Mathematical_Recreations-Kraitchik-2e

96 Mathematical Recreations
of the other two; that is, z may be taken as the length of the
hypotenuse.
A further possible restriction is useful. If the three numbers
have a common factor the Pythagorean relation continues
to hold when this common factor has been removed.
Unless the contrary is stipulated we shall assume that all
common factors have been removed. The right triangle is
then said to be primitive.
2. Since neither x nor y is zero, z must be larger than either.
Furthermore x ~ y, for if x = y = a, then Z2 = 2a2 and
z = av2, that is, z is not rational.
If the triangle is primitive, the length of one of the legs
must be even and that of the other must be odd. If both
were even, the hypotenuse would also be even and the
triangle would not be primitive. Suppose, contrariwise,
that both were odd, say x = 2h + 1, y = 2k + 1. Then
X2 = 4h2 + 4h + 1 = 4(h2 + h) + 1. Whether h is even or
odd, h2 + h = h(h + 1) is even, say h2 + h = 2u, since either
h or h + 1 is even. Hence X2 = 4· 2u + 1 = 8u + 1. Similarly,
y2 can be written 8v + 1.
Then
Z2 = 8(u+ v) + 2 = 2[4(u+ v) + IJ.
Since the second factor is odd, z cannot be rational.
If the right triangle is primitive we shall therefore assume
that x is the even number. If the right triangle is not primitive,
its numbers may be expressed as the multiples of a
primitive right triangle (x', y', z') by' a common integral multiplier
n, and we shall assume that x is n times the even
number x'.
3. THE FUNDAMENTAL PROBLEM. Find three relatively
prime numbers x, y, z, rational integers satisfying the relation
X2 + y2 = Z2. (x, y, z are relatively prime if they have no
common factor.)
We have seen that we may assume x even and y and z odd.