Let , be two positive integers and denote the finite field with elements. An S-box is a vectorial Boolean function from to , also called an -function. The security of most modern block ciphers deeply relies on cryptographic properties of their S-boxes since S-boxes usually are the only nonlinear elements of these cryptosystems. It is therefore significant to employ S-boxes with good cryptographic properties in order to resist various kinds of cryptanalytic attacks.
Differential attack  is one of the most fundamental cryptanalytic approaches targeting symmetric-key primitives and is the first statistical attack for breaking iterated block ciphers. The differential uniformity of S-boxes, which was introduced by Nyberg in , can be used to measure how well the S-box used in a cipher could resist the differential attack.
Let be the finite field of elements. A function defined over is called differentially -uniform, where and
We call the function perfect nonlinear (PN) or almost perfect nonlinear (APN), if or
, respectively. It is well-known that PN functions only exists for an odd prime power. Thus, when is even, APN functions have the best resistance to differential attacks. To analyze the ciphers using modular multiplication as primitive operations more effectively, the authors in  proposed the concept of multiplicative differential. Very recently, based on this new type of differential, Ellingsen, Felke, Riera, Stǎnicǎ and Tkachenko gave the definition of -differential uniformity in :
Let be a prime power and be the finite field with elements. Given a function , the (multiplicative) -derivative of with respect to is defined as
Then is called differentially -uniform.
Note that if or , then is just a shift of or trivially . If , then becomes the usual derivative and the -differential uniformity becomes differential uniformity in Definition 1. Similarly, we call a function perfect -nonlinear (PcN) or almost perfect -nonlinear (APcN), if or , respectively. It is worth noting that PcN functions exist for even , which is a big difference between PN and PcN properties. So far as we know, there are only very few results about PcN and APcN functions. The -differential property of some power functions including Inverse functions, Gold functions, etc., have been investigated [6, 11, 13, 14]. In  the authors give a necessary and sufficient condition for the Gold functions to be PcN, they further conjectured that all the PcN functions in are linear functions, Gold functions and their inverses. Several ideas including the AGW criterion, cyclotomic method, the perturbing and swapping method [3, 8, 10] have been used to construct functions with low -differential uniformity.
In this paper, we prove that this special power permutation over is APcN on satisfying , where . By introducing two parameters and satisfying , we transform the APcN problem into solving a two-equation system on and . Then a new variable is used to induce an equation with algebraic degree four, which help us give the final proof. To the best of our knowledge, there are only two classes of APcN power functions over the finite fields with even characteristic, the first one is the well-known Inverse functions, the second one is the power functions proposed in this paper.
Let be a prime power, and are two finite fields with . Then can be seen as a subfield of and the relative trace from to is defined as
If , we call the above trace Absolute. Given a finite field , let be a positive integer and , define
which is constituted by all th root of unity in . A very important such set is the unit circle of when , which is exactly defined as
The following lemma describes exactly the conditions that a quadratic equation has one or two solutions in the unit circle.
 Let be an even positive integer and satisfy . Then the quadratic equation has
(1) both solutions in the unit circle if and only if
(2) exactly one solution in the unit circle, if and only if
An important fact about the unit circle is the polar-decomposition of elements, i.e., each element in can be uniquely written as
where and . Let be a power function on . Note that , where is defined as Definition 2. Hence the differential characteristics of are completely determined by the values of for . Let be defined as follows:
The -differential spectrum of at the point is the set of with :
3 The APcN property of the power functions
From now on, we always assume that is a positive integer, and . We further assume that , and we will prove that at all such ’s the power function is APcN. Note that for such , we have , due to that and are two solutions in of the quadratic equation . The polar decomposition for any , with and , will be frequently used in later discussions, and we always denote for convenience. To be preparations, several useful propositions are given as follows:
(1) For , if , then
(2) For satisfying , indicates .
(1) Let for some , then if and only if . The denominator
where due to . The numerator equals
(2) From (1), if and only if
in which the middle part equals
then due to the facts and . ∎
For all , define
If for some , then .
The proof is proceeded as follows:
If , then gives that , due to for . The assumptions and also indicate that . Now assume that and . If there exists some such that , then , which means that
satisfies . Note that if and only if
by expanding the above equality we have
which implies . By substituting with , we have
We substitute into and obtain that
Since all the coefficients belong to and , we have that if and only if
which means that and then . That is to say, we have . Then (3.3) gives
This completes the proof. ∎
Assume that . Define
Then for any , and are not zero at the same time.
Firstly observe that if , then , and . Similarly gives that . If and , it suffices to show due to . Since means , which gives that , and then
due to . Now assume that , the decomposition indicates that and . Note that gives
If , then we have
By squaring the second equality and plugging into which with (3.5), we obtain that
and then , which contradicts with the assumption . ∎
For , if , then
By Proposition 1, it suffices to show . The proposition obviously holds if . Assume with and . The conditions give that
which is equivalent to
Denote , from and , we have
for some . Obviously . Then
together with , we get , i.e,
If , the condition implies