Cryptography Reference
In-Depth Information
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. T hen n can be expressed as the
area ofa right trianglewithrationalsides ifand onlyifthe num ber ofinteger
solutions to
2 x 2 + y 2 +8 z 2 = n
with z even equalsthe num ber of solutions w ith z odd.
Let n =2 m with m odd, squarefree, and positive. T hen n can be expressed
as the area of a right trianglewithrationalsides ifand onlyifthe num ber of
integer solutions to
4 x 2 + y 2 +8 z 2 = m
with z even equalsthe num ber of integer solutions w ith 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 2 x 2 + y 2 +8 z 2 =5.
Since 0 = 0, the condition is trivially satisfied and the existence of a triangle
of area 5 is predicted. Now consider n =1. Thesolutionsto2 x 2 + y 2 +8 z 2 =1
are ( x, y, z )=(0 , 1 , 0) and (0 ,
=0,there
is no rational right triangle of area 1. This was first proved by Fermat by his
method of descent (see Chapter 8).
For a nontrivial example, consider n = 41. The solutions to 2 x 2 + y 2 +8 z 2 =
41 are
1 , 0), and both have z even. Since 2
( ± 4 , ± 3 , 0) , ( ± 4 , ± 1 , ± 1) , ( ± 2 , ± 5 , ± 1) , ( ± 2 , ± 1 , ± 2) , (0 , ± 3 , ± 2)
Search WWH ::




Custom Search