Wilder'' s Algebraic Solution. 277 



Proposition fourth. Let cp{xypq etc.) be a factor of 

 ^'{xypq etc.) -{•cp"{xi/pq etc.), independently of x, y,p, etc. ; 

 then I say that li' cp"'{xypq etc.) written for ^/, makes 

 cp'{xypq etc.) =0, it will also make (p"{xypq etc.) =0, and 

 <p{xypq etc.)=0 ; for writing y'=cp"\xpy etc.), we shall have 

 (p'ixy'pq etc.)-{-(p"{xy'pq etc.) 



^{y'pq etc.) ~^^ 



(p"[xij'pq etc.) 

 ^^ ^^"^*' ~^{x^~Qt^^^' ^'"^^ therefore (^{xy'pq etc.) di- 

 vides the whole (p'(^?/.?'</6tc.)+cp"(xy'p^ etc.), and the part 

 <p"{xy'pq etc.), it must also divide the remainder q)'{xy'pq etc.) ; 



(p'(xy'pq etc.) 



consequently, ^(^^-^=^(^^^' and 



((>{xy'pq etc.)=0 ; but (p{xy'pq etc.) is a factor of(p"{xy'pq etc.), 

 therefore, cp"{xy'pq etc.)— 0. 



Further, it may not be amiss to observe, that a function is 

 equal to the continued product of all its factors; and if a is 

 aroot of (^'' + 1)=0, different from unity, then the roots of 

 this equation are, 



1 a a- a^ . . . a"-J. 

 Let us now return to the second example, and continue to 

 denote by x^, (p{bc), we have then, 

 a:=cp(6c), 

 x=acp{bc), 

 x=a^(pibc), and consequently 



(x^+p) 

 '/ = "^"~' 



•'~ ax ' 



(«*x2+p) 

 «= ^ ; thus (c) admits of but 



Q/~' CO 



three simple factors. 



The continued product {y-\ ) {y + —- ) {y -{- 



a'^x^-\-p 



~aFF~) °"S^* ^^ reproduce (c) ; and accordingly we have 



( a"x^-\-px^{\ -{- a^-)-\.p^ "1 



I ax"^ j 



?/^+< --^^7^ >2/4— ^7~= (by havmg 



I a'5a;4+pa;2(a2+a'')+p2 j 



L a^x^ j 



