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Pythagorean triples and the congruent number problem
As everyone knows, the triangle with sides 3, 4 and 5 is a right-
angled triangle, because 32+42 = 52 satisfies Pythagoras's
theorem. In fact, there is a general formula for triples of integers
satisfying Pythagoras's theorem which we prove next:
Theorem. Suppose that a, b and c are integers which have
no common factor, and such that a2+b2 = c2. Then there exist
integers p and q such that (after possibly interchanging
a and b)
Proof. To prove the theorem, one argues as follows.
If a and b were both even, then c must also be even, as
c2 = a2+b2. But then a, b and c have a common factor of 2. If
a and b were both odd, then both a2 and b2 would leave a
remainder of 1 after dividing by 4, so that c2 = a2+b2 would have
remainder 2. But this cannot happen for any integer c.
So precisely one of a and b is even, and the other is odd. Thus
c is also odd. Suppose b is even, and a is odd. Now
and the only possible common factor of c+a and c-a would have to
divide (c+a)+(c-a) = 2c. Clearly both c+a and c-a are both even,
but it is easy to see that they cannot have any other common factor.
As c+a and c-a have highest common factor 2, and their product is
a square, it follows that each is itself twice a square. So put
Solving these simultaneously leads to the solution in the theorem.
Then substitute back to find b.
Of course, if (a,b,c) is a triple satisfying a2+b2 = c2, then so
does any multiple (la,lb,lc). Other integer
solutions are obtained as multiples of the ones given by the theorem.
So (6,8,10) and (30,40,50) also work. Other examples are
(5,12,13) (where p = 3, q = 2), or (15,8,17) (where p = 4, q = 1).
Also, rational solutions to the Pythagoras equation can be obtained
in the same way; given any rational solution, we can multiply to clear
denominators, and obtain an integer solution. Thus all rational
solutions are multiples of integer solutions. For example,
(5/7,12/7,13/7) is a solution.
A classical problem in elementary number theory, going back to the
Greeks, is the congruent number problem.
It asks for a characterisation of the possible integers which may
arise as the area of a right-angled triangle with rational
sides. We say that an integer n is a congruent number if it is the
area of such a right-angled triangle. So 6 is a congruent number,
as it is the area of the (3,4,5) triangle. The next smallest integer
which is the area of a right-angled triangle with integer sides is
24, the area of the (6,8,10) triangle, followed by 30, the area
of (5,12,13).
But the question asks for areas of right-angled triangles with
rational sides. Can we find a right-angled triangle with rational
sides and area 1?
Scaling up, this is equivalent to the question: can we find a right-
angled triangle with integer sides and area a square number?
We'll explain that this is not possible.
Theorem. There is no right-angled triangle with integer sides
whose area is a square number.
Proof. We argue by contradiction. If (a,b,c) is a Pythagorean
triple with square area, then
(the second equation coming from the formula for the area of a right-
angled triangle
1/2×base×height). Write y = 2k.
Adding four times the second equation to the first gives:
|
a2+b2+2ab = (a+b)2 = c2+y2. |
|
Similarly, subtracting four times the second equation from the first
gives:
|
a2+b2-2ab = (a-b)2 = c2-y2. |
|
So both c2+y2 and c2-y2 are squares, and so their product is
also. Thus
|
c4-y4 = (c2+y2)(c2-y2) = x2 |
|
for some x. We thus get a solution to the equation
Fermat proved that there were no solutions to this equation.
(Note that Fermat's Last Theorem for exponent 4 is a special case of
this result, when x is a square!) In PMA305, you will see
Fermat's proof for the equation x4+y4 = z2, which is very similar.
It follows that 1 is not a congruent number. One might then wonder
whether 6 is the smallest integer which is a congruent number. It
turns out that this is also false. The smallest congruent number is 5,
the area of the right-angled triangle with sides
(3/2,20/3,41/6).
It's much harder to check when a given number is congruent. For
example, 157 is congruent, but to actually construct a right-angled
triangle with area 157 is difficult. Indeed, the simplest (!) solution
is:
|
a = |
6803298487826435051217540
411340519227716149383203
|
|
|
|
b = |
411340519227716149383203
21666555693714761309610
|
|
|
|
c = |
224403517704336969924557513090674863160948472041
8912332268928859588025535178967163570016480830
|
|
|
Remarkably, the problem of characterising congruent numbers was
solved in 1983 by J.Tunnell, using a similar circle of ideas to
Wiles's work on Fermat's Last Theorem. There are two
theorems, similar to each other, but slightly different in the cases
when n is odd or even. His answer looks rather surprising:
Theorem 1. If n is odd and squarefree, then there is a right-
angled triangle with rational sides and area n if and only if the
number of integer solutions to 2x2+y2+8z2 = n is exactly twice the
number of solutions to 2x2+y2+32z2 = n.
Theorem 2. If n = 2m is even and squarefree, then there is a
right-angled triangle with rational sides and area n if and only if
the number of integer solutions to 4x2+y2+8z2 = m is exactly twice
the number of solutions to 4x2+y2+32z2 = m.
Strictly speaking, one direction of each of these ``if and only if''
statements depends an unproven result, for which much numerical
evidence exists, and which all experts believe should be true.
Further details and references, with a complete account of Tunnell's
proof, can be found in the book ``Introduction to Elliptic Curves and
Modular Forms'' by N.Koblitz.
File translated from TEX by TTH, version 2.00.
On 11 Dec 2000, 12:30.
