Correction to: On generalizations of differential uniform permutations over finite fields based on 2-to-1 mappings (Applicable Algebra in Engineering, Communication and Computing, (2025), 10.1007/s00200-025-00691-9)

A careful re-examination of the original work indicates that certain mathematical imprecisions influenced the accuracy of some results. The following corrections are given in detail, using the same notation as in the original paper [1]. Theorem [Thoerem 3.1 [1]], demonstrates that t(x) does not possess the PcN property when g(x) is nonlinear. To address both linear and nonlinear aspects of g(x), it is necessary that t(x) is a permutation. By imposing this permutation condition on t(x), the following result is established as Theorem 1. Let be linear polynomials, and such that with is constant. Let is PP over. Then, t(x) is PcN for any in if permutes. We know that, t(x) is PcN if and only if permutes for any and. For any and, (Formula presented.) Since, for any, let and denoting that the is constant k. We can then write: (Formula presented.) Let. Since is linear, we have (Formula presented.) Then, for, (Formula presented.) The above expression becomes (Formula presented.) Set since is permutation we can write also. Therefore, (Formula presented.) Let us call it thus L is affine map on. Then for every x,, which implies As and t(x) permutes, we can conclude that L permutes. Hence, t(x) is PcN for all in. Following the modifications to the preceding theorem, the subsequent corollaries have been revised to include the necessary conditions. Let T(x) and Q(x) be linear polynomials over. Let, and R(x) in be such that. Let permutes. Then, t(x) is PcN for any in if Q(x) permutes. Let Q(x) and Tr(x) are linear polynomials over. Let, and be such that. Let permutes. Then, t(x) is PcN for any in if Q(x) permutes. In [1], Theorem 3.5 overlaps with Theorem 0.1 presented above; hence, Theorem 3.5 in [1] should be removed. Consequently, proof of Corollary 3.6, which was originally derived from Theorem 3.5, can be reformulated since it directly follows from Theorem 0.1 by substituting,, and. In [1], Theorem 3.7 states that the function K is surjective, it is not enough to prove the PcN property, hence we considered. The statement of the theorem should accordingly be corrected to reflect this. The following is the corrected form of Theorem 3.7. Let and let be surjective. Let be a linear translator with respect to A for the function K. For any polynomial such that, then the function is PcN for any in. We know that, t(x) is PcN if and only if permutes for any and. For any and, (Formula presented.) Denote, then the last term in the above expression becomes. But both belong to so belong to. Then the above expression becomes (Formula presented.) Now we have to prove permutes. By the AGW criteria, we have to prove that induces a permutation on A and is injective on for each. Because K is surjective and is 0-translator, for any x,. Hence. Therefore, in particular, the induced map on A is, where is any preimage of a. Because, this induced map is a permutation of A. To prove that is injective on, let fix and suppose satisfy. Subtracting,. If, then because. Similarly, depends only on the values of. Using the fact that the induced map on A is injective, we can get and all the u-terms cancel, thus we get. Since we get. So is injective on. Even if, the above arguments hold based on the translator property together with. Therefore, is a permutation on. Therefore, is PcN for any in. In [1], Proposition 3.13 refers to s(x) as a Boolean function, although the proof relies on the stronger assumption that s(x) is linear. For clarity and correctness, the statement should thus be understood with s(x) being linear. We present below a corrected formulation of Proposition 3.13. Let be a linearized permutation polynomial. Let s(x) be a linear Boolean function over, and let be a non-zero linear translator of s(x). Then, the function is APcN for any where. Theorem 3.15, as stated in [1], does not explicitly distinguish between the cases where the characteristic q is even or odd. Determining whether the parameter q alone suffices to ensure that the differential equation admits at most two solutions is, in general, insufficient. Consequently, Theorem 0.6 provides a comprehensive analysis by distinguishing and examining the two cases separately. Let q be a prime power, and with such that. if then t(x) is APcN, if and only if, if, t(x) is APcN, if and only if, equivalently, if and only if either t is injective, or and. if then t(x) is APcN, if and only if, if, t(x) is APcN, if and only if, equivalently, if and only if either t is injective, or and. t(x) can be expressed as (Formula presented.) For any and, (Formula presented.) To prove t(x) is APcN, we need to show that the equation has atmost two solutions in for any. If, then, which implies that every x in is a solution. Hence has 2 solutions if or 0 solution if. Hence, t(x) is APcN, if and only if. If implies. Here t(x) is a linear map, hence the solution set is either 0 or a coset of ker t, i.e., 0 or for. If only possiblity is, i.e., t(x) is injective. if either or. Hence, t(x) is APcN, if and only if, equivalently, if and only if either t is injective, or and.