156 Solutions of two diophantine Problems. 



Art. XIII. — Solutions of two diophantine Problems ; by Prof. 

 Theodore 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 z, 

 p-\-q, 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 - ?) ^ =^ ^ +2^=* + 



4-z^-2p^ 24p-6pz^-l2p^ , 

 6pq? =4, which gives ^=^ = g- = gg-^ , hence, 



6 



we must make 24p — 6pz^ -12p*=to a sq. (1.) Put p = ^—yx, 



4 144 24 144, 



z=-^-\-vx then (1) becomes 625+][25(3233/ — 72tj)a;+ ^g- (2uy - 



36 



Sv"" -18i/2)a;2 4--^(8yH%^'" -v^)x''-{-6 (yv^ -2y^)x^= sq. = 



/12 (32Sy-12v\ \ ^ 144 24 ■ 



(25 + i-V- ) -+-V =6^5 + 1-2-5 (323y - 72.) X + 



//323v-72uV 24a\ /646w-144v\ 



((-^5—) +^y+[-^—]ax^+a^cc^, by as- 



sumption ; .'.by reduction, and putting the coefficients of x^ equal 



46800t)w-5616t;2 -1069212/2 

 to each other, we shall have a = ^ ,(«) 



288m3+72p^ -36y3-646«?/ + 144at? 



then x= 5(a2_|_i2z/4-6p^) '^^)' ^^^^^ equations 



will enable us to solve the problem as required. Assume y = l, 



17 318049 



v=i-^, then by (a) we get a=- gQ„ , and by (6) we have x = 



29739653520 ^ _4 505413181012 , _ 6 _ 

 505372948805' '*• ^"5 + ^"^ "505372948805' ^"^ P" 5"^^- 

 576707885046 , . ^/12 /323?/-72«\ , \ 

 505372948805'^^'°'''"^^^=-l25+l 5 j^+«^^ j-^6p, by 



2 9264421879328262529155 

 takingthes.gn-weeas.ly get ^=5Xg^g^Q^gg5Q^g ^ 101074589761' 



566270943093850697 _ 506506070263707307 



hence,p+ ?=470040809835 1 96035' ^^"^"470040809835196035' 

 these and the value of z found above are the roots of the required 

 cubes. 



