14 Z. Tu et al. / Finite Fields and Their Applications 31 (2015) 12–24
(i) x + y = x + y and xy = xy for all x, y ∈ F
2
2m
, and
(ii) x + x ∈ F
2
m
and xx ∈ F
2
m
for all x ∈ F
2
2m
.
The unit circle of F
2
2m
is the set
U =
λ ∈ F
2
2m
: λ
2
m
+1
= λλ =1
. (2)
Lemma 1. (See [1].) For a positive integer n, the quadratic equation x
2
+ ax + b =0,
a, b ∈ F
2
n
, a =0, has solutions in F
2
n
if and only if Tr
n
1
(
b
a
2
) =0.
Lemma 2. For a positive integer m, let c, d ∈ F
2
2m
and c =0. Then the equation x +
cx
+ d =0has a unique solution
cd+d
cc+1
in F
2
2m
if cc =1. Otherwise, the equation has no
solution in F
2
2m
for cd + d =0, and it has 2
m
solutions in F
2
2m
for cd + d =0.
Proof. By x = cx + d, we have x = cx + d = c(cx + d) + d, and then
(cc +1)x = cd + d.
Thus, the equation has a unique solution if cc =1, and has no solution if cc =1and
cd + d =0.
In the case of cc =1and cd + d =0. By cc =1, c is in the unit circle of F
2
2m
. Let
β be a primitive element of F
2
2m
, then c = β
(2
m
−1)i
for some integer i, i.e., c =
β
i
β
i
. By
substituting x with β
i
x, the equation x + cx + d =0is reduced to x + x + dβ
−i
=0,
i.e., Tr
2m
m
(x) = dβ
−i
. Note that cd + d =0, that is to say, β
(2
m
−1)i
d = d, i.e., dβ
−i
=
(d
β
−i
)
2
m
. Thus, dβ
−i
∈ F
2
m
and then Tr
2m
m
(x) = dβ
−i
has 2
m
solutions. 2
Lemma 3. For any θ ∈ F
2
2m
\ F
2
m
, the mapping Φ: x →
θ+x
θ+x
from F
2
m
to U \{1} is a
bijection, where U is given by Eq. (2).
Proof. For any x ∈ F
2
m
and θ ∈ F
2
2m
\ F
2
m
, we have
θ+x
θ+x
=
θ+x
θ+x
=(θ + x)
2
m
−1
and
then (
θ+x
θ+x
)
2
m
+1
=(θ + x)
(2
m
−1)(2
m
+1)
=1. Further,
θ+x
θ+x
= 1 due to θ/∈ F
2
m
. Thus,
θ+x
θ+x
∈ U \{1}. Since F
2
m
and U \{1} have the same cardinality, to prove that Φ is a
bijection, it is sufficient to verify that the mapping Φ is injective.
If
θ+x
1
θ+x
1
=
θ+x
2
θ+x
2
for x
1
and x
2
∈ F
2
m
, then (θ + x
1
)(θ + x
2
) =(θ + x
1
)(θ + x
2
). Thus,
we have θθ + θx
2
+ θx
1
+ x
1
x
2
= θθ + θx
1
+ θx
2
+ x
1
x
2
, i.e., (θ + θ)(x
1
+ x
2
) =0. The
fact θ = θ implies x
1
= x
2
.
Therefore, Φ is injective and the proof is completed. 2
Lemma
4. (See [1].) For a positive integer n and a ∈ F
∗
2
n
, the cubic equation x
3
+x +a =0
has
1. a
unique solution in F
2
n
if and only if Tr
n
1
(a
−1
+1) =1;
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