174 HISTORY OF THE THEORY OF NUMBERS. [CHAP, iv 



thus determining m rationally. Hence we get two new rational right 

 triangles. For any k, let a, b, h be the legs and hypotenuse of one right 

 triangle; another of equal area has the sides 



' 2abh 2b 2 - h 2 h 4 



2b 2 -h 2 ' 2h ' 2h(2b 2 - h 2 ) 



From this, we obtain a third, etc. To find any number of rational squares 

 h 2 , h' z , - and a number N which if added to or subtracted from each of 

 the squares yields sums and differences which are rational squares, use 

 right triangles of equal area and take h 2 = a 2 + b 2 , h' 2 = a' 2 + b' 2 , , 

 N = 2ab = 2a!V = . Cf. Ch. XVI. 



D. S. Hart 71 repeated the method of Diophantus V, 8. 



A. Martin, 72 using the 3 triangles of Collins, 69 concluded that the 

 conditions reduce to x = z y, v 2 = z 2 zy + y 2 , which is satisfied if 

 y = m 2 n 2 , z = 2mn + m 2 , v = m 2 + mn + n 2 . 



C. E. Hillyer 73 noted that equal right triangles are formed 7 from 



k 2 + kl + l\ k 2 - I 2 ; k 2 + kl + I 2 , 2kl + I 2 ; k 2 + 2kl, k 2 + kl + I 2 . 



C. Tweedie, 74 to find all rational right triangles of area A, discussed 



a 2 + p 2 = 7 2 , afi = 2A, whence x\ + y\ = 1, 7 2 i2/i = 2A. Thus 



1 ~ 2) = 2A(1 + m2)2< 



Write x = m, y = (1 + m^/y. Hence we seek the rational points on 

 (2) x(l - x 2 ) = Ay 2 . 



To apply Cauchy's tangential method (papers 287, 296, etc. of Ch. XXI), 

 start with any right triangle with sides a, /5, 7 and derive the corresponding 

 rational point (x, y). The tangent there cuts the cubic at a new rational 

 point, which corresponds to a new right triangle with the legs 2aj3'y/(a 2 /3 2 ), 

 (a 2 /3 2 )/(27). From it we get a third right triangle. The problem is also 

 treated by Cauchy's second method (the line joining two rational points 

 of a cubic determines a third). 



E. Bahier, 62 pp. 149-168, treated the subject. 



> 



TWO RIGHT TRIANGLES WHOSE AREAS HAVE A GIVEN RATIO. 



Diophantus, V, 24, asked for three squares x] such that x\x\x\ + x\ 

 are squares for i = 1, 2, 3. A solution will be x { = sbi/pi if three right 

 triangles (p f , b i} hi) are found such that pip^ps = S 2 bib 2 b 3 , since 



71 Math. Visitor, 2, 1882, 17-18. 



72 Math. Quest. Educ. Times, 48, 1888, 118-9. 



73 Math. Quest. Educ. Times, 72, 1900, 30. 



74 Proc. Edinb. Math. Soc., 24, 1905-6, 7-19. He quoted from "Life and Letters of Lewis 

 Carroll," p. 343, that the triangles (20, 21, 29) and (12, 35, 37) are equal, but failed to 

 find three. 



