454 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xv 



u + (7 2 m 2 + n 2 ) 2 are squares . It remains to make (I 2 m 2 + n 2 } 2 I 2 (m 2 + n 2 ) 2 = D . 

 Take 1 = 2. Then 3(4m 4 n 4 ) = D. From the case m = n=l, we get the 

 new solution ?w = 37, n = 23 by Euler's 67 method of Ch. XXII. 



W. Wright 53 found three numbers v 2 1, z 2 1, ?/ 2 1 whose sum is a 

 square, each plus unity is a square, and the sum of the roots of the latter 

 squares is a square. Take v 2 +x 2 +y 2 3 = (v+p} 2 , v+x+y = q 2 . 



To find three numbers such that the sum of the first and second and 

 difference of first and third are squares, the sum of whose roots shall be 

 a square and equal to the sum of the required three numbers, F. N. Bene- 

 dict 54 took the latter to be a 2 x 2 a; 2 , x 2 , (6 2 +a 2 l)x 2 . Then ax-\-bx = cx 2 

 determines x, where c = 2a 2 +& 2 1. Finally, c= D = (b m} 2 gives 6. 



Several 55 found three numbers x 2 !, y 2 !, z 2 ! in arithmetical pro- 

 gression, whose sum is a square and each plus unity is a square. Use the 

 known solution x, Z=(m 2 n 2 ')- s r2mn; y = m?-{-n 2 of x 2 -\-z 2 = 2y 2 . To 

 make z 2 +?/ 2 +2 2 -3 = 3(m 2 +n 2 ) 2 -3= D, take n = l and solve 3m 2 +6=D 

 as usual. 



W. Wright 56 found three integers x, y, z, double the difference of any two 

 being a square, also double the difference of the sum of any two and the 

 third. First, solve n(a 2 b 2 ) = p 2 , n(c 2 b 2 ) = q 2 . Since p 2 q 2 = n(a 2 c 2 ), 

 take a-\-c=(p-\-q)t/(vri), a c= (p q)vft, which give a, c. Then p 2 q 2 = D 

 if p = 2tvn(d 2 -\-e 2 }, q = 2tvn-2de. For brevity, set r = t 2 +nv 2 , s = t 2 nv 2 . 

 Then a = r(d 2 +e 2 ')+2des, c = s(d 2 +e 2 )+2der. Then n(c 2 -b 2 )=q 2 or 

 c 2 q 2 /n= D, becomes a quartic in d, which is satisfied if d = 2rse/(4:t 2 v 2 n s 2 ). 

 The case n = l/2 leads to a solution of the initial problem. Set 

 2(x-\-y z}=a? } 2(y-\-z x)=b'\ 2(x-\-z y}=c 2 , which give x, y, z. Then 

 the init'al three conditions require that %(c 2 b 2 ), - be squares. 



J. R. Young 57 treated Diophantus III, 7, 9 somewhat as had Prestet. 15 

 To make (pp. 347-51) xy, xz, yz all squares, take x-\-y = u?, x-\-z = v 2 , 

 Then x y = v 2 w 2 , xz = u 2 w 2 are squares if u = ac-\-bd, 

 w 2 = 4:abcd. Thmy-z=(a 2 -b 2 )(c 2 -d 2 ). Fora = 9, 6 = 4 c = 81, 

 d = 49, we get Euler's 28 first answer, believed to give the smallest possible 

 numbers. Or we may make a 2 b 2 = D by taking a = m 2 +n 2 , b = 2mn and 

 similarly for c 2 d 2 . Other methods are based on the choice 



u 2 = f a 2 +6 2 )(c 2 +d 2 ), v = acbd, w = adbc. 



He (p. 345) treated Diophantus IV, 14. 



F. T. Poselger 58 treated A = D , 5=0 for the case in which A B is 

 factorable into pq (cf. Diophantus II, 12). We may set 



A,B 

 since 



63 The Gentleman's Math. Companion, London, 5, No. 28, 1825, 369-71. 



64 The Math. Diary, New York, 1, 1825, 27. 



66 The Gentleman's Math. Companion, London, 5, No. 29, 1826, 361-2. 

 M Ibid., 5, No. 30, 1827, 574-5. 



67 Algebra, 1816. American edition by S. Ward, 1832, 324-6, 335-6. 



68 Abh. Akad. Wiss. Berlin (Math.), 1832, 1. 



