Mathematical_Recreations-Kraitchik-2e

100 Mathematical Recreations
6. Given the odd leg of a primitive right triangle, find the
other two sides.
Solution: Since y = a2 - b2 = (a + b)(a - b), we factor y
into a product of two unequal, relatively prime factors in all
possible ways, and in each case take the larger factor as
a+ b = u, the smaller as a - b = v. Then
a=u+v b=u-v
2 ' 2 '
and both are integers since both u and v are odd.
Examples:
y a+b a-b a b (x, y, z)
3 3 1 2 1 (4, 3, 5)
5 5 1 3 2 (12, 5, 13)
15 15 1 8 7 (112, 15, 113)
15 5 3 4 1 (8, 15, 17)
7. Find a right triangle whose legs are consecutive integers.
Solution: y - x is either +1 or -1, so (a2 - b2) - 2ab =
(a - b)2 - 2b2 = ± 1. We may set a - b = u and b = v (then
a = u + v), and our problem now is to solve the equation
u 2 - 2V2 = ± 1 in integers. (This is a particular case of what
is usually called Pell's equation.) In the theory of numbers
it is shown that this equation can be solved by expressing v/2
as an infinite continued fraction,
V2=I+ 1
2 + 2 + ...
1
(1,2,2,2· .. ).
If we cut off this continued fraction at any point the value
of the resulting fraction is an irreducible simple fraction
called a convergent. Thus 1 = t is the first convergent,
1 + t = 1- is the second,
1 7
1+ 2+! =.5