CHAP. XV] LINEAR FUNCTIONS MADE SQUARES. 445 



C. G. Bachet 8 treated ax+b=\H, ax-\-c=\H, by finding two rational 

 squares whose difference equals b c. To solve &r+4=D, 6+4=D, 

 take the double 4 of the side 2 of the common square 4, and the difference 

 2x of the left members, and one fourth of 2x. Then the square of f (fx+4) 

 equals 8z+4 and the square of |(f x 4) equals 6x+4. By either condition, 

 x = 112. Next, let the constant terms be distinct squares, as in 10x4-9 = D, 

 5x+4 = D. Seek two numbers (5 and 1) whose sum is double the root 3 

 of the larger square and whose difference is double the root 2 of the smaller 

 square. Take one of these numbers 1 and 5 as one of two factors whose 

 product gives the difference 5x+5 of the given functions. From x+ 1 and 5, 

 we get 



But the factors 5x+5, 1 give {K5x+6)} 2 >10z+9. Next, for 65-6z = D, 

 65 24x = D, multiply the first by 4 and we have a problem of the first type. 

 For 16 x=D, 16 5x=D, seek two squares whose difference is the 

 quadruple of x. Take 4 N as the side of the larger square. Then 

 (4-A0 2 -4{16-(4-A0 2 }=16-40N+5N 2 is the smaller square, say 

 (4-7N) 2 , whence N = 4fll. Thus the squares are (40/11) 2 and (16/11) 2 , 

 one fourth of whose difference gives a; = 336/121. [Bachet here used the 

 same letter for x and N and put 4 6N erroneously for 4 7^.] 



Fermat, 9 commenting on Diophantus III, 10 and V, 30, desired four 

 numbers such that the sum of any pair increased by a given number a 

 gives a square. Let a = 15. The three squares 9, 1/100, 529/225 are such 

 that the sum of any pair increased by 15 gives a square (as found by 

 Diophantus, V, 30, who took 9 as one square and solved x 2 +24=D, 

 ?/ 2 +24= D, x 2 4-2/ 2 4-15 = D)- Take as the four numbers 



x 2 -15, 6x49, 



(the last three being of the form 2nx+n~). Then three of the conditions 

 are satisfied identically. The remaining three conditions are 



a " triple equation " in which each constant term is a square. To treat 10 

 such a problem, z+4 = D , 2x4-4 = D , 5x44 = D , replace x by an expression, 

 like x"4-4x, which if increased by 4 gives a square. Then it remains only 

 to solve the " double equation " 2x 2 4-8x+4 = D , 5x 2 +20x+4 = D , from one 

 solution x = c of which we can deduce a second, by replacing x by x+c. 

 Fermat 11 later explained this method in detail. It is stated (9-11) 

 that the method fails for 



(1) 



8 Diophanti Alexandrini Arith., 1621, 435-9. Comment on Diop., VI, 24 (p. 177 above). 



9 Oeuvres, I, 292, 326-7; French transl., Ill, 242, 263-4. 



"Oeuvres, I, 334-5; 111,269-270. Comment on Diophantus VI, 24. For further examples, 



Fermat 91 - 100 of Ch. XIX. 

 11 J. de Billy, Inventum Novum, Toulouse, 1670, Part II, 1-28; German transl. by P. 



von Schaewen, Berlin, 1910; Oeuvres de Fermat, III, 360-374 (p. 329 for 2x + 12 = D, 



2x+5=D). 



