Introduction

Suppose a collection of cannonballs is piled in a square pyramid with one ball

on the top layer, four on the second layer, nine on the third layer, etc. If the

pile collapses, is it possible to rearrange the balls into a square array?

Figure 1.1

A Pyramid of Cannonballs

If the pyramid has three layers, then this cannot be done since there are

1 + 4 + 9 = 14 balls, which is not a perfect square. Of course, if there is only

one ball, it forms a height one pyramid and also a one-by-one square. If there

are no cannonballs, we have a height zero pyramid and a zero-by-zero square.

Besides theses trivial cases, are there any others? We propose to ¬nd another

example, using a method that goes back to Diophantus (around 250 A.D.).

If the pyramid has height x, then there are

x(x + 1)(2x + 1)

12 + 22 + 32 + · · · + x2 =

6

balls (see Exercise 1.1). We want this to be a perfect square, which means

that we want to ¬nd a solution to

x(x + 1)(2x + 1)

y2 =

6

1

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2 CHAPTER 1 INTRODUCTION

Figure 1.2

y 2 = x(x + 1)(2x + 1)/6

in positive integers x, y. An equation of this type represents an elliptic curve.

The graph is given in Figure 1.2.

The method of Diophantus uses the points we already know to produce new

points. Let™s start with the points (0,0) and (1,1). The line through these two

points is y = x. Intersecting with the curve gives the equation

x(x + 1)(2x + 1) 1 1 1

x2 = = x3 + x2 + x.

6 3 2 6

Rearranging yields

3 1

x3 ’ x2 + x = 0.

2 2

Fortunately, we already know two roots of this equation: x = 0 and x = 1.

This is because the roots are the x-coordinates of the intersections between

the line and the curve. We could factor the polynomial to ¬nd the third root,

but there is a better way. Note that for any numbers a, b, c, we have

(x ’ a)(x ’ b)(x ’ c) = x3 ’ (a + b + c)x2 + (ab + ac + bc)x ’ abc.

Therefore, when the coe¬cient of x3 is 1, the negative of the coe¬cient of x2

is the sum of the roots.

In our case, we have roots 0, 1, and x, so

3

0+1+x= .

2

Therefore, x = 1/2. Since the line was y = x, we have y = 1/2, too. It™s hard

to say what this means in terms of piles of cannonballs, but at least we have

found another point on the curve. In fact, we automatically have even one

more point, namely (1/2, ’1/2), because of the symmetry of the curve.

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3

INTRODUCTION

Let™s repeat the above procedure using the points (1/2, ’1/2) and (1, 1).

Why do we use these points? We are looking for a point of intersection

somewhere in the ¬rst quadrant, and the line through these two points seems

to be the best choice. The line is easily seen to be y = 3x ’ 2. Intersecting

with the curve yields

x(x + 1)(2x + 1)

(3x ’ 2)2 = .

6

This can be rearranged to obtain

51 2

x3 ’ x + · · · = 0.

2

(By the above trick, we will not need the lower terms.) We already know the

roots 1/2 and 1, so we obtain

51

1

+1+x= ,

2 2

or x = 24. Since y = 3x ’ 2, we ¬nd that y = 70. This means that

12 + 22 + 32 + · · · + 242 = 702 .

If we have 4900 cannonballs, we can arrange them in a pyramid of height 24,

or put them in a 70-by-70 square. If we keep repeating the above procedure,

for example, using the point just found as one of our points, we™ll obtain

in¬nitely many rational solutions to our equation. However, it can be shown

that (24, 70) is the only solution to our problem in positive integers other than

the trivial solution with x = 1. This requires more sophisticated techniques

and we omit the details. See [5].

Here is another example of Diophantus™s method. Is there a right triangle

with rational sides with area equal to 5? The smallest Pythagorean triple

(3,4,5) yields a triangle with area 6, so we see that we cannot restrict our

attention to integers. Now look at the triangle with sides (8, 15, 17). This

yields a triangle with area 60. If we divide the sides by 2, we end up with

a triangle with sides (4, 15/2, 17/2) and area 15. So it is possible to have

nonintegral sides but integral area.

Let the triangle we are looking for have sides a, b, c, as in Figure 1.3. Since

the area is ab/2 = 5, we are looking for rational numbers a, b, c such that

a2 + b2 = c2 , ab = 10.

A little manipulation yields

2

a2 + 2ab + b2 c2 + 20 2

a+b c

= = = + 5,

2 4 4 2

2

a2 ’ 2ab + b2 c2 ’ 20

a’b 2

c

’ 5.

= = =

2 4 4 2

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4 CHAPTER 1 INTRODUCTION

c

b

a

Figure 1.3

Let x = (c/2)2 . Then we have

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, 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 di¬erence 5.

Suppose we have such a number x. Then the product (x ’ 5)(x)(x + 5) =

3

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

y 2 = x3 ’ 25x.

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

(a2 ’ b2 )c

(a ’ b)(c)(a + b)

2

c 1/2

y = ((x ’ 5)(x)(x + 5))

, = = .

x=

2 8 8

There are three “obvious” points on the curve: (’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 di¬erentiation yields

3x2 ’ 25 23

2

2yy = 3x ’ 25, y= = .

2y 12

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5

INTRODUCTION

The tangent line is therefore

23 41

y= x+ .

12 3

Intersecting with the curve yields

2

23 41

= x3 ’ 25x,

x+

12 3

which implies

2

23

3

x2 + · · · = 0.

x’

12

Since the line is tangent to the curve at (’4, 6), the root x = ’4 is a double

root. Therefore the sum of the roots is

2

23

’4 ’ 4 + x = .

12

We obtain x = 1681/144 = (41/12)2 . The equation of the line yields y =

62279/1728.

Since x = (c/2)2 , we obtain c = 41/6. Therefore,

(a2 ’ b2 )c 41(a2 ’ b2 )

62279

=y= = .

1728 8 48

This yields

1519

a2 ’ b2 = .

36

Since

a2 + b2 = c2 = (41/6)2 ,

we solve to obtain a2 = 400/9 and b2 = 9/4. We obtain a triangle (see

Figure 1.4) with

20 3 41

a= , b= , c= ,

3 2 6

which has area 5. This is, of course, the (40, 9, 41) triangle rescaled by a factor

of 6.

There are in¬nitely many other solutions. These can be obtained by suc-

cessively repeating the above procedure, for example, starting with the point

just found (see Exercise 1.4).

The question of which integers n can occur as areas of right triangles with

rational sides is known as the congruent number problem. Another for-

mulation, as we saw above, is whether there are three rational squares in

arithmetic progression with di¬erence n. It appears in Arab manuscripts

around 900 A.D. A conjectural answer to the problem was proved by Tunnell

in the 1980s [122]. Recall that an integer n is called squarefree if n is not

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6 CHAPTER 1 INTRODUCTION

41

6 3

2

20

3

Figure 1.4

a multiple of any perfect square other than 1. For example, 5 and 15 are

squarefree, while 24 and 75 are not.

CONJECTURE 1.1

Let n be an odd, squarefree, positive integer. Then n can be expressed as the

area of a right triangle with rational sides if and only if the number of integer

solutions to

2x2 + y 2 + 8z 2 = n

with z even equals the number of solutions with z odd.

Let n = 2m with m odd, squarefree, and positive. Then n can be expressed

as the area of a right triangle with rational sides if and only if the number of

integer solutions to

4x2 + y 2 + 8z 2 = m

with z even equals the number of integer solutions with z odd.

Tunnell [122] proved that if there is a triangle with area n, then the number

of odd solutions equals the number of even solutions. However, the proof of

the converse, namely that the condition on the number of solutions implies the

existence of a triangle of area n, uses the Conjecture of Birch and Swinnerton-

Dyer, which is not yet proved (see Chapter 14).

For example, consider n = 5. There are no solutions to 2x2 + y 2 + 8z 2 = 5.

Since 0 = 0, the condition is trivially satis¬ed and the existence of a triangle

of area 5 is predicted. Now consider n = 1. The solutions to 2x2 +y 2 +8z 2 = 1

are (x, y, z) = (0, 1, 0) and (0, ’1, 0), and both have z even. Since 2 = 0, there

is no rational right triangle of area 1. This was ¬rst proved by Fermat by his

method of descent (see Chapter 8).

For a nontrivial example, consider n = 41. The solutions to 2x2 +y 2 +8z 2 =

41 are

(±4, ±3, 0), (±4, ±1, ±1), (±2, ±5, ±1), (±2, ±1, ±2), (0, ±3, ±2)

© 2008 by Taylor & Francis Group, LLC

7

INTRODUCTION

(all possible combinations of plus and minus signs are allowed). There are

32 solutions in all. There are 16 solutions with z even and 16 with z odd.

Therefore, we expect a triangle with area 41. The same method as above,

using the tangent line at the point (’9, 120) to the curve y 2 = x3 ’ 412 x,

yields the triangle with sides (40/3, 123/20, 881/60) and area 41.

For much more on the congruent number problem, see [64].

Finally, let™s consider the quartic Fermat equation. We want to show that

a4 + b4 = c4 (1.1)

has no solutions in nonzero integers a, b, c. This equation represents the easiest

case of Fermat™s Last Theorem, which asserts that the sum of two nonzero

nth powers of integers cannot be a nonzero nth power when n ≥ 3. This

general result was proved by Wiles (using work of Frey, Ribet, Serre, Mazur,

Taylor, ...) in 1994 using properties of elliptic curves. We™ll discuss some of

these ideas in Chapter 15, but, for the moment, we restrict our attention to

the much easier case of n = 4. The ¬rst proof in this case was due to Fermat.

Suppose a4 + b4 = c4 with a = 0. Let

b2 + c2 b(b2 + c2 )

x=2 , y=4

a2 a3

(see Example 2.2). A straightforward calculation shows that

y 2 = x3 ’ 4x.

In Chapter 8 we™ll show that the only rational solutions to this equation are

(x, y) = (0, 0), (2, 0), (’2, 0).

These all correspond to b = 0, so there are no nontrivial integer solutions of

(1.1).

The cubic Fermat equation also can be changed to an elliptic curve. Suppose

that a3 + b3 = c3 and abc = 0. Since a3 + b3 = (a + b)(a2 ’ ab + b2 ), we must

have a + b = 0. Let

a’b

c

, y = 36 .

x = 12

a+b a+b

Then

y 2 = x3 ’ 432.

(Where did this change of variables come from? See Section 2.5.2.) It can be

shown (but this is not easy) that the only rational solutions to this equation

are (x, y) = (12, ±36). The case y = 36 yields a’b = a+b, so b = 0. Similarly,

y = ’36 yields a = 0. Therefore, there are no solutions to a3 + b3 = c3 when

abc = 0.

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8 CHAPTER 1 INTRODUCTION

Exercises

1.1 Use induction to show that

x(x + 1)(2x + 1)

12 + 22 + 32 + · · · + x2 =

6

for all integers x ≥ 0.

1.2 (a) Show that if x, y are rational numbers satisfying y 2 = x3 ’ 25x and

x is a square of a rational number, then this does not imply that

x + 5 and x ’ 5 are squares. (Hint: Let x = 25/4.)

(b) Let n be an integer. Show that if x, y are rational numbers sat-

isfying y 2 = x3 ’ n2 x, and x = 0, ±n, then the tangent line to

this curve at (x, y) intersects the curve in a point (x1 , y1 ) such that

x1 , x1 ’ n, x1 + n are squares of rational numbers. (For a more

general statement, see Theorem 8.14.) This shows that the method

used in the text is guaranteed to produce a triangle of area n if we

can ¬nd a starting point with x = 0, ±n.

1.3 Diophantus did not work with analytic geometry and certainly did not

know how to use implicit di¬erentiation to ¬nd the slope of the tangent

line. Here is how he could ¬nd the tangent to y 2 = x3 ’ 25x at the

point (’4, 6). It appears that Diophantus regarded this simply as an

algebraic trick. Newton seems to have been the ¬rst to recognize the

connection with ¬nding tangent lines.

(a) Let x = ’4 + t, y = 6 + mt. Substitute into y 2 = x3 ’ 25x. This

yields a cubic equation in t that has t = 0 as a root.

(b) Show that choosing m = 23/12 makes t = 0 a double root.

(c) Find the nonzero root t of the cubic and use this to produce x =

1681/144 and y = 62279/1728.

1.4 Use the tangent line at (x, y) = (1681/144, 62279/1728) to ¬nd another

right triangle with area 5.

1.5 Show that the change of variables x1 = 12x + 6, y1 = 72y changes the

curve y1 = x3 ’ 36x1 to y 2 = x(x + 1)(2x + 1)/6.

2

1

© 2008 by Taylor & Francis Group, LLC