42G Strong's Problems, 



= bz =^ sialic, and Oy = Oz = costc, we shall have, by sub- 



,., ,. , ,. cosa. cos6+ sina. sin6 

 stitution, cos(a— 6) = , = (ifr =l)cosa. 



cost -f- sina. sin6. 



From what has been said it appears, that if A and B be any 

 two arcs, of which A is the greatest, then 



cjp/'A -I- p's sinA-cosB ± sinB. cosA 



T ' 



f^ r » I T.\ cosA* cosB zr: sinA* sinB 

 Cos(A ± B) = — . 



When the radius r is supposed = 1, the denominators in 

 these formulae disappear. In the latter, A and B are used for 

 a and b, 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 6, the first being 

 supposed in each case the greatest. 



The following Diophantine Problem was proposed for solution 

 some months ago in a Periodical Journal, which has since 

 been discontinued. To those who are interested in spe* 

 culations of this nature, we presume that the following 

 solution, forwarded by Professor Strong, of Hamilton Col- 

 lege, will not be unacceptable. 



Problem. 



To find three positive rational Numbers, x, y, and z, such that 



x^~y, x^—z, y^—x, and y^—z may all be squares. 



Assume x — ay for the root of the square x^—y: thenx^—y 



= (x—ayY, whence x = ^ ■ . In like manner, by assum- 



2bx— 1 

 ing x—bz for the root of the square x^—z, we find z = — p — 



But ii^~xz=if ~ fL01_ (since x t= "^"' ) : and as this is 



