2 ( i ) ( τ i , n ) 2 ( n 2 ) 2 i = 1 n 3 n 2 i ( i ) c n 2 x D ( c 2 ) .

For the ease of writing, we will use the following equalities

i = 1 n 3 n 2 i ( n 2 i ) = i = 1 n 3 n 2 i ( i ) = ( 2 n 3 1 ) ( n 2 ) ,

i = 1 n 2 ( n 1 i ) ( τ i , n ) = i = 1 n 2 ( i ) ( τ i , n ) = ( n 1 ) ( 2 n 3 ( n 4 ) + 1 ) .

Equation (44) can now be rewritten as

0 = 4 2 n 3 ( n 2 ) c D ( c 2 x ) + 2 2 n 3 ( n 2 ) D ( c 3 ) x + ( 2 2 n 3 ( n 2 ) ( 3 n ) ) c 2 D ( c x ) + 2 2 n 3 ( n 2 ) x D ( c 3 ) + 2 2 n 3 ( n 2 ) ( n 1 ) c 3 D ( x ) 2 n 3 ( n 2 ) ( n 1 ) c 2 D ( c ) x 2 n 3 ( n 2 ) ( n 1 ) c 2 x D ( c ) + ( 2 n 3 ( n 2 ) ( n 3 ) ) c D ( c 2 ) x + ( 2 n 3 ( n 2 ) ( n 3 ) ) c x D ( c 2 )

and so

(45) 0 = 4 c D ( c 2 x ) + 2 D ( c 3 ) x + 2 ( 3 n ) c 2 D ( c x ) + 2 x D ( c 3 ) + 2 ( n 1 ) c 3 D ( x ) ( n 1 ) c 2 D ( c ) x ( n 1 ) c 2 x D ( c ) + ( n 3 ) c D ( c 2 ) x + ( n 3 ) c x D ( c 2 ) .

Setting c 2 x instead of x in (45) we obtain

(46) 0 = 4 c D ( c 4 x ) + 2 c 2 D ( c 3 ) x + 2 ( 3 n ) c 2 D ( c 3 x ) + 2 c 2 x D ( c 3 ) + 2 ( n 1 ) c 3 D ( c 2 x ) ( n 1 ) c 4 D ( c ) x ( n 1 ) c 4 x D ( c ) + ( n 3 ) c 3 D ( c 2 ) x + ( n 3 ) c 3 x D ( c 2 ) .

On other hand, setting c 2 instead of c in (45) we obtain

(47) 0 = 4 c 2 D ( c 4 x ) + 2 D ( c 6 ) x + 2 ( 3 n ) c 4 D ( c 2 x ) + 2 x D ( c 6 ) + 2 ( n 1 ) c 6 D ( x ) ( n 1 ) c 4 D ( c 2 ) x ( n 1 ) c 4 x D ( c 2 ) + ( n 3 ) c 2 D ( c 4 ) x + ( n 3 ) c 2 x D ( c 4 ) .

Comparing (subtracting) equations (46) and (47) we obtain

(48) 0 = 2 c 3 D ( c 3 ) x + 2 ( 3 n ) c 3 D ( c 3 x ) + 2 c 3 x D ( c 3 ) + 2 ( 2 n 4 ) c 4 D ( c 2 x ) ( n 1 ) c 5 D ( c ) x ( n 1 ) c 5 x D ( c ) + ( 2 n 4 ) c 4 D ( c 2 ) x + ( 2 n 4 ) c 4 x D ( c 2 ) 2 D ( c 6 ) x 2 x D ( c 6 ) 2 ( n 1 ) c 6 D ( x ) ( n 3 ) c 2 D ( c 4 ) x ( n 3 ) c 2 x D ( c 4 ) .

Using x = c in the last identity we obtain

(49) D ( c 6 ) = ( 3 n ) c 2 D ( c 4 ) + ( n 1 ) c 3 D ( c 3 ) + ( n 2 ) c 4 D ( c 2 ) ( n 1 ) c 5 D ( c ) .

Now setting c x instead of x in (45) we obtain

(50) 4 D ( c 3 x ) = 2 D ( c 3 ) x + 2 ( 3 n ) c D ( c 2 x ) + 2 x D ( c 3 ) + 2 ( n 1 ) c 2 D ( c x ) ( n 1 ) c 2 D ( c ) x ( n 1 ) c 2 x D ( c ) + ( n 3 ) c D ( c 2 ) x + ( n 3 ) c x D ( c 2 ) .

Putting x = c in the last equation we obtain

(51) 4 D ( c 4 ) = ( 10 2 n ) c D ( c 3 ) + ( 4 n 8 ) c 2 D ( c 2 ) ( 2 n 2 ) c 3 D ( c ) .

Putting equations (49), (50), and (51) into equation (48) we obtain

(52) 0 = 4 ( n 1 ) c D ( c 2 x ) 2 ( n 1 ) D ( c 3 ) x 4 ( n 3 ) c 2 D ( c x ) 2 ( n 1 ) x D ( c 3 ) 8 c 3 D ( x ) + 4 c 2 D ( c ) x + 4 c 2 x D ( c ) + 2 ( n 3 ) c D ( c 2 ) x + 2 ( n 3 ) c x D ( c 2 ) .

Comparing (addition) equations (45) and (52) we obtain

(53) 0 = 2 ( n 3 ) ( n + 1 ) c D ( c x ) + 2 ( n 3 ) ( n + 1 ) c 2 D ( x ) ( n 3 ) ( n + 1 ) c D ( c ) x ( n 3 ) ( n + 1 ) c x D ( c ) + ( n 3 ) ( n + 1 ) D ( c 2 ) x + ( n 3 ) ( n + 1 ) x D ( c 2 )

and so if n > 3 we obtain

(54) 2 c D ( c x ) = 2 c 2 D ( x ) c D ( c ) x c x D ( c ) + D ( c 2 ) x + x D ( c 2 ) .

On the other way, if n = 3 we obtain from equation (52)

(55) 2 c D ( c 2 x ) = D ( c 3 ) x + x D ( c 3 ) + 2 c 3 D ( x ) c 2 D ( c ) x c 2 x D ( c ) .

If n = 3 , equation (6) is equal to 2 D ( x 3 ) = D ( x 2 ) x + x 2 D ( x ) + D ( x ) x 2 + x D ( x 2 ) . Setting x 1 = x 2 = c and x 3 = x in the complete linearization of the last equation, we obtain

(56) 6 D ( c 2 x ) = D ( c 2 ) x + x D ( c 2 ) + 4 c D ( c x ) + 2 c 2 D ( x ) + 2 c x D ( c ) + 2 c D ( c ) x .

If x = c we obtain D ( c 3 ) = c D ( c 2 ) + c 2 D ( c ) . Comparing (subtracting) equations (55) and (56) and using D ( c 3 ) = c D ( c 2 ) + c 2 D ( c ) , we obtain

(57) 2 c D ( c x ) = 2 c 2 D ( x ) c D ( c ) x c x D ( c ) + D ( c 2 ) x + x D ( c 2 ) .

Equation (54) is now equal to equation (57).

Now putting c 2 instead of x in equation (54) we obtain

D ( c 3 ) = 2 c D ( c 2 ) c 2 D ( c ) .

Now setting c 3 instead of x in (54) we obtain

D ( c 4 ) = c D ( c 3 ) c 3 D ( c ) + c 2 D ( c 2 ) = c ( 2 c D ( c 2 ) c 2 D ( c ) ) c 3 D ( c ) + c 2 D ( c 2 ) = 3 c 2 D ( c 2 ) 2 c 3 D ( c ) .

Similarly setting c 4 instead of x in (54) we obtain

D ( c 5 ) = c D ( c 4 ) c 4 D ( c ) + c 3 D ( c 2 ) = c ( 3 c 2 D ( c 2 ) 2 c 3 D ( c ) ) c 4 D ( c ) + c 3 D ( c 2 ) = 4 c 3 D ( c 2 ) 3 c 4 D ( c ) .

After n steps we can conclude that

D ( c n ) = ( n 1 ) c n 2 D ( c 2 ) ( n 2 ) c n 1 D ( c ) .

On the other hand, setting x = c in (41) we obtain

D ( c n ) = c n 2 D ( c 2 ) + ( n 2 ) c n 1 D ( c ) .

Subtracting last two equations we obtain 0 = ( n 2 ) D ( c 2 ) 2 ( n 2 ) c D ( c ) and so D ( c 2 ) = 2 c D ( c ) . Equation (54) can now be rewritten as

(58) 2 D ( c x ) = 2 c D ( x ) + D ( c ) x + x D ( c ) .

Equation (41) can now be rewritten as

(59) n ( n 1 ) 2 n 3 D ( c n 2 x 2 ) = 2 + i = 1 n 3 n 2 i c n 2 D ( x 2 ) + 2 ( ( n 2 ) 2 n 3 ) c n 3 D ( c x ) x + 2 ( ( n 2 ) 2 n 3 ) c n 3 x D ( c x ) + 2 n 2 2 + i = 1 n 3 n 2 i n 2 i 2 c n 3 D ( c ) x 2 + 2 n 2 2 + i = 1 n 3 n 2 i i 2 c n 3 x 2 D ( c ) + 2 i = 1 n 3 n 2 i i ( n 2 i ) c n 3 x D ( c ) x + ( ( n 1 ) ( n 2 ) 2 n 3 ) c n 2 D ( x ) x + ( ( n 1 ) ( n 2 ) 2 n 3 ) c n 2 x D ( x ) + n 1 2 ( 2 n 2 1 ) + i = 1 n 2 n 1 i 2 ( τ i , n ) c n 3 D ( c ) x 2 + n 1 2 ( 2 n 2 1 ) + i = 1 n 2 i 2 ( τ i , n ) c n 3 x 2 D ( c ) + i = 1 n 2 i ( n 1 i ) ( τ i , n ) c n 3 x D ( c ) x .

Now using (58) in the last equation we obtain

(60) n ( n 1 ) 2 n 3 D ( c n 2 x 2 ) = 2 + i = 1 n 3 n 2 i c n 2 D ( x 2 ) + ( ( n 2 ) 2 n 3 ) c n 3 ( 2 c D ( x ) + D ( c ) x + x D ( c ) ) x + ( ( n 2 ) 2 n 3 ) c n 3 x ( 2 c D ( x ) + D ( c ) x + x D ( c ) ) + 2 n 2 2 + i = 1 n 3 n 2 i n 2 i 2 c n 3 D ( c ) x 2 + 2 n 2 2 + i = 1 n 3 n 2 i i 2 c n 3 x 2 D ( c ) + 2 i = 1 n 3 n 2 i i ( n 2 i ) c n 3 x D ( c ) x + ( ( n 1 ) ( n 2 ) 2 n 3 ) c n 2 D ( x ) x + ( ( n 1 ) ( n 2 ) 2 n 3 ) c n 2 x D ( x ) + n 1 2 ( 2 n 2 1 ) + i = 1 n 2 n 1 i 2 ( τ i , n ) c n 3 D ( c ) x 2 + n 1 2 ( 2 n 2 1 ) + i = 1 n 2 i 2 ( τ i , n ) c n 3 x 2 D ( c ) + i = 1 n 2 i ( n 1 i ) ( τ i , n ) c n 3 x D ( c ) x .

Now setting c n 2 instead of c and x 2 instead of x in (58) and using ( D ( c n 2 ) = ( n 2 ) c n 3 D ( c ) ) we obtain

2 D ( c n 2 x 2 ) = 2 c n 2 D ( x 2 ) + D ( c n 2 ) x 2 + x 2 D ( c n 2 ) = 2 c n 2 D ( x 2 ) + ( n 2 ) c n 3 D ( c ) x 2 + ( n 2 ) c n 3 x 2 D ( c )

and so

(61) n ( n 1 ) 2 n 3 D ( c n 2 x 2 ) = n ( n 1 ) 2 n 3 c n 2 D ( x 2 ) + ( n ( n 1 ) 2 n 4 ) ( n 2 ) c n 3 D ( c ) x 2 + ( n ( n 1 ) 2 n 4 ) ( n 2 ) c n 3 x 2 D ( c ) .

Comparing equations (60) and (61) we obtain

(62) 0 = 2 + i = 1 n 3 n 2 i n ( n 1 ) 2 n 3 c D ( x 2 ) + ( ( n + 1 ) ( n 2 ) 2 n 3 ) c D ( x ) x + ( ( n + 1 ) ( n 2 ) 2 n 3 ) c x D ( x ) + ( ( n 2 ) 2 n 3 ) + 2 n 2 2 + i = 1 n 3 n 2 i n 2 i 2 + n 1 2 ( 2 n 2 1 ) + i = 1 n 2 n 1 i 2 ( τ i , n ) ( n ( n 1 ) 2 n 4 ) ( n 2 ) D ( c ) x 2 + ( ( n 2 ) 2 n 3 ) + 2 n 2 2 + i = 1 n 3 n 2 i i 2 + n 1 2 ( 2 n 2 1 ) + i = 1 n 2 i 2 ( τ i , n ) ( n ( n 1 ) 2 n 4 ) ( n 2 ) x 2 D ( c ) + 2 ( ( n 2 ) 2 n 3 ) + 2 i = 1 n 3 n 2 i i ( n 2 i ) + i = 1 n 2 i ( n 1 i ) ( τ i , n ) x D ( c ) x .

For the ease of writing and presenting results we will use the following equalities:

i = 1 n 3 n 2 i = 2 n 2 2 , i = 1 n 3 n 2 i i = ( 2 n 3 1 ) ( n 2 ) , i = 1 n 3 n 2 i i 2 = ( n 2 ) ( 2 n 4 ( n 1 ) + ( 2 n ) ) , i = 1 n 3 n 2 i i 2 = i = 1 n 3 n 2 i n 2 i 2 = ( 2 n 5 2 1 ) ( n 3 ) ( n 2 ) .

i = 1 n 3 n 2 i i ( n 2 i ) = ( n 2 ) i = 1 n 3 n 2 i i i = 1 n 3 n 2 i i 2 = ( n 2 ) ( n 3 ) 2 n 4 .

Equation (62) can now be rewritten as

(63) 0 = ( 2 n 3 ( n 2 ) ( n + 1 ) ) c D ( x 2 ) + ( ( n + 1 ) ( n 2 ) 2 n 3 ) c D ( x ) x + ( ( n + 1 ) ( n 2 ) 2 n 3 ) c x D ( x ) + ( ( n 2 ) 2 n 3 ) + 2 1 2 ( n 3 ) ( n 2 ) + ( ( 2 n 5 2 1 ) ( n 3 ) ( n 2 ) ) + 1 2 ( n 2 ) ( n 1 ) ( 2 n 2 1 ) + i = 1 n 2 n 1 i 2 ( τ i , n ) ( n ( n 1 ) 2 n 4 ) ( n 2 ) D ( c ) x 2 + ( ( n 2 ) 2 n 3 ) + 2 1 2 ( n 3 ) ( n 2 ) + ( ( 2 n 5 2 1 ) ( n 3 ) ( n 2 ) ) + 1 2 ( n 2 ) ( n 1 ) ( 2 n 2 1 ) + i = 1 n 2 i 2 ( τ i , n ) ( n ( n 1 ) 2 n 4 ) ( n 2 ) x 2 D ( c ) + 2 ( ( n 2 ) 2 n 3 ) + 2 ( ( n 2 ) ( n 3 ) 2 n 4 ) + i = 1 n 2 i ( n 1 i ) ( τ i , n ) x D ( c ) x .

For the ease of writing and presenting results we will use the following equalities:

(64) i = 1 n 2 ( n 1 i ) ( τ i , n ) = i = 1 n 2 ( i ) ( τ i , n ) = ( n 1 ) ( 2 n 3 ( n 4 ) + 1 )

(65) i = 1 n 2 n 1 i 2 ( τ i , n ) = i = 1 n 2 i 2 ( τ i , n ) = i = 1 n 2 1 2 ( i 2 i ) ( τ i , n ) = 1 2 i = 1 n 2 i 2 ( τ i , n ) 1 2 i = 1 n 2 ( i ) ( τ i , n ) = 1 2 i = 1 n 2 i 2 ( τ i , n ) 1 2 ( ( n 1 ) ( 2 n 3 ( n 4 ) + 1 ) )

(66) i = 1 n 2 i ( n 1 i ) ( τ i , n ) = ( n 1 ) i = 1 n 2 ( i ) ( τ i , n ) i = 1 n 2 i 2 ( τ i , n ) = ( n 1 ) ( ( n 1 ) ( 2 n 3 ( n 4 ) + 1 ) ) i = 1 n 2 i 2 ( τ i , n ) .

Now using (64), (65), and (66) in equation (63) we obtain

(67) 0 = ( 2 n 3 ( n 2 ) ( n + 1 ) ) c D ( x 2 ) + ( 2 n 3 ( n 2 ) ( n + 1 ) ) c D ( x ) x + ( 2 n 3 ( n 2 ) ( n + 1 ) ) c x D ( x ) + 1 2 ( ( 1 n ) ( 2 n 3 ( n 2 4 n + 2 ) + ( n 1 ) ) + i = 1 n 2 i 2 ( τ i , n ) ) D ( c ) x 2 + 1 2 ( ( 1 n ) ( 2 n 3 ( n 2 4 n + 2 ) + ( n 1 ) ) + i = 1 n 2 i 2 ( τ i , n ) ) x 2 D ( c ) ( 1 n ) ( 2 n 3 ( n 2 4 n + 2 ) + ( n 1 ) ) + i = 1 n 2 i 2 ( τ i , n ) x D ( c ) x .

Now setting c 2 instead of c in the last equation and using D ( c 2 ) = 2 c D ( c ) we obtain

(68) 0 = ( 2 n 3 ( n 2 ) ( n + 1 ) ) c 2 D ( x 2 ) + ( 2 n 3 ( n 2 ) ( n + 1 ) ) c 2 D ( x ) x + ( 2 n 3 ( n 2 ) ( n + 1 ) ) c 2 x D ( x ) + ( 1 n ) ( 2 n 3 ( n 2 4 n + 2 ) + ( n 1 ) ) + i = 1 n 2 i 2 ( τ i , n ) c D ( c ) x 2 + ( 1 n ) ( 2 n 3 ( n 2 4 n + 2 ) + ( n 1 ) ) + i = 1 n 2 i 2 ( τ i , n ) c x 2 D ( c ) 2 ( 1 n ) ( 2 n 3 ( n 2 4 n + 2 ) + ( n 1 ) ) + i = 1 n 2 i 2 ( τ i , n ) c x D ( c ) x .

Subtracting equation (68) from (67) we obtain

( 2 n 3 ( n 2 ) ( n + 1 ) ) c 2 D ( x 2 ) = ( 2 n 3 ( n 2 ) ( n + 1 ) ) c 2 D ( x ) x + ( 2 n 3 ( n 2 ) ( n + 1 ) ) c 2 x D ( x )

and so

D ( x 2 ) = D ( x ) x + x D ( x ) .

Now Herstein theorem completes the proof.□

Acknowledgements

The authors would like to thank the reviewers for their thoughtful comments and efforts toward improving our manuscript.

  1. Conflict of interest: The authors state no conflict of interest.

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Received: 2022-02-04
Revised: 2023-12-01
Accepted: 2023-12-03
Published Online: 2024-02-06

© 2024 the author(s), published by De Gruyter

This work is licensed under the Creative Commons Attribution 4.0 International License.

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