620 HISTORY OF THE THEORY OF NUMBERS. [CHAP, xxn 



like the proposed equation, but with v\ < VSc. Rychlik 232 of Ch. XXVI 

 gave a proof. 



J. Sommer 33 reproduced Euler's 13 proof of the impossibility of 

 x *+if = z* in integers and Hilbert's 153 proof (Ch. XXVI) of its impossibility 

 in complex integers a+H. 



A. Bottari 34 proved 2 4 +7/ 4 = 2 2 impossible by use of an unnecessarily 

 complicated set of solutions of x 2 +?/ 2 = 2 2 . 



F. Nutzhorn 35 gave a complicated proof of the impossibility of x*+y* = z 4 . 



R. D. Carmichael 36 gave a new proof that neither of the equations 

 m 4 4n 4 ==t 2 is possible in integers each +0. Hence the system 

 p 2 2q 2 = km 2 , p 2 +2<? 2 = db&n 2 is impossible in integers each 4=0. Thus the 

 area of a right triangle is not the double of a square. Hence m 4 +n 4 = a 2 

 is impossible in integers each 



SOLUTION OF 2x 4 ?/ 4 =D; RIGHT TRIANGLE WHOSE HYPOTENUSE AND SUM 

 OF LEGS ARE SQUARES; x 2 +y 2 =B*, x+y = A 2 . ALSO, z 4 2?/ 4 =n, 



Fermat 37 proposed to St. Martin and Frenicle, May 31, 1643, the problem 

 to find a rational right triangle whose hypotenuse and the sum of whose 

 legs are squares. Fermat 38 affirmed that the smallest such triangle with 

 rational sides is that with the sides 39 



(1) 4 687 298 610 289, 4 565 486 027 761, 1 061 652 293 520. 



Fermat 's 40 method consists in forming the right triangle from x-\-l, x; its 

 sides are 2x z +2x+l, 2x+l, 2x*+2x. The first and the sum 2z 2 +4o;+l of 

 the last two shall be squares. By the usual method of Diophantus, we get 

 a; =12/7. The triangle is therefore formed from 5/7, 12/7. Employ- 

 ing 5, 12 instead, we get 41 (169, 119, 120). When a negative result is 

 obtained it is in accord with a general procedure of Fermat to repeat the 

 operation and to form the triangle from x+5, 12. Its sides are (+5) 2 =b 12 2 

 and 24(z+5). Hence 2 +10x+169 and 2 -f 34z+l are to be squares, say 

 a 2 and 6 2 /169. Then 6 2 -a 2 = 168x 2 +5736rr. Taking 



b-a = 14x, b+a=12x+2868/7, 

 we get a= x+ 1434/7. Comparing its square with the earlier a 2 , we get 



= 1343J. 525 = 2048075 

 X 7-2938 20566 



33 Vorlesungen iiber Zahlentheorie, 1907, 176-193. French transl. by A. Levy, 1911, 184-199. 



34 Periodico di Mat., 23, 1908, 109. 



36 Nyt Tidsskrift for Mat., 23, B, 1912, 33-38. 

 36 Amer. Math. Monthly, 20, 1913, 213-21. 

 87 Oeuvres, II, 259-63. 



38 Oeuvres, I, 336; III, 270, observation on Bachet's comment on Diophanlus VI, 24. Also, 



Oeuvres, II, 261 (259, 263), letter to Mersenne, Aug. 1, 1643. 



39 Cited by Frenicle, M6m. Acad. Sc., 5, 1666-99; ed. Paris, 1729, 56-71. Since his numerical 



search was fruitless, he doubtless learned of Format's solution from Mersenne. 

 Inventum Novum, I, 25, 45; III, 32; Oeuvres, III, 340, 353, 388. 

 41 Whence the hypotenuse and leg difference of (169, 119, 120) are squares. 



