Mathematical_Recreations-Kraitchik-2e

Arithmetico-Geometrical Questions 101
is the third, and so on. The next two convergents are Hand
t§-. If we set the numerator of any convergent equal to u
and its denominator equal to v, then'U and v satisfy u 2 - 2V2 =
+1 or -1 according to whether ~ is an even or odd converv
gent. The first few solutions of our problem are, then,
a-b b a (x, y, z)
---------
1 1 2 (4, 3, 5)
3 2 5 (20, 21, 29)
7 5 12 (120, 119, 169)
17 12 29 (696, 697, 985)
41
8. Find a primitive right triangle whose hypotenuse exceeds
its even leg by 1.
Solution: From z - x = 1 we have a2 + b2 - 2ab =
(a - b)2 = 1. Since a > b, we must have a - b = 1; that
is, b and a must be consecutive integers.
9. Find a right triangle whose area is equal to its perimeter.
Solution: We must have xi = x + y + z, that is,
ab(a2 - b2) = 2ab+ a2+ b 2 + a 2 - b2 = 2a(a+ b),
so b(a - b) = 2. Hence we must have either b = 2, a - b = 1,
a = 3, or b = 1, a - b = 2, a = 3. Thus there are just two
such triangles: (12,5, 13) and (6, 8, 10), with perimeters and
areas 30 and 24 respectively.
10. FURTHER PROPERTIES OF PRIMITIVE RIGHT TRI­
ANGLES.
First: One leg of a primitive right triangle is divisible by 3.
For if a or b is divisible by 3, x = 2ab is divisible by 3. If
neither a nor b is divisible by 3, then y = a 2 - b 2 is, since the
squares of both a and b are of the form 3k + 1.