Chapter 1
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.




© 2008 by Taylor & Francis Group, LLC
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




© 2008 by Taylor & Francis Group, LLC
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




© 2008 by Taylor & Francis Group, LLC
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




© 2008 by Taylor & Francis Group, LLC
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.




© 2008 by Taylor & Francis Group, LLC
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




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