This research is motivated by the recent work's of Ali-Dar [1], Qi-Zhang [5] and Hou-Wang [3]. However, our approach is different from that of the authors of [5] and [3]. A ring R with an involution '*' is called a *-ring or ring with involution '*'. Throughout, we let R be a ring with involution '*' and Z(R), the center of the ring R. Moreover, the sets of all hermitian and skew-hermitian elements of R will be denoted by H(R) and S(R), respectively. The involution is called the first kind if $Z(R)\subseteq H(R)$, otherwise $S(R)\cap Z(R)\ne (0)$(see [2] for details). A ring R is said to be 2-torsion free if 2x=0 (where x ∈ R) implies x=0. A ring R is called prime if aRb=(0) (where a,b ∈ R) implies a=0 or b=0. A derivation on R is an additive mapping $d:R\to R$ such that d(xy)=d(x)y+xd(y) for all x,y ∈ R.

For any x,y ∈ R, the symbol [x,y] will denote the Lie product xy-yx and the symbol ${}_{*}[x,y]$ will denote the skew Lie product xy-yx^*, where '*' is an involution on R. In a recent paper, Hou and Wang [3] extended the concept of skew Lie product as follows: for an integer n ≥ 1, the n-skew Lie product of any two elements x and y is defined by ${}_{*}{[x,y]}_n{=}_{*}[x{,}_{*}{[x,y]}_{n-1}]$, where ${}_{*}{[x,y]}_0=y$, ${}_{*}[x,y]=xy-y{x}^{*}$ and ${}_{*}{[x,y]}_2=x^{2}y-2xy{x}^{*}+y{(x^{*})}^2$. Obviously, for n=1, the skew Lie product and n-skew Lie product coincides. Note that, for n=2, we call it 2-skew Lie product. In [3], Hou and Wang studied the strong 2-skew commutativity preserving maps in prime rings with involution. In fact, they described the form of strong 2-skew commutativity preserving maps on a unital prime ring with involution that contains a non-trivial symmetric idempotent. In [5], Qi and Zhang studied the properties of n-skew Lie product on prime rings with involution and as an application, they characterized n-skew commuting additive maps, i.e.; an additive mapping f on R into itself such that _*[x,f(x)]_n=0 for all x∈ R. In definition of n-skew commuting mapping (defined in [5]), if we consider that f is any map (not necessarily additive) then it is more reasonable to call f a n-skew commuting. To give its precise definition, we make a slight modification in Qi and Zhang's definition for n-skew commuting mapping. For an integer n ≥ 1, a map f of a *-ring R into itself is called n-skew commuting mapping on R if ${}_{*}{[x,f(x)]}_n=0$ for every x∈ R. For an integer n ≥ 1, a map f of a *-ring R into itself is called n-skew centralizing mapping on R if ${}_{*}{[x,f(x)]}_n\in Z(R)$ for every x ∈ R. In particular, for n = 1, 2, we call them 1-skew commuting (resp. 1-skew centralizing) and 2-skew commuting (resp. 2-skew centralizing) mapping.

The objective of this paper is to introduce the notion of n-skew centralizing mappings on *-rings. Further, we investigate the impact of these mappings and describe the nature of prime *-rings which satisfy certain *-differential identities. In particular, for an integer n ≥ 1 we prove that if a 2-torsion free prime ring R with involution '*' of the second kind which admits a nonzero derivation d such that ${}_{*}{[x,d(x)]}_n\in Z(R)$ for all $x\in R (n=1,2)$, then R is commutative. Moreover, some more related results are obtained. Further more, examples prove that, the assumed curtailment can not be relaxed as given.

In order to study the effect of n-skew centralizing mappings, we need the following two lemmas for developing the proofs of our main results. We begin our discussion with the following lemmas:

Lemma 2.1. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If $x^2{x}^{*}\in Z(R)$ for all x∈ R, then R is commutative.

Proof. Linearization of $x^2{x}^{*}\in Z(R)$ gives that $x^2{y}^{*}+xy{x}^{*}+xy{y}^{*}+yx{x}^{*}+yx{y}^{*}+y^2{x}^{*}\in Z(R)$ for all $x,y\in R.$ Taking x=-x in the last expression and combine it with the above relation, we get

(2.1)$$x^2{y}^{*}+xy{x}^{*}+yx{x}^{*}\in Z(R) \text{for all} x,y\in R.$$

Substituting ky for y, where $k\in S(R)\cap Z(R)$, we obtain $(-x^2{y}^{*}+xy{x}^{*}+yx{x}^{*})k\in Z(R)$ for all $x,y\in R.$ Invoking the primeness of R and using the fact that $S(R)\cap Z(R)\ne (0)$, we get

(2.2)$$-x^2{y}^{*}+xy{x}^{*}+yx{x}^{*}\in Z(R) \text{for all} x,y\in R.$$

Combining (2.1) and (2.2), we conclude that $x^2{y}^{*}\in Z(R).$ Replacing y by y^*, we get $x^{2}y\in Z(R)$ for all $x,y\in R.$ This can be further written as $[x^{2}y,w]=0$ for all x,y,w∈ R. Replacing y by ry, we get $x^{2}r[y,w]=0$ for all $x,y,w,r\in R$. Hence by the primeness of the ring R, we are force to conclude that R is commutative.

Lemma 2.2. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If ${}_{*}{[x,x^{*}]}_2\in Z(R)$ for all x ∈ R, then R is commutative.

Proof. By the hypothesis, we have

(2.3)$$x^2{x}^{*}-2x{(x^{*})}^2+{(x^{*})}^3\in Z(R) \text{for all} x\in R.$$

Replacing x by kx in (2.3) where $k\in S(R)\cap Z(R)$, we get

(2.4)$$-x^2{x}^{*}{k}^3-2x{(x^{*})}^2{k}^3-{(x^{*})}^3{k}^3\in Z(R) \text{for all} x\in R.$$

By (2.3) and (2.4), we conclude that $-4x{(x^{*})}^2{k}^3\in Z(R)$ for all x∈ R. Since R is 2-torsion free prime ring and $S(R)\cap Z(R)\ne (0)$, we obtain $x{(x^{*})}^2\in Z(R)$ for all x∈ R. On linearizing we get

(2.5)$$x{x}^{*}{y}^{*}+x{y}^{*}{x}^{*}+x{(y^{*})}^2+y{(x^{*})}^2+y{x}^{*}{y}^{*}+y{y}^{*}{x}^{*}\in Z(R) \text{for all} x,y\in R.$$

Taking x=-x in (2.5) and using (2.5), we obtain

(2.6)$$x{x}^{*}{y}^{*}+x{y}^{*}{x}^{*}+y{(x^{*})}^2\in Z(R) \text{for all} x,y\in R.$$

Substitute ky for y, where $k\in S(R)\cap Z(R)$ in (2.6), we get $2y{(x^{*})}^{2}k\in Z(R)$ for all $x,y\in R.$ This implies that $y{x}^2\in Z(R)$ for all $x,y\in R.$ Henceforth, using the same arguments as we have used in Lemma 2.1, we conclude that R is commutative. This proves the lemma.

Theorem 2.3. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If R admits a nonzero derivation d such that ${}_{*}{[x,d(x)]}_n\in Z(R)$ for all $x\in R(n=1,2)$, then R is commutative.

proof. Case (i) First we discuss the case, "when n=1" and "i.e.",

$${}_{*}[x,d(x)]\in Z(R) \text{for all} x\in R.$$

Linearizing the above expression, we get

$${}_{*}[x,d(y)]+*[y,d(x)]\in Z(R) \text{for all} x,y\in R.$$

That is,

$$xd(y)-d(y)x^{*}+yd(x)-d(x)y^{*}\in Z(R) \text{for all} x,y\in R.$$

This further implies that

$$[xd(y),r]-[d(y)x^{*},r]+[yd(x),r]-[d(x)y^{*},r]=0 \text{for all} x,y,r\in R.$$

Hence

(2.7)$$\begin{array}{l}x[d(y),r]+[x,r]d(y)-d(y)[x^{*},r]-[d(y),r]x^{*}\\ +y[d(x),r]+[y,r]d(x)-d(x)[y^{*},r]-[d(x),r]y^{*}=0 \text{for all} x,y,r\in R.\end{array}$$

Replacing y by hy in (2.7), where $h\in H(R)\cap Z(R)$ and using it, we have

$$(x[y,r]+[x,r]y-y[x^{*},r]-[y,r]x^{*})d(h)=0 \text{for all} x,y,r\in R.$$

Using the primeness of R, we obtain either d(h)=0 or

$$x[y,r]+[x,r]y-y[x^{*},r]-[y,r]x^{*}=0 \text{for all} x,y,r\in R.$$

First we consider the situation

(2.8)$$x[y,r]+[x,r]y-y[x^{*},r]-[y,r]x^{*}=0 \text{for all} x,y,r\in R.$$

Substituting kx for x in (2.8), where $k\in S(R)\cap Z(R)$ and combining it with (2.8), we get

$$2(x[y,r]+[x,r]y)k=0 \text{for all} x,y,r\in R.$$

Since R is 2-torsion free prime ring, we deduce that

$$x[y,r]+[x,r]y=0 \text{for all} x,y,r\in R.$$

Replacing x by z, where z ∈ Z(R), we get [y,r]z=0 for all y,r ∈ R. Henceforth, we conclude that R is commutative. Now consider the case d(h)=0 for all $h\in H(R)\cap Z(R)$. This implies that d(k)=0 for all $k\in S(R)\cap Z(R).$ Replacing y by ky in (2.7), where $k\in S(R)\cap Z(R)$ with d(k)=0 and adding with (2.7), we get

$$2(x[d(y),r]+[x,r]d(y)-d(y)[x^{*},r]-[d(y),r]x^{*}+y[d(x),r]+[y,r]d(x))k=0$$

for all x,y,r ∈ R. Since R is 2-torsion free ring and $S(R)\cap Z(R)\ne (0),$ the above relation implies that

$$x[d(y),r]+[x,r]d(y)-d(y)[x^{*},r]-[d(y),r]x^{*}+y[d(x),r]+[y,r]d(x)=0 \text{for all}$$

$x,y\in R.$ Taking y=h, where $h\in H(R)\cap Z(R)$ and using the fact that d(h)=0, we get [d(x),r]h=0 for all x,r ∈ R and $h\in H(R)\cap Z(R)$. This yields that [d(x),r]=0 for all x,r ∈ R. Hence in view of Posner's [4] first theorem, R is commutative.

Case (ii) Now, we prove the result for n=2 i.e.,

$${}_{*}{[x,d(x)]}_2\in Z(R) \text{for all} x\in R.$$

On expansion we acquire

(2.9)$$x^{2}d(x)-2xd(x)x^{*}+d(x){(x^{*})}^2\in Z(R) \text{for all} x\in R.$$

Replacing x by xh in (2.9), where $h\in H(R)\cap Z(R)$, we obtain

$$(x^3-2x^2{x}^{*}+x{(x^{*})}^2)d(h)h^2\in Z(R) \text{for all} x\in R.$$

Then by the primeness of R we are force to conclude that either $d(h)h^2=0$ or $x^3-2x^2{x}^{*}+x{(x^{*})}^2\in Z(R)$ for all x∈ R. First we consider the case

(2.10)$$x^3-2x^2{x}^{*}+x{(x^{*})}^2\in Z(R) \text{for all} x\in R.$$

Substituting kx for x in (2.10), where $k\in S(R)\cap Z(R)$, we get $(x^3+2x^2{x}^{*}+x{(x^{*})}^2)k^3\in Z(R) \text{for all} x\in R.$ This further implies that

(2.11)$$x^3+2x^2{x}^{*}+x{(x^{*})}^2\in Z(R) \text{for all} x\in R.$$

Subtracting (2.10) from (2.11) and using 2-torsion freeness of R, we obtain $x^2{x}^{*}\in Z(R) \text{for all} x\in R.$ Therefore, by Lemma 2.1, R is commutative. Now consider the second case $d(h)h^2=0$ for all $h\in H(R)\cap Z(R)$. This implies that d(h)=0 for all $h\in H(R)\cap Z(R)$. Therefore, d(k)=0 for all $k\in S(R)\cap Z(R)$. Replacing x by kx in (2.9), where $k\in S(R)\cap Z(R)$ and using the fact that d(k)=0, we obtain

(2.12)$$x^{2}d(x)k^3+2xd(x)x^{*}{k}^3+d(x){(x^{*})}^2{k}^3\in Z(R) \text{for all} x\in R.$$

Application of (2.9) yields $4xd(x)x^{*}{k}^3\in Z(R)$ for all x ∈ R. Since R is 2-torsion free ring and $S(R)\cap Z(R)\ne (0)$, we get

(2.13)$$xd(x)x^{*}\in Z(R) \text{for all} x\in R.$$

Putting x+h in place of x, where $h\in H(R)\cap Z(R)$, we arrive at

(2.14)$$xd(x)h+d(x)x^{*}h+d(x)h^2\in Z(R) \text{for all} x\in R.$$

Taking x=-x in (2.14) and then combining it with the obtained relation, we get $2d(x)h^2\in Z(R)$. This implies that $d(x)\in Z(R)$ for all x∈ R, since the involution '*' is of the second kind. Hence, by Posner's [4] first theorem, R is commutative.

Theorem 2.4. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If R admits a nonzero derivation d such that $d{(}_{*}{[x,x^{*}]}_n)\in Z(R)$ for all x ∈ R (n=1,2), then R is commutative.

proof. Case (i) Firstly we are focus to discuss the case when n=1 i.e.,

$$d{(}_{*}[x,x^{*}])\in Z(R) \text{for all} x\in R.$$

Linearizing this, we get

$$d{(}_{*}[x,y^{*}])+d(*[y,x^{*}])\in Z(R) \text{for all} x,y\in R.$$

This implies that

(2.15)$$\begin{array}{l}d(x)[y^{*},r]+[d(x),r]y^{*}+x[d(y^{*}),r]+[x,r]d(y^{*})-d(y^{*})[x^{*},r]-[d(y^{*}),r]x^{*}-y^{*}[d(x^{*}),r]\\ -[y^{*},r]d(x^{*})+d(y)[x^{*},r]+[d(y),r]x^{*}+y[d(x^{*}),r]+[y,r]d(x^{*})-d(x^{*})[y^{*},r]-\\ [d(x^{*}),r]y^{*}-x^{*}[d(y^{*}),r]-[x^{*},r]d(y^{*})=0 \text{for all} x,y,r\in R.\end{array}$$

Replacing y by hy, where $h\in H(R)\cap Z(R)$ in (2.15), we obtain

$$\begin{array}{l}(x[y^{*},r]+[x,r]y^{*}-y^{*}[x^{*},r]-[y^{*},r]x^{*}+y[x^{*},r]+\\ [y,r]x^{*}-x^{*}[y^{*},r]-[x^{*},r]y^{*})d(h)=0 \text{for all} x,y,r\in R.\end{array}$$

Using the primeness of R, we have either d(h)=0 or

(2.16)$$x[y^{*},r]+[x,r]y^{*}-y^{*}[x^{*},r]-[y^{*},r]x^{*}+y[x^{*},r]+[y,r]x^{*}-x^{*}[y^{*},r]-[x^{*},r]y^{*}=0$$

for all x,y,r ∈ R. We first consider the relation (2.16). Replacing y by ky, where $k\in S(R)\cap Z(R)$ in (2.16), we get $2(y[x^{*},r]+[y,r]x^{*})k=0$ for all $x,y,r\in R$. Since R is 2-torsion free ring and $S(R)\cap Z(R)\ne (0)$, we obtain $y[x^{*},r]+[y,r]x^{*}=0$ for all $x,y,r\in R.$ Taking x=k, where $k\in S(R)\cap Z(R)$, we get -[y,r]k=0 for all y,r∈ R. Thus -[y,r]=0 for all y,r∈ R. That is, R is commutative. Now consider d(h)=0 for all $h\in H(R)\cap Z(R)$. This implies that d(k)=0 for all $k\in S(R)\cap Z(R).$ Replacing y by ky, where $k\in S(R)\cap Z(R)$ in (2.15) and making use of (2.15), we get

$$2(d(y)[x^{*},r]+[d(y),r]x^{*}+y[d(x^{*}),r]+[y,r]d(x^{*}))k=0 \text{for all} x,y,r\in R.$$

This implies that

$$d(y)[x^{*},r]+[d(y),r]x^{*}+y[d(x^{*}),r]+[y,r]d(x^{*})=0 \text{for all} x,y,r\in R.$$

Taking x=h where $h\in H(R)\cap Z(R)$ and using d(h)=0, we arrive at h[d(y),r]=0 for all y,r ∈ R. Then by the primeness of R and the fact that $S(R)\cap Z(R)\ne (0)$, we obtain [d(y),r]=0 for all y,r∈ R. Hence by Posner's [4] first theorem, R is commutative.

Case (ii) Next, for n=2 we have

$$d{(}_{*}{[x,x^{*}]}_2)\in Z(R) \text{for all } x\in R.$$

On expansion we get

(2.17)$$d(x^2{x}^{*})-2d(x{(x^{*})}^2)+d({(x^{*})}^3)\in Z(R) \text{for all} x\in R.$$

Linearization of (2.17) yields

(2.18)$$\begin{array}{l}d(x^2{y}^{*})+d(xy{x}^{*})+d(xy{y}^{*})+d(yx{x}^{*})+d(yx{y}^{*})+d(y^2{x}^{*})-2d(x{x}^{*}{y}^{*})-2d(x{y}^{*}{x}^{*})\\ -2d(x{(y^{*})}^2)-2d(y{(x^{*})}^2)-2d(y{x}^{*}{y}^{*})-2d(y{y}^{*}{x}^{*})+d({(x^{*})}^2{y}^{*})+d(x^{*}{y}^{*}{x}^{*})\\ +d(x^{*}{(y^{*})}^2)+d(y^{*}{(x^{*})}^2)+d(y^{*}{x}^{*}{y}^{*})+d({(y^{*})}^2{x}^{*})\in Z(R) \text{for all} x,y\in R.\end{array}$$

Substituting -x for x in (2.18) and combining the obtained relation with (2.18), we obtain

$$\begin{array}{l}2(d(x^2{y}^{*})+d(xy{x}^{*})+d(yx{x}^{*})-2d(x{x}^{*}{y}^{*})-2d(x{y}^{*}{x}^{*})-2d(y{(x^{*})}^2)\\ +d({(x^{*})}^2{y}^{*})+d(x^{*}{y}^{*}{x}^{*})+d(y^{*}{(x^{*})}^2))\in Z(R) \text{for all} x,y\in R.\end{array}$$

Since R is 2-torsion free, the last relation gives

(2.19)$$\begin{array}{l}d(x^2{y}^{*})+d(xy{x}^{*})+d(yx{x}^{*})-2d(x{x}^{*}{y}^{*})-2d(x{y}^{*}{x}^{*})-2d(y{(x^{*})}^2)\\ +d({(x^{*})}^2{y}^{*})+d(x^{*}{y}^{*}{x}^{*})+d(y^{*}{(x^{*})}^2)\in Z(R) \text{for all} x,y\in R.\end{array}$$

Replacing y by hy, where $h\in H(R)\cap Z(R)$ in (2.19) and intermix it with (2.19), we come to

$$\begin{array}{l}(x^2{y}^{*}+xy{x}^{*}+yx{x}^{*}-2x{x}^{*}{y}^{*}-2x{y}^{*}{x}^{*}-2y{(x^{*})}^2\\ +{(x^{*})}^2{y}^{*}+x^{*}{y}^{*}{x}^{*}+y^{*}{(x^{*})}^2)d(h)\in Z(R) \text{for all} x,y\in R.\end{array}$$

By the primeness of the ring R, we get either d(h)=0 or

(2.20)$$x^2{y}^{*}+xy{x}^{*}+yx{x}^{*}-2x{x}^{*}{y}^{*}-2x{y}^{*}{x}^{*}-2y{(x^{*})}^2+{(x^{*})}^2{y}^{*}+x^{*}{y}^{*}{x}^{*}+y^{*}{(x^{*})}^2\in Z(R)$$

for all x,y ∈ R. Replacing y by ky, where $k\in S(R)\cap Z(R)$ in (2.20), we arrive at

$$2(xy{x}^{*}+yx{x}^{*}-2y{(x^{*})}^2)\in Z(R) \text{for all} x,y\in R.$$

This implies that

$$xy{x}^{*}+yx{x}^{*}-2y{(x^{*})}^2\in Z(R) \text{for all} x,y\in R.$$

Substituting kx for x in the last relation, we conclude that $y{(x^{*})}^2\in Z(R)$ for all x,y∈ R. Now proceed as we have already done in Lemma 2.1, we conclude that R is commutative. "Considering the second case in which we have d(h)=0 for all $h\in H(R)\cap Z(R).$" This implies that d(k)=0 for all $k\in S(R)\cap Z(R).$ Now replacing x by kx in (2.19) and using "d(k)=0, we get"

$$\begin{array}{l}(-d(x^2{y}^{*})+d(xy{x}^{*})+d(yx{x}^{*})+2d(x{x}^{*}{y}^{*})+2d(x{y}^{*}{x}^{*})-2d(y{(x^{*})}^2)\\ -d({(x^{*})}^2{y}^{*})-d(x^{*}{y}^{*}{x}^{*})-d(y^{*}{(x^{*})}^2))k\in Z(R) \text{for all} x,y\in R.\end{array}$$

Since $S(R)\cap Z(R)\ne (0)$, the last expression implies that

$$\begin{array}{l}-d(x^2{y}^{*})+d(xy{x}^{*})+d(yx{x}^{*})+2d(x{x}^{*}{y}^{*})+2d(x{y}^{*}{x}^{*})-2d(y{(x^{*})}^2)\\ -d({(x^{*})}^2{y}^{*})-d(x^{*}{y}^{*}{x}^{*})-d(y^{*}{(x^{*})}^2)\in Z(R) \text{for all} x,y\in R.\end{array}$$

Combining this with (2.19) and using the fact that R is 2-torsion free ring, we arrive at

$$d(xy{x}^{*})+d(yx{x}^{*})-2d(y{(x^{*})}^2)\in Z(R) \text{for all} x,y\in R.$$

Replacing x by kx, where $k\in S(R)\cap Z(R)$ and combining it with previous expression, we obtain $2d(y{(x^{*})}^2)\in Z(R)$ for all x,y ∈ R. Replacing x by h, where $h\in H(R)\cap Z(R)$ we come to $d(y)h^2\in Z(R)$ for all y ∈ R. This implies that $d(y)\in Z(R)$ for all y ∈ R. Hence, R is commutative by Posner's [4] first theorem.

Theorem 2.5. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If R admits a derivation d such that ${}_{*}{[x,d(x)]}_2{+}_{*}{[x,x^{*}]}_2\in Z(R)$ for all x ∈ R, then R is commutative.

Proof. By the hypothesis we assume that

(2.21)$${}_{*}{[x,d(x)]}_2{+}_{*}{[x,x^{*}]}_2\in Z(R) \text{for all} x\in R.$$

If we take d=0. Then, application of Lemma 2.2, yields the required result. Now consider the case d ≠ 0 and on expansion of (2.21), we get

(2.22)$$x^{2}d(x)-2xd(x)x^{*}+d(x){(x^{*})}^2+x^2{x}^{*}-2x{(x^{*})}^2+{(x^{*})}^3\in Z(R) \text{for all} x\in R.$$

Replacing x by xh in (2.21), where $h\in H(R)\cap Z(R)$, we obtain $(x^3-x^2{x}^{*}+x{(x^{*})}^2)d(h)h^2\in Z(R)$ for all x ∈ R. Now by the primeness of R we get either $x^3-2x^2{x}^{*}+x{(x^{*})}^2\in Z(R)$ for all x ∈ R or $d(h)h^2=0$. Now, we suppose that

(2.23)$$x^3-2x^2{x}^{*}+x{(x^{*})}^2\in Z(R) \text{for all} x\in R.$$

This is same as the relation (2.10) in Theorem 2.3 and hence we conclude that R is commutative. Now we consider the case $d(h)h^2=0$ for all $h\in H(R)\cap Z(R)$. Since R is prime ring, so we get d(h)=0. This also implies that d(k)=0 for all $k\in S(R)\cap Z(R).$ Replacing x by xk in (2.22) and combining with (2.22) we arrive at $(2xd(x)x^{*}-x^2{x}^{*}-{(x^{*})}^3)k^3\in Z(R)$ for all x∈ R. Since $S(R)\cap Z(R)\ne (0)$, so by the primeness of R, we get

(2.24)$$2xd(x)x^{*}-x^2{x}^{*}-{(x^{*})}^3\in Z(R) \text{for all} x\in R.$$

Linearization of (2.24) gives

(2.25)$$\begin{array}{l}2xd(x)y^{*}+2xd(y)x^{*}+2xd(y)y^{*}+2yd(x)x^{*}+2yd(x)y^{*}+2yd(y)x^{*}-x^2{y}^{*}-xy{x}^{*}\\ -xy{y}^{*}-yx{x}^{*}-yx{y}^{*}-y^2{x}^{*}-{(x^{*})}^2{y}^{*}-x^{*}{y}^{*}{x}^{*}-x^{*}{(y^{*})}^2-y^{*}{(x^{*})}^2-y^{*}{x}^{*}{y}^{*}-{(y^{*})}^2{x}^{*}\end{array}$$

$\in Z(R)$ for all x,y ∈ R.Replacing x by -x in (2.25) and combining the obtained relation with (2.25), we obtain

$$2xd(x)y^{*}+2xd(y)x^{*}+2yd(x)x^{*}-x^2{y}^{*}-xy{x}^{*}-yx{x}^{*}-{(x^{*})}^2{y}^{*}-x^{*}{y}^{*}{x}^{*}-y^{*}{(x^{*})}^2$$

$\in Z(R)$ for all x,y ∈ R. Taking x=h, where $h\in H(R)\cap Z(R)$, we get

$$(2d(y)-4y^{*}-2y)h^2\in Z(R) \text{for all} y\in R.$$

The primeness of R yields that $d(y)-2y^{*}-y\in Z(R)$ for all y ∈ R. Replacing y by ky and on solving, we get y ∈ Z(R) for all y ∈ R. Hence, this conclude that R is commutative.

Theorem 2.6. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If R admit two distinct derivations d₁ and d₂ such that ${}_{*}{[x,d_1(x)]}_2{-}_{*}{[x,d_2(x)]}_2\in Z(R)$ for all x ∈ R, then R is commutative.

Proof. We assume that

(2.26)$${}_{*}{[x,d_1(x)]}_2{-}_{*}{[x,d_2(x)]}_2\in Z(R) \text{for all} x\in R.$$

If either d₁ or d₂ is zero, then we get the required result by Theorem 2.3 above. Now consider both d₁, d₂ are non-zero. Expansion of (2.26) yields that

(2.27)$$x^2{d}_1(x)-2x{d}_1(x)x^{*}+d_1(x){(x^{*})}^2-x^2{d}_2(x)+2x{d}_2(x)x^{*}-d_2(x){(x^{*})}^2\in Z(R)$$

for all x ∈ R. Replacing x by xh in (2.27), where $h\in H(R)\cap Z(R)$ and on simplifying with the help of (2.27), we get

$$(x^3-2x^2{x}^{*}+x{(x^{*})}^2)(d_1(h)-d_2(h))h^2\in Z(R) \text{for all} x\in R.$$

This implies either $x^3-2x^2{x}^{*}+x{(x^{*})}^2\in Z(R)$ for all x ∈ R or $(d_1(h)-d_2(h))h^2=0$. If $x^3-2x^2{x}^{*}+x{(x^{*})}^2\in Z(R)$ for all x∈ R, then by using the same steps as we have used after (2.10), we arrive at $x^2{x}^{*}\in Z(R)$ for all x ∈ R. Thus R is commutative, by Lemma 2.1. On the other hand, if $(d_1(h)-d_2(h))h^2=0$ for all $h\in H(R)\cap Z(R).$ Then we are force to conclude that $d_1(h)=d_2(h)$ and hence $d_1(k)=d_2(k)$ for all $k\in S(R)\cap Z(R).$ Replacing x by kx in (2.27), and combining with (2.27) by using the fact that $d_1(k)=d_2(k)$, we get

$$4(x{d}_1(x)x^{*}-x{d}_2(x)x^{*})k^3\in Z(R) \text{for all} x\in R.$$

Since R is 2-torsion free and $S(R)\cap Z(R)\ne (0),$ the last relation gives

(2.28)$$x{d}_1(x)x^{*}-x{d}_2(x)x^{*}\in Z(R) \text{for all} x\in R.$$

Linearizing (2.28), we obtain

(2.29)$$\begin{array}{l}x{d}_1(x)y^{*}+y{d}_1(x)x^{*}+y{d}_1(x)y^{*}+x{d}_1(y)x^{*}+x{d}_1(y)y^{*}+y{d}_1(y)x^{*}-x{d}_2(x)y^{*}-x{d}_2(y)x^{*}\\ -x{d}_2(y)y^{*}-y{d}_2(x)x^{*}-y{d}_2(x)y^{*}-y{d}_2(y)x^{*}\in Z(R) \text{for all} x,y\in R.\end{array}$$

Replacing x by -x in (2.29) and combining the obtained result with (2.29), we get

$$2(x{d}_1(x)y^{*}+y{d}_1(x)x^{*}+x{d}_1(y)x^{*}-x{d}_2(x)y^{*}-x{d}_2(y)x^{*}-y{d}_2(x)x^{*})\in Z(R)$$

for all x,y ∈ R. Since R is 2-torsion free ring, the above expression yields

(2.30)$$x{d}_1(x)y^{*}+y{d}_1(x)x^{*}+x{d}_1(y)x^{*}-x{d}_2(x)y^{*}-x{d}_2(y)x^{*}-y{d}_2(x)x^{*}\in Z(R)$$

for all x,y∈ R. Replacing x by kx in (2.30) and on solving with the help (2.30) and using the fact that $d_1(k)=d_2(k)$, we get $(x{d}_1(x)-x{d}_2(x))y^{*}\in Z(R)$ for all x,y ∈ R. Replacing y by h, where $h\in H(R)\cap Z(R)$. Then by the primeness of R and $S(R)\cap Z(R)\ne (0)$ condition force that $x{d}_1(x)-x{d}_2(x)\in Z(R)$ for all x ∈ R. Linearizing this we get $x{d}_1(y)+y{d}_1(x)-x{d}_2(y)-y{d}_2(x)\in Z(R)$ for all x,y ∈ R. Taking y by h where $h\in H(R)\cap Z(R)$ and using $d_1(h)=d_2(h),$ we obtain $d_1(x)-d_2(x)\in Z(R)$ for all x ∈ R. This can be further written as

(2.31)$$[d_1(x),r]-[d_2(x),r]=0 \text{for all} x,r\in R.$$

Replacing x by xr in (2.31), we get $[x,r](d_1(r)-d_2(r))=0$ for all x,r ∈ R. Substitute xu for x in the last relation, we obtain $[x,r]u(d_1(r)-d_2(r))=0$ for all x,r,u ∈ R. Then by the primeness of R, for each fixed r ∈ R, we get either [x,r]=0 for all x ∈ R or $d_1(r)-d_2(r)=0$. Define $A=\{r\in R | [x,r]=0 \text{for all} x\in R\}$ and $B=\{r\in R | d_1(r)-d_2(r)=0\}$. Clearly, A and B are additive subgroups of R whose union is R. Hence by Brauer's trick, either A=R or B=R. If A=R, then [x,r]=0 for all x,r∈ R. This implies that R is commutative. If B=R, then $d_1(r)=d_2(r)$ for all r∈ R, which is a contradiction to our assumption. Hence, we conclude that R is commutative.

Theorem 2.7. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If R admit derivations d₁, d₂ such that at least one of them is nonzero and satisfies $d_1{(}_{*}{[x,x^{*}]}_2){+}_{*}{[x,d_2(x^{*})]}_2\in Z(R)$ for all x ∈ R, then R is commutative.

Proof. We are given that d₁ and d₂ are derivations of R such that

(2.32)$$d_1{(}_{*}{[x,x^{*}]}_2){+}_{*}{[x,d_2(x^{*})]}_2\in Z(R) \text{for all} x\in R.$$

If d₂ is zero then by Theorem 2.4, we get R is commutative. If d₁ is zero then we have ${}_{*}{[x,d_2(x^{*})]}_2\in Z(R)$ for all x∈ R. Expansion of last relation gives

(2.33)$$x^2{d}_2(x^{*})-2x{d}_2(x^{*})x^{*}+d_2(x^{*}){(x^{*})}^2\in Z(R) \text{for all} x\in R.$$

Replacing x by hx, where $h\in H(R)\cap Z(R)$ in (2.33) and combining the obtained expression, we get ${}_{*}{[x,x^{*}]}_2{d}_2(h)h^2\in Z(R)$ for all x ∈ R. Now applying the primeness of the ring R, we get either ${}_{*}{[x,x^{*}]}_2\in R$ or $d(h)h^2=0$. If ${}_{*}{[x,x^{*}]}_2\in Z(R)$ for all x∈ R, then by Lemma 2.2, we get R is commutative. Now consider the second case in which we have $d_2(h)h^2=0$ for all $h\in H(R)\cap Z(R).$ This implies that $d_2(h)=0,$ from here we get $d_2(k)=0$ for all $k\in S(R)\cap Z(R).$ Replacing x by kx in (2.30) and using the fact that $d_2(k)=0$, we get $4x{d}_2(x^{*})x^{*}{k}^3\in Z(R)$ for all x∈ R. This implies that $x{d}_2(x^{*})x^{*}\in Z(R)$ for all x ∈ R. Arguing as above after (2.13), we conclude that R is commutative.

Now consider the second case in which both d₁ and d₂ are nonzero. On expansion of (2.32), we have

(2.34)$$d_1(x^2{x}^{*})-2d_1(x{(x^{*})}^2)+d_1({(x^{*})}^3)+x^2{d}_2(x^{*})-2x{d}_2(x^{*})x^{*}+d_2(x^{*}){(x^{*})}^2\in Z(R)$$

for all x ∈ R. Replacing x by hx, where $h\in H(R)\cap Z(R)$ in (2.34) and solving with the help of (2.34), we get

$${}_{*}{[x,x^{*}]}_2(3d_1(h)+d_2(h))h^2\in Z(R) \text{for all} x\in R.$$

By the primeness of the ring R, we get either ${}_{*}{[x,x^{*}]}_2\in Z(R)$ for all x∈ R or $(3d_1(h)+d_2(h))h^2=0$ for all $h\in H(R)\cap Z(R).$ If ${}_{*}{[x,x^{*}]}_2\in Z(R)$ for all x ∈ R, then by Lemma 2.2, we get R is commutative. Now consider the case $(3d_1(h)+d_2(h))h^2=0$. This implies that $d_2(h)=-3d_1(h)$ and hence $d_2(k)=-3d_1(k)$ for all $k\in S(R)\cap Z(R).$ Now substituting kx for x, where $k\in S(R)\cap Z(R)$ in (2.34) and combining the obtained result with (2.34), we get $4(d_1(x{(x^{*})}^2)+x{d}_2(x^{*})x^{*})k^3\in Z(R)$ for all x ∈ R. Since R is 2-torsion free ring and $S(R)\cap Z(R)\ne (0)$, then invoking the primeness of R we obtain $d_1(x{(x^{*})}^2)+x{d}_2(x^{*})x^{*}\in Z(R)$ for all x∈ Z(R). Linearization to the last expression gives

(2.35)$$\begin{array}{l}{d}_1(x{x}^{*}{y}^{*})+d_1(x{y}^{*}{x}^{*})+d_1(x{(y^{*})}^2)+d_1(y{(x^{*})}^2)+d_1(y{x}^{*}{y}^{*})+d_1(y{y}^{*}{x}^{*})+x{d}_2(x^{*})y^{*}\\ +x{d}_2(y^{*})x^{*}+x{d}_2(y^{*})y^{*}+y{d}_2(x^{*})x^{*}+y{d}_2(x^{*})y^{*}+y{d}_2(y^{*})x^{*}\in Z(R) \text{ for all} x,y\in R.\end{array}$$

Replacing x by -x in (2.35), we get

$$2(d_1(x{x}^{*}{y}^{*})+d_1(x{y}^{*}{x}^{*})+d_1(y{(x^{*})}^2)+x{d}_2(x^{*})y^{*}+x{d}_2(y^{*})x^{*}+y{d}_2(x^{*})x^{*})\in Z(R)$$

for all x,y ∈ R. Since R is 2-torsion free ring, we get

(2.36)$$d_1(x{x}^{*}{y}^{*})+d_1(x{y}^{*}{x}^{*})+d_1(y{(x^{*})}^2)+x{d}_2(x^{*})y^{*}+x{d}_2(y^{*})x^{*}+y{d}_2(x^{*})x^{*}\in Z(R)$$

for all x,y ∈ R. Substituting ky for y, where $k\in S(R)\cap Z(R)$ in (2.36) and combining with (2.36) with use of $d_2(k)=-3d_1(k)$, we arrive at

(2.37)$$2d_1(y{(x^{*})}^2)k+2y{d}_2(x^{*})x^{*}k-x{x}^{*}{y}^{*}{d}_1(k)+y{(x^{*})}^2{d}_1(k)+2x{y}^{*}{x}^{*}{d}_1(k)\in Z(R)$$

for all x,y ∈ R and $k\in S(R)\cap Z(R)$. Substitute ky for y in (2.37) yields

(2.38)$$\begin{array}{l}2d_1(y{(x^{*})}^2)k^2+2y{d}_2(x^{*})x^{*}{k}^2+x{x}^{*}{y}^{*}{d}_1(k)k+y{(x^{*})}^2{d}_1(k)k\\ -2x{y}^{*}{x}^{*}{d}_1(k)k+2y{(x^{*})}^2{d}_1(k)k\in Z(R) \text{for all} x,y\in R.\end{array}$$

Subtracting (2.38) form (2.37), we get $(-2x{x}^{*}{y}^{*}+4x{y}^{*}{x}^{*}-2y{(x^{*})}^2)d_1(k)k\in Z(R)$ for all $x,y\in R$. Since R is 2-torsion free prime ring and $S(R)\cap Z(R)\ne (0)$, the last expression forces that either $x{x}^{*}{y}^{*}-2x{y}^{*}{x}^{*}+y{(x^{*})}^2\in Z(R)$ for all x,y ∈ R or $d_1(k)k=0$. Suppose

(2.39)$$x{x}^{*}{y}^{*}-2x{y}^{*}{x}^{*}+y{(x^{*})}^2\in Z(R) \text{for all} x,y\in R.$$

Substituting ky for y, where $k\in S(R)\cap Z(R)$ in (2.39) and combining with (2.39), we get $2y{(x^{*})}^{2}k\in Z(R)$ for all x,y ∈ R. Taking x=k, we obtain $2y{k}^3\in Z(R)$ for all y∈ R. Since R is 2-torsion free prime ring and $S(R)\cap Z(R)\ne (0)$, we conclude that R is commutative. Now consider the case in which we have d₁(k)k=0 for all $k\in S(R)\cap Z(R).$ This implies that d₁(k)=0 for all $k\in S(R)\cap Z(R)$. This further implies that $d_2(k)=0$. Substitute k for x in (2.36), to get

(2.40)$$-2d_1(y^{*})k^2+d_1(y)k^2-d_2(y^{*})k^2\in Z(R) \text{for all} y\in R.$$

Replacing y by ky, where $k\in S(R)\cap Z(R)$ in (2.40) and combining the obtained relation with (2.40), finally we get $2d_1(y)k^3\in Z(R)$ for all y∈ R. Since R is 2-torsion free ring and $S(R)\cap Z(R)\ne (0)$, we obtain $d_1(y)\in Z(R)$ for all y ∈ R. Hence, by Posner's [4] first theorem, R is commutative.

As an immediate consequence of the above theorem, we get the following corollary:

Corollary 2.8. Let R be a 2-torsion free prime ring with involution '*' of the second kind. If R admits a nonzero derivation d such that $d{(}_{*}{[x,x^{*}]}_2){+}_{*}{[x,d(x^{*})]}_2\in Z(R)$ for all x ∈ R, then R is commutative.

The following example shows that the second kind involution assumption is essential in Theorem 2.3 and Theorem 2.4.

Example 2.9. Let $R=\left\{\left(\begin{array}{cc}{\beta }_1& {\beta }_2\\ {\beta }_3& {\beta }_4\end{array}\right)| {\beta }_1,{\beta }_2,{\beta }_3,{\beta }_4\in \mathbb{Z}\right\}.$ Of course, R with matrix addition and matrix multiplication is a noncommutative prime ring. Define mappings $*,d_1,d_2:R\to R$ such that ${\left(\begin{array}{cc}{\beta }_1& {\beta }_2\\ {\beta }_3& {\beta }_4\end{array}\right)}^{*}=\left(\begin{array}{cc}{\beta }_4& -{\beta }_2\\ -{\beta }_3& {\beta }_1\end{array}\right)$, $d_1\left(\begin{array}{cc}{\beta }_1& {\beta }_2\\ {\beta }_3& {\beta }_4\end{array}\right)=\left(\begin{array}{cc}0& -{\beta }_2\\ {\beta }_3& 0\end{array}\right)$ and $d_2\left(\begin{array}{cc}{\beta }_1& {\beta }_2\\ {\beta }_3& {\beta }_4\end{array}\right)=\left(\begin{array}{cc}0& {\beta }_2\\ -{\beta }_3& 0\end{array}\right)$

Obviously, $Z(R)=\left\{\left(\begin{array}{cc}{\beta }_1& 0\\ 0& {\beta }_1\end{array}\right)| {\beta }_1\in \mathbb{Z}\right\}.$ Then $x^{*}=x$ for all $x\in Z(R)$, and hence $Z(R)\subseteq H(R)$, which shows that the involution * is of the first kind. Moreover, d₁ and d₂ are nonzero derivations of R such that ${}_{*}{[x,d_1(x)]}_2\in Z(R)$ and ${}_{*}{[x,d_1(x)]}_2{-}_{*}{[x,d_2(x)]}_2\in Z(R)$ for all x∈ R. However, R is not commutative. Hence, the hypothesis of second kind involution is crucial in Theorem 2.3 & 2.4. Our next example shows that Theorems 2.3 and 2.4 are not true for semiprime rings.

Example 2.10. Let $S=R\times ℂ,$ where R is same as in Example 2.9 with involution '*' and derivations d₁ and d₂ same as in above example, $ℂ$ is the ring of complex numbers with conjugate involution τ. Hence, S is a 2-torsion free noncommutative semiprime ring. Now define an involution α on S, as ${(x,y)}^{\alpha }=(x^{*},y^{\tau })$. Clearly, α is an involution of the second kind. Further, we define the mappings D₁ and D₂ from S to S such that $D_1(x,y)=(d_1(x),0)$ and $D_2(x,y)=(d_2(x),0)$ for all (x,y)∈ S. It can be easily checked that D₁, D₂ are derivations on S and satisfying $\alpha {[X,D_1(X)]}_2\in Z(S)$ and $\alpha {[X,D_1(X)]}_2-\alpha {[X,D_2(X)]}_2\in Z(S)$ for all X ∈ S, but S is not commutative. Hence, in Theorems 2.3 & 2.4, the hypothesis of primeness is essential.

We conclude the paper with the following Conjectures.

Conjecture 2.11. Let $n>2$ be an integer, R be a prime ring with involution '*' of the second kind and with suitable torsion restrictions on R. Next, let d be a nonzero derivation on R such that ${}_{*}{[x,d(x)]}_n\in Z(R)$ for all x∈ R. Then what we can say about the structure of R or the form of d?

Conjecture 2.12. Let $n>2$ be an integer, R be a prime ring with involution '*' of the second kind and with suitable torsion restrictions on R. Next, let d be a nonzero derivation on R such that $d{(}_{*}{[x,x^{*}]}_n)\in Z(R)$ for all x∈ R. Then what we can say about the structure of R or the form of d?

Conjecture 2.13. Let $n>2$ be an integer, R be a prime ring with involution '*' of the second kind and with suitable torsion restrictions on R. Next, let d be a nonzero derivation on R such that $d{(}_{*}{[x,x^{*}]}_n){+}_{*}{[x,d(x^{*})]}_n\in Z(R)$ for all x∈ R. Then what we can say about the structure of R or the form of d?