110 CHAPTER 9

exponent. Just one for starters and, boom, you get stuck right away.

Voil `a—elliptic curves.

The Congruent Number Problem

Another interesting Greek problem that turns out to be tied up with

elliptic curves is what might be called a Diophantine–Pythagorean

problem: Find all right triangles whose side lengths are rational

numbers and whose area is the integer D. This is also called the

“congruent number problem.” It does not sound so difﬁcult, does it?

But it is difﬁcult, and it has not yet been fully solved.

If you want to do a little algebra, you can see how elliptic curves

come up. We look at the case where D = 1. Using only rational

numbers, we want to solve the system of equations:

1. x

2

+ y

2

= z

2

.

2. xy/2 = 1.

These equations correspond to a right triangle whose sides have

lengths |x| and |y|, with hypotenuse |z|. Divide equation (1) through

by z

2

and set X = x/z, Y = y/z. Then our system is equivalent to

1

. X

2

+ Y

2

= 1.

2

. XY/2 = 1/z

2

.

On page 54 in the subsection on Z-equations, we pointed out

that if we set w = 1 + t

2

, the solutions to (1

) are all of the form

X = (1 − t

2

)/w and Y = 2t/w, where t can take the value of any

rational number. (Oh yes, there is also the solution X =−1, Y = 0.)

Plugging this into (2

), multiplying through by w

2

, and cancelling

the 2’s, we get

3. t − t

3

= (w/z)

2

.

You may think this looks bad, but remember that we can make z

anything we need it to be to make item (3) hold. But we do have

to be able to take the square root of the left side of (3) in order to

solve for z. In other words, the Diophantine–Pythagorean problem

we started with is equivalent to ﬁnding all rational numbers t such

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