SERIES—SHAPES AND SQUARES 23

tenuse of a right-angled triangle, and a and & are the other
two sides.
Thus (@) 3%+ 4%=5?
(b) 7°+24%=25"
(¢) 1124602=612
If we know one of the above figures, it is possible to cal-
culate the other two. In the above examples, taken in turn,

(a) (i) 3%*=9 (i) 4+5=9 (i) 5—4=I
(b) (i) 72=49 (i) 24+25=49 (ili) 25—24 =TI
(¢) (i) 112=121 (ii) 60+61=121 (ili) 61 —60=1
From these we can see that where a%-4-b%2=¢?, then
a®=>b-+4c and ¢c—b=1. If a% is known, ¢ and b can be calcu-
lated.
If a2=121, then a=11

also b4c=121
and c—b= 1
Subtracting: 2b=120

and therefore 5 =60, and ¢=61.
The above applies only where a is an odd number.

The following is a formula for generating all Pythagorean
numbers. If x, y and z be the sides of a right-angled triangle,
then for any values of @ and & such that 2ab is a perfect
square, then:

x=a+\/2_al;
y=b+"'2ab
z:a+b+\/2—ab
Thus if a=1; b= 8; then 2ab=16=42
Then x=1+44=5
y=8+44=12
2=9+4=13
and x%24-yt=22

becomes 52+122=132%=169)