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4.5. Experimental results for m even 125Proposition 4.5.2. Let n = 2m with m ≥ 3. Let a ∈ F ∗ 2 and b ∈ m F∗ 4. Let f a,b be the functiondefined on F 2 n by Equation (4.2). Then f a,b is bent if and only if f a 2 ,b2 is bent.Proof.̂χ fa,b (ω) = ∑x∈F 2 n= ∑x∈F 2 nx∈F 2 n(−1) f a,b(x)+Tr n 1 (ωx)(−1) Trn 1 (ax 2m −1)+Tr 2 1= ∑ ( )(−1) Trn 1 a 2 x 22m −1+Tr 2 1= ∑x∈F 2 n(−1) Trn 1 (a 2 x 2m −1)+Tr 2 1= ∑(−1) f a 2 ,b 2 (x)+Trn 1 (ω 2 x)x∈F 2 n= ̂χ fa 2 ,b 2(ω 2 ) .(bx 2n −13((b 2 x 2 2n −13b 2 x 2n −13)+Tr n 1 (ωx))+Tr n 1 (ω 2 x 2 ))+Tr n 1 (ω 2 x)In the specific case b = 1 that we are interested in, it gives that f a,1 is bent if and only if f a 2 ,1is, which proves that checking one element of each cyclotomic class is enough.Finally, as mentioned in Section 4.4, finding all the a’s in F 2 m giving a specific value isa different problem from finding one such a ∈ F 2 m. One can compute the Walsh–Hadamardtransform of the trace of the inverse function using a fast Walsh–Hadamard transform. As longas the basis of F 2 m considered as a vector space over F 2 is correctly chosen so that the tracecorresponds to the scalar product, the implementation is straightforward.The algorithm that we implemented is described in Algorithm 4.2.Algorithm 4.2: Testing bentness for m evenInput: An even integer m ≥ 3Output: A list of couples made of one representative for each cyclotomic class of elementsa ∈ F 2 m such that K m (a) = 4 together with 1 if the corresponding Booleanfunctions f a,b are bent, 0 otherwise1 Build the Boolean function f : x ∈ F 2 n ↦→ Tr n 1 (1/x) ∈ F 22 Compute the Walsh–Hadamard transform of f3 Build a list A made of one a ∈ F 2 m for each cyclotomic class such that K m (a) = 44 Initialize an empty list R5 foreach a ∈ A do6 Build the Boolean function f a,17 Compute the Walsh–Hadamard transform of f a,18 if f a,1 is bent then9 Append (a, 1) to R10 else11 Append (a, 0) to R12 return R