Correction to: Aequat. Math. 91 (2017), 1025–1040  https://doi.org/10.1007/s00010-017-0514-7

We correct an error in [1, Theorem 4.2(b)]. The correct statement of Theorem 4.2(b) is:

Let \(N\in \mathbb {N}{\setminus } \{1\}\) and \(\gamma \in K^*\). If \(f,g:K \rightarrow K\) are additive solutions of (2) with \(\phi (x)=\gamma x^N\), then there exist a derivation \(D:K \rightarrow K\) and a constant \(a\in K\) such that

$$\begin{aligned} f(x) = \frac{1}{N}D(x) + \left( a - \frac{D(\gamma )}{\gamma N}\right) x\; \text { and }\; g(x) = D(x) + ax \quad \text { for all } x \in K. \end{aligned}$$

The converse is also true.

In the original proof the mistake occurs at the bottom of p. 1030, where an N is missing from the last line of the calculation for f(t). It should read

$$\begin{aligned} f(t) = \frac{D(t)}{N} + at - \frac{D(\gamma )}{\gamma N}t, \quad t \in K. \end{aligned}$$

This gives the form of f shown above (and the formula for g is already established).

For the converse we show that such f and g satisfy (2) with \(\phi (x) = \gamma x^N\) as follows.

$$\begin{aligned} xf&(\gamma x^N) - \gamma x^N g(x)\\&= x\left( \frac{D(\gamma x^N)}{N} + a\gamma x^N - \frac{D(\gamma )}{\gamma N}\gamma x^N \right) - \gamma x^N (D(x) + ax)\\&= \frac{1}{N} \left( xD(\gamma x^N) - D(\gamma ) x^{N + 1}\right) - \gamma x^N D(x)\\&= \frac{1}{N} \left( x[D(\gamma )x^N + \gamma N x^{N-1} D(x)] - D(\gamma ) x^{N + 1}\right) - \gamma x^N D(x)\\&= 0 \end{aligned}$$

for all \(x \in K\).