Mathematical_Recreations-Kraitchik-2e

Since
Arithmetico-Geometrical Questions
y2 = Z2 - X2 = (z + x)(z - x),
both z + x and z - x are odd. Further, they have no common
divisor. For every common divisor of z + x and z - x
is a common divisor of their sum, 2z, and their difference, 2x,
and hence is a divisor of the greatest common divisor of 2z
and 2x. If z and x had a common divisor it would divide y
and the triangle would not be primitive. Hence 2z and 2x
have only the common divisor 2. Hence z + x and z - x
could have only the common divisor 2, which is impossible
since both are odd.
From this it follows that each of the factors z + x and z - x
is itself a perfect square, so we may write z + x = m 2 , z - x =
n2. Since z + x and z - x are both odd, so are m and n.
Hence m + nand m - n are both even, and we may set
m+n
--2- = a,
that is, m = a + b, n = a-b.
m-n
-2-= b,
Then
z+ x = a 2 + b 2 + 2ab,
z - x = a 2 + b2 - 2ab,
so z = a 2 + b 2 , x = 2ab, and y = a2 - b 2 • It is also easy to
see that x and y (and hence x, y, and z) will have a common
factor if and only if either a and b have a common factor, or
both a and b are odd.
The results of the preceding paragraphs may be summed
up thus: (x, y, z) is a primitive right triangle if and only if
x, y, and z can be expressed by the formulas x = 2ab, y =
a2 - b2 , z = a 2 + b2, where a and b are any two relatively
prime integers such that a > b and one is even and the other
odd. (It is understood that in the notation (x, y, z) the largest
number is z and the even number is x.) It is also true that
under these conditions no two distinct pairs (a, b) yield the
same set (x, y, z).
Now we can solve some problems about right triangles.
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