Strong’s Problems. 427 
1 pee? ye 
aa i ) for its ro = 
to be made a square, assume y —-(S 
Z 2a-+c? es ag 
whence, by proceeding as before, we find y=7-7—Gaza- But 
a*y+1 ree: 06. So 
2=——3 = (by substituting for y its value) 4ca—c?a*" 
: 2bz —1 
Again z=- >; = (by substituting for x its value) 
a?+2c 
M(icr= eat) ~! : i A 
; hence 
j re (= 2a+c? este) anc a0h (st a? +2c y+ 
b? 
Aca—c?a? 4ea—c?a? 
(by substituting for y and z their values 3) and as this also is 
to be made a square, assume for its root S Then we 
2 2 
shall have (=) = xb2—2b (a) +1=(be-1)? 5 
from which, by reduction, 
e(4ca—c?a?)? —(a* +-9c)(4ca — c?a?) 
b=2X-—“i(icg- tap ate ys 
Hence the values of the required numbers are as follows : 
2bx— 
20 (in which the value of b is to be found from the 
+2 2a+c* 
last equation,) T=777 aq?’ and y=7-5 oa?" 
The numbers a, c, and e, are to be so assumed that 2, y, 
and z, may come out positive. Ifa=1, c=2, and c=2, then 
will z=, y=2, andz='; , which numbers. will be found 
upon trial to satisfy the tian. It may also be observed. 
that c and a being positive, ca must not exceed 4; but the 
form of the above expressions for x, ¥, 5 will be sufficient to 
direct us how a, ¢, and e, are to be assumed. 
