both 
The sum fsj, and the product fpj of any two numbers 
being given; to find the sum of the squares, cubes, biqua- 
drates, <§-c. of those numbers. 
If the two numbers be denoted by x and y; then will 
x y ~~ S ]■ by the problem. 
and xy — p y J 1 
The former of which, squared, gives xx 4- Qxy 4- yy\ 
from whence subtracting the double of the latter, we 
have x x 4- y z zz s z — 2 p, the sum of the squares. 
Let this equation be multiplied by x + y — s; so 
shall x 3 4- xy x x + y + y 3 - s 3 — <2sp, that is, x 3 + 
p x s + y 3 ~s 3 — 2sp (because xy — p, and x + y — s) ; 
and therefore x 3 + y 3 — s' — 3sp, the sum of the cubes. 
Multiply, again, by x + y — s, then will x 4 + xy 
3s 1 p, or a ,+ + p x s 1 
— 2p). Conse- 
the sum of the 
X x + y i + y — s‘ 
i 4 — 3s x p (because x 2 4- y 2 zz s' 
quently a? 4 4- y 4 — i 4 — 4s-p 4- 2p' 
biquadrates. 
Hence the law of continuation is manifest, being 
such, that the sum of the next superior powers will be 
always obtained by multiplying the sum of the powers 
last found by s and subtracting from the product, the 
sum of the preceding ones multiplied by p. And the 
sum of the nth powers, expressed in a general manner, 
will be s n — ns n ~~p 4- n . H —. s n ~ 4 p x — n . n —_ 
s n ~ 6 p 3 
+ n 
n 6 
s n ~ 3 p\ 
PROBLEM LXIX, 
The sum of the squares f a J, and the excess (b) of the 
product above the sum of two numbers being given-, to find 
the yumbers. 
Let the sum of the numbers be denoted by .9, and 
¡their product by r: then the sum of their squares will be 
1 4 
•sp?