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24 Chapter 2. On a conjecture about addition modulo 2 k − 12.2 The case ɛ = −12.2.1 Proof of the conjecture of Tang, Carlet and TangIn this subsection we prove Conjecture 1.2.7 of Tang, Carlet and Tang [259], and so its extensionfor u equal to any power of 2, that is Conjecture 1.2.8 of the same authors for u = 2 i and ɛ = −1,according to Lemma 2.1.7, thus yielding the following theorem which additionally includes thecase t = 0.Theorem 2.2.1. Let k ≥ 2 be an integer, t ∈ Z/(2 k − 1)Z and u = 2 i where i is any integer.Then#S t,−1,u,k ≤ 2 k−1 .Proof. First, we note that for u = 1 and v = −1, Lemma 2.1.10 becomes#S t,−1,1,k = #S −t,−1,1,k .Second, for the specific cases a = 0 and a = t, we have that• w H (0) + w H (−t) = w H (−t) ≤ k − 1,• and w H (t) + w H (0) = w H (t) ≤ k − 1,so that we always have{0, t} ⊂ { a ∈ Z/(2 k − 1)Z | w H (a) + w H (−(t − a)) ≤ k − 1 } .Finally, for a ≠ 0 and a ≠ t, we have thatw H (a) + w H (−(t − a)) = k − w H (−a) + k − w H (t − a)= 2k − (w H (−a) + w H (t − a)) .Now suppose that t ≠ 0. Then, using the fact that a ↦→ −a is a permutation of Z/(2 k − 1)Z,we can prove that#S t,−1,1,k = 2 + # { a ∈ Z/(2 k − 1)Z \ {0, t} | w H (a) + w H (−(t − a)) ≤ k − 1 }= 2 + # { a ∈ Z/(2 k − 1)Z \ {0, t} | w H (−a) + w H (t − a) ≥ k + 1 }= 2 + # { a ∈ Z/(2 k − 1)Z | w H (−a) + w H (t − a) ≥ k + 1 }= 2 + # { a ∈ Z/(2 k − 1)Z | w H (a) + w H (t + a) ≥ k + 1 }= 2 + (2 k − 1 − # { a ∈ Z/(2 k − 1)Z | w H (a) + w H (t + a) ≤ k } )≤ 2 + (2 k − 1 − # { a ∈ Z/(2 k − 1)Z | w H (a) + w H (t + a) ≤ k − 1 } )≤ 2 k + 1 − #S −t,−1,1,k≤ 2 k + 1 − #S t,−1,1,k .We already mentioned that #S t,−1,1,k = #S −t,−1,1,k , hence we get the inequality2#S t,−1,1,k ≤ 2 k + 1 .But we also know that #S t,−1,1,k is an integer, which concludes the proof for t ≠ 0.