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4.4. Finding specific values of binary Kloosterman sums 121Proposition 4.3.5. Let a ∈ F ∗ 2 m.• If m is odd, then K m (a) ≡ 1 (mod 3) if and only if t<strong>here</strong> exists t ∈ F 2 m such that a = t 4 +t 3 .• If m is even, then:– K m (a) ≡ 0 (mod 3) if and only if t<strong>here</strong> exists t ∈ F 2 m such that a = t 4 + t 3 andTr m 1 (t) = 0;– K m (a) ≡ −1 (mod 3) if and only if t<strong>here</strong> exists t ∈ F 2 m such that a = t 4 + t 3 andTr m 1 (t) = 1.Proof. According to Proposition 4.3.3 we only have to show that, if a verifies the given congruence,it can be written as a = t 4 + t 3 .• We begin with the case m odd, so that 2 m ≡ −1 (mod 3). Then K m (a) ≡ 1 (mod 3) if andonly if #E a ≡ 0 (mod 3), i.e. if E a has a non-trivial rational point of 3-torsion. It impliesthat the 3-division polynomial of E a given by f 3 (x) = x 4 + x 3 + a has a rational solution,so that t<strong>here</strong> exists t ∈ F 2 m such that a = t 4 + t 3 .• Suppose now that m is even, so that 2 m ≡ 1 (mod 3).– If K m (a) ≡ −1 (mod 3), then #E a ≡ 0 (mod 3), and as in the previous case we canfind t ∈ F 2 m such that a = t 4 + t 3 .– If K m (a) ≡ 0 (mod 3), then #E a ≡ 1 (mod 3), but #Ẽa ≡ 0 (mod 3). The 3-divisionpolynomial of Ẽa is also given by f 3 (x) = x 4 + x 3 + a, so that t<strong>here</strong> exists t ∈ F 2 msuch that a = t 4 + t 3 .4.4 Finding specific values of binary Kloosterman sums4.4.1 Generic strategyIn this subsection we present the most generic method to find specific values of binary Kloostermansums. To this end, one picks random elements of F 2 m and computes the corresponding values untila correct one is found. Before performing any complicated computations, divisibility conditions asthose stated in the previous section can be used to restrict the pool of elements to those satisfyingcertain conditions (but without missing any element giving the value searched for) or to filter outelements which will give ina<strong>de</strong>quate values.Then, the most naive method to check the value of a binary Kloosterman sum is to computeit as a sum. However, one test would need O(2 m m log 2 m log log m) bit operations and this isobviously highly inefficient. Theorem 4.2.1 tells that this costly computation can be replaced bythe computation of the cardinality of an elliptic curve over a finite field of even characteristic.Using p-adic methods à la Satoh [228], also known as canonical lift methods, this can be donequite efficiently in O(m 2 log 2 m log log m) bit operations and O(m 2 ) memory [126, 275, 274, 174].Working with elliptic curves also has the advantage that one can check that the current curve isa good candidate before computing its cardinality as follows: one picks a random point on thecurve and multiplies it by the targeted or<strong>de</strong>r; if it does not give the point at infinity, the curvedoes not have the targeted cardinality.Finally, it should be noted that, if ones looks for all the elements giving a specific value,a different strategy can be adopted as noted in the paper of Ahmadi and Granger [3]. Recall

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