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§2. Illustrations of the Elliptic Curve Group Law 31

2
2
The calculation is an exercise; it is helpful to note that y − y = z − 1/4 when
y = z + 1/2.
The transformation is not defined at the solution (0, 0, 0) of the Fermat equa-
tion, but the solutions (1, 0, 1) and (−1, 0, −1) map to (x, y) = (3, 5) on E(Q)
while (0, 1, 1) and (0, −1, −1) map to (x, y) = (3, −4) on E(Q). Thus E(Q) is
the cyclic group of three elements {0,(3, 5), (3, −4)}, for if E(Q) contained other
rational points (x, y) there would be corresponding (u,v,w) solutions to the Fermat
3
3
3
equation u + v = w .
Now let us check the 3P = 0 for P = (3, 5) by calculating −2P from the
2
tangent line to E at P. From the derivative (2y − 1)y = 3x we calculate the slope

2
of the tangent line to E at (3, 5) to be 3 · 3 /(2 · 5 − 1) = 27/9 = 3. The tangent line
2
3
is y = 5 + 3(x − 3) = 3x − 4, and this line intersects the cubic y − y = x − 7at
points whose x coordinates are roots of
3
2
x − 7 = (3x − 4)(3x − 5) = 9x − 27x + 20
or
3
2
0 = x − 9x + 27x − 27
3
= (x − 3) .
Since the roots are the x-coordinates of the intersection of the tangent line to E at
(3, 5) with E at least two roots are equal to 3, and the third root, which is again 3,
is the x-coordinate of −2P. Hence we must have −2P = P,sothat3P = 0. The
2
3
graph of y − y = x − 7 has just two rational points (3, 5) and (3, −4).

4
(2.6) Example. The example is related to the fourth-power Fermat equation w =
4
4
u + v . Again it is known that the only rational solutions of this equation are