156 Solutions of two diophantine Problems. 
Arr. XIII.—Solutions of two diophantine Problems ; 3 by Prof. 
Tueopore STRONG. 
Qu. 1. To divide unity into three positive parts, such that if each 
is increased by unity, each sum shall be a rational cube.. Assume 2, 
p+4q p—q, for the roots of three cubes, such that their sum shall 
equal 4, and each cube shall be greater than unity, then by subtract- 
ing unity from each of these cubes, we shall evidently have the num- 
bers required. Hence, we have z*+(p+¢q)*+-(p—q)*=z*+2p?+ 
—I9n3 rae tea 4 
6p = — Te - , hence, 
: 6 
‘we must make 24p —6pz* —12p*=to a sq. (1.) Put P=5—Yt 
6pq* =4, which gives q? a 
4 144 24 144 
z=p tur then (1) becomes 625 + [a5 (8239 — Ta)e+ os (Qvy — 
36 
Bv2 — 18y?)a? +5 (8y° +2yv? —v*)x° +6 - —2y4)at= sq. = 
(12 — (323y—72v 144 
35 + Omg ER star?) =, = 625 +. Seo (323y — 72v) @+ 
323y—72v\? 24a  (646y—144e 
(= 5 )**+ a) an? ate, by as- 
sumption ; .’. by reduction, and putting the coefficients of x? equal 
46800vy — 561602 — 106921 y? 
; (0) 
to each other, we shall have a= 
QB8y? + 72yv? — 36v3—646ay+ 144av 
a a 5(a? + 12y* — 6yv?) ,(b), these equations 
will oe us to solve the =rirview as required. Assume y=1, 
31804 
v= e, then by (a) we ig =a and by (6) we have r= 
eo . a Stee 505413181012 
505372048805" 505372948805" 29d P= o—ye= 
576707885046 3231-72 
505372948805’ ; also, since g=+ ae = tac - = atax? ) 6p, by 
taking the sign—we easily get =X 9264421 8793282625291 55 
4 JU40 X LULU 14009 
566270943093850697 kp eo 
hence, p+ y 
P 
nence, p+ 
these and the value of z found above are the roots of the required 
cubes. ' 
