426 : Strong’s Problems. 
BD=cosbE, Ox=cosA or AB=sinBD=sinbE, By. 
=sinbc, and Oy=Oz=cosbc, we shall have, by sub- 
See ~_.... Cosa. cosb--sina. sind 
stitution, cos(a—b= = a 
_ ==cos 
, =(if r=1)cosa. 
cosb-+ sina. sinb. . 
From what has been said itappears, that if A and B be any 
- two arcs, of which A is the greatest, then 
sinA‘cosBtsinB. cosA 
r 
Sin(AtB) = 
cosA‘cosBzsinA: sinB 
Cos(AtB)= i 
When the radius r is supposed=1, the denominators in 
these formulz disappear. In the latter, A and B are used for 
_ @and 6, for the sake of homogeneity. The propriety of this 
~~ is manifest; for as. a and 6 denote two indefinite arcs, the same 
reasoning will apply to A and B, as to a and 8, the first be- 
ing supposed in each case the greatest. 
(cB 
The following Diophantine Problem was proposed for solu- 
tion some months ago in @ Periodical Journal, which has 
since been discontinued. ‘To those who are interested in 
speculations of this nature, we presume that the following 
solution, forwarded by Professor Srrone, of Hamilton 
College, will not be unacceptable. 
Prosiem. 
To find three positive rational Numbers, x, y, and z, such 
that x* —y, x* —z, y2 — x, and y? —z may all. be squares. 
Assume x —ay forthe root of thesquare x? —y: thena? -y 
a*?y+l 
=(x—ay)?, whence r= 4 = In like manner, by assum- 
: “— Qba—1 
ing x —bz for the root of square #? —z, we find z=—yz—* 
a*y+1 
2a 
a? 1 : 
But y? -2=y*— oe (since z= ) and as this 1s 
