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Theorem 6 yields the complete characterization of PP of type X 2 k + X + Tr ( X s ).
2 n
Corollary 2. Let 1
k
n
1 and 1
s
2 .Then
X 2 k + X + Tr ( X s )
is PP over
F 2 n
if and only if the following conditions are satisfied:
- n is odd
- gcd ( k, n )=1
- s has binary weight 1 or 2.
Proof. Firstly observe that the polynomial X 2 k + X has at least two zeros, 0 and
1. Hence from Claim 3 it follows that if X 2 k + X + Tr ( X s ) is PP then necessarily
the mapping L ( x )= x 2 k + x is 2- to -1. This holds if and only if gcd ( k, n )=1.
Further note that the image set of such an L is the hyperplane H 1 (0). Hence
γ = 1 does not belong to the image set of L if and only if Tr (1) = 1, equivalently
if n is odd. The rest of the proof follows from Lemma 4 and Theorem 6 with
α = δ =1and β =0.
Remark 2. Some results of this paper are valid also in the finite fields of odd
characteristic. In a forthcoming paper we will report more accurately on that.
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