In this paper we prove the following result. Let m ≥ 0 and n ≥ 0 be integers with m+n ≠ 0 and let R be a prime ring with char(R)=0 or m+n+1 ≤ char(R) ≠ 2. Suppose there exists a nonzero additive mapping D:R → R satisfying the relation D(xm+n+1)=(m+n+1)xmD(x)xn for all x R. In this case D is a derivation and R is commutative.
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机译：在本文中，我们证明以下结果。令m≥0和n≥0为m + n≠0的整数，令R为char（R）= 0或m + n + 1≤char（R）≠2的质环。假设存在一个非零加性对于所有x R，满足关系D（xm + n + 1）=（m + n + 1）xmD（x）xn的映射D：R→R。在这种情况下，D是导数，R是可交换的。
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