CHAP, xix] PRODUCTS BY Twos PLUS UNITY MADE SQUARES. 519 



Then five of the conditions are satisfied. There remains CD 1 = D. 

 Replacing A+B by its value (A 2 +p 2 +l)/A, we see that the condition 

 becomes A 4 +2A 3 /H ---- + 2 +l) 2 =n, where fc=(o+6)p+a+0. The 

 quartic is the square of A*+Ak p 2 1 if 



This solution is of course not general. For instance, if a = /3 = 0, a=l, 

 b= 1, then the preceding A is zero, whereas we may obtain solutions as 

 follows. We have, in this case, C = A+B+2p, D = A+B-2p. Then 



Thus=(r 2 +3)/(2r). Also A+B = 2p+s if p = (g 2 +l-s 2 )/(4s). Setr = 2, 

 s = 15/4. Then g = 7/4, p=-2/3, A = C = 13/12, 5 = 4/3, Z> = 15/4. For 

 r = 2, s = 7/2, we get 



289 7?_2_3_3 /^_6._5 r) _ 7. 



224> -D 224j ^~ 56j U o,. 



For 6= -a, |8= -a, Euler found C, D= {a(A+)(4a+2) }/(4a) and noted 

 that all the resulting solutions are fractional. He cited the solution 

 A=D = 1, B = 2, (7 = 5, and asked if there are other integral solutions. 



PRODUCT OF ANY TWO OF FOUR NUMBERS INCREASED BY n A SQUARE. 



C. G. Bachet 103 proposed the problem and took n = 3. From (N+2) 2 

 and (N+6) 2 subtract 3 and divide the remainders by the difference 4 of 

 the roots of the squares; we get 



a = i]V 2 + AH- 1, b = l 

 As the third number, he took 



Hence by a general canon, ab+3, ac+3, 6c+3 are squares. Take the fourth 

 number to be d = 4. Then ad+3 and 6d+3 are squares. Finally, 



cd+3 = (2]V-10) 2 if # = 5/8. 



He gave also a second method of solution. 



Fermat 104 remarked that it is easy to deduce a solution from 

 Diophantus 88 V, 3. As three of the numbers take solutions x\, x 2 , x 3 of 

 the latter problem. As the fourth number, take x+1. We then have a 

 " triple equation " XiX+Xi+n= D, whose constant terms Xt+n are squares, 

 and hence easily solved (Ch. XV). 



P. lacobo de Billy 105 took w = 4, R as the first number, and R+2, 

 2R-\-2, 3R+2 as the roots of the squares obtained when R is one factor. 

 Thus the remaining three numbers are E+4, 4#+8, 9^+12. Then 

 (#+4)(9E + 12)+4 is the square of 3^-8 if R = 1/8. The other two condi- 

 tions are seen to be satisfied. 



N. Saunderson 87 (p. 398) took any number a> Vn, subtracted 4a 2 3w 

 from any larger square 6 2 , and called d the quotient obtained on dividing 



103 Dioph. Alex., 1621, 150. 



104 Oeuvres, III, 254. 



106 Diophantvs Geometria, Paris, 1660, 100. 



