176 HISTORY OF THE THEORY OF NUMBERS. [CHAP. IV 



and solved the problem when g = 6. Fermat (Oeuvres, III, 265) pointed 

 out the probable origin of Vieta's unnecessary assumption on g. Let the 

 triangle be formed from ax 2 , a; its area x 2 a 4 (x 4 1) increased by 5z 2 shall 

 give a square. Since 5 is a sum of two squares, we can determine y so 

 that 5y 2 1 = D . Take y = x + 1 ; then x* - 1 + 5?/ 2 can readily be 

 made a square. But Vieta did not observe that the problem can be solved 

 when x 4 1 is replaced by 1 a; 4 since we can solve gy z + 1 = D . Fermat 

 found the triangle (9/3, 40/3, 41/3) whose area 20 increased by 5 gives 5 2 . 



The history of the theorem that the area of a rational right triangle is 

 never a square or double a square is given in Ch. XXII, where are given 

 Bachet's and Vieta's comments on the problem to find a right triangle 

 with a given area. 



Fermat 80 stated that the area of the right triangle with the sides 

 2896804, 7216803, 7776485 is of the form 6tt 2 ; likewise for the triangle with 

 the sides 3, 4, 5. E. Lucas 81 obtained these triangles and that with the 

 sides 49, 1200, 1201 and area 6(70) 2 . He noted that the area of a right 

 triangle is never a square, nor the double, triple or quintuple of a square. 



Fermat 's problem to find three right triangles the sum of whose areas 

 by twos are sides of a right triangle was solved by Gillot at the request of 

 Descartes. 82 The triangles 



/24 35 337\ /8 21 65\ / 7 25\ 



\5 ' 12' 60 /' \3' 2 ' 6 /' \ ' 2' 2 ) 



have the areas 7, 14, 21, whose sums by twos are the sides 35, 28, 21 of a 

 right triangle. Gillot gave also the areas 15, 30, 45 and 7 more sets. 



MISCELLANEOUS PROBLEMS INVOLVING THE AREA AND OTHER ELEMENTS. 



In an early Greek manuscript 79 there occurs the problem to find the 

 integral legs a, 6 and hypotenuse c of a right triangle such that the sum of 

 the area T and perimeter 2s is a given number A. The solution given for 

 A = 8-35, 6-45, 5-20, 5-18 is made clear if we introduce the radius r of 

 the inscribed circle, whence T = rs = ab/2, r + s = a + b, c = s r. 

 Separate A into two factors s, r + 2 such that (r + s) 2 Srs is a perfect 

 square n 2 . Then 2a, 2b = r + s db n. Cf. E. Bahier, 62 pp. 190-9. 



Diophantus VI, 6-9 relate to right triangles whose areas increased or 

 diminished by one leg or by the sum of both legs shall be a given number g. 

 To solve the first two problems, Fermat formed the triangle from g, 1 

 and divided the sides by g + 1 or g - 1 ; he enunciated the problems to 

 find a right triangle such that one leg or the sum of the legs diminished by 

 the area is a given number. Cf. E. Bahier, 62 pp. 170-190. 



Diophantus, VI, 10 [11], found a right triangle (283, 45x, 53z) whose 

 area increased [diminished] by the sum of the hypotenuse and one leg is 4. 



80 Oeuvres, III, 256, 348; comment on Diophantus V, 8 and Inventum Novum, I, 38, 



Cf. A. Genocchi, Annali Sc. Mat. Fis., 6, 1855, 319-20. 



81 Bull. Bibl. Storia Sc. Mat., 10, 1877, 290. 



82 Oeuvres, II, 179; letter from Descartes to Mersenne, June 29, 1638. Cf. Oeuvres de 



Fermat, IV, 1912, 56. 



