Introduction - Elliptic Curves: Number Theory and Cryptography - page 4

c

b

a

Figure 1.3

Let x =( c/ 2) 2 .Thenwehave

x − 5=(( a − b ) / 2) 2 x +5=(( a + b ) / 2) 2 .

and

We are therefore looking for a rational number x such that

x − 5 , , x +5

are simultaneously squares of rational numbers. Another way to say this
is that we want three squares of rational numbers to be in an arithmetical
progression with difference 5.
Suppose we have such a number x . Then the product ( x − 5)( x )( x +5)=

x 3

− 25 x must also be a square, so we need a rational solution to

y 2 = x 3 − 25 x.

As above, this is the equation of an elliptic curve. Of course, if we have such
a rational solution, we are not guaranteed that there will be a corresponding
rational triangle (see Exercise 1.2). However, once we have a rational solution
with y = 0, we can use it to obtain another solution that does correspond to
a rational triangle (see Exercise 1.2). This is what we'll do below.
For future use, we record that
x = c

2 2

= ( a 2

b 2 ) c

y =(( x − 5)( x )( x +5)) 1 / 2 = ( a

− b )( c )( a + b )

8

−

, .

8

5 , 0) , (0 , 0) , (5 , 0). These
do not help us much. They do not yield triangles and the line through any
two of them intersects the curve in the remaining point. A small search yields
the point ( − 4 , 6). The line through this point and any one of the three other
points yields nothing useful. The only remaining possibility is to take the
line through ( − 4 , 6) and itself, namely, the tangent line to the curve at the
( − 4 , 6). Implicit differentiation yields

There are three “obvious” points on the curve: ( −

y = 3 x 2

− 25
2 y

= 23

2 yy =3 x 2 − 25 ,

12 .