Mathematical_Recreations-Kraitchik-2e

98 Mathematical Recreations
4. Given the hypotenuse z of a right triangle, find the legs
of the triangle.
Solution: The relation z = a2 + b2 determines a and b.
From them we can calculate x = 2ab and y = a 2 - b2•
It is shown in the theory of numbers that if z is composite
then z can be represented as the sum of two squares without
a common factor if and only if each of its factors can be so
represented. Hence we must first examine the cases in which
z is a prime.
First: If z = 2 we must have a = b = 1. This solution
does not give us a primitive triangle, but it will be useful later.
Second: If z is a prime of the form 4k + 1 it can be shown
that there is just one solution for the equation a + 2 b 2 = z.
Third: If z is a prime of the form 4k - 1 there are no
solutions.
If z is composite there are two more cases to consider:
Fourth: If z has a prime factor of the form 4k - 1 there
are no solutions unless every such prime factor occurs to an
even degree. In the latter case we can write z = t 2 w, where
every prime factor of t is of the form 4k - 1 and no prime
factor of w is of that form. The solutions will then be of the
form (ta', tb'), where (a', b') are solutions of w = X2 + y2.
Fifth: If z has no prime factors of the form 4k - 1 there
will be several solutions, which may be derived from the expressions
of the separate prime factors as sums of squares.
To see how this is done let z = uv, and suppose that
u = a 2 + b2, v = c2 + d 2 • Then
z = uv = (a 2 + b 2 ) (c 2 + d 2 )
= a 2 c 2 + b 2 d 2 + a 2 d 2 + b 2 c 2
= (a 2 c2 ± 2aebd + b 2 d2) + (a 2 d2 =1= 2adbc + b 2 c 2 )
= (ae + bd)2 + (ad - bC)2 = (ae - bd)2 + (ad + bC)2.
Thus from two prime factors we can find solutions for their
product; from these and the solutions for another prime we