Elliptic Curves over Finite Fields - Elliptic Curves: Number Theory and Cryptography - page 106

Example 4.6
Let E be the curve y 2 = x 3 +7 x +1 over F 101 . It is possible to show that the
point (0 , 1) has order 116, so N 101 =# E ( F 101 ) is a multiple of 116. Hasse's
theorem says that
101 + 1 − 2 √ 101 ≤ N 101 ≤ 101+1+2 √ 101 ,

which means that 82 122. The only multiple of 116 in this range is

116, so N 101 = 116. As a corollary, we find that the group of points is cyclic
of order 116, generated by (0,1).

≤ N 101 ≤

Example 4.7
Let E be the elliptic curve y 2 = x 3

− 10 x +21 over F 557 . The point (2 , 3) can
be shown to have order 189. Hasse's theorem implies that 511 ≤ N 557 ≤ 605.
The only multiple of 189 in this range is 3 · 189 = 567. Therefore N 557 = 567.

Example 4.8
Let E be the elliptic curve y 2 = x 3 +7 x +12 over F 103 .Thepoint( − 1 , 2)
has order 13 and the point (19 , 0) has order 2. Therefore the order N 103 of
E ( F 103 ) is a multiple of 26. Hasse's theorem implies that 84 ≤ N 103 ≤ 124.
The only multiple of 26 in that range is 104, so N 103 = 104.

Example 4.9
Let E be the elliptic curve y 2 = x 3 +2 over F 7 , as in Example 4.2. The group
of points E ( F 7 ) is isomorphic to Z 3 ⊕ , has order

3, so the best we can conclude with the present method is that the order N 7
of the group is a multiple of 3. Hasse's theorem says that 3

Z 3 . Every point, except ∞

13, so the
order is 3, 6, 9, or 12. Of course, if we find two independent points of order 3
(that is, one is not a multiple of the other), then they generate a subgroup of
order 9. This means that the order of the full group is a multiple of 9, hence
is 9.

≤ N 7 ≤

Z n ⊕ Z n ,makesitmore
di cult to find the order of the group of points, but is fairly rare, as the next
result shows.

The situation of the last example, where E ( F q )

PROPOSITION 4.16
Let E be an elliptic curve over F q and suppose

E ( F q ) Z n ⊕ Z n

for som e integer n .Theneither q = n 2 +1 or q = n 2

± n +1 or q =( n ± 1) 2 .