CHAP, xxil] AREA RIGHT TRIANGLE 4= D , re 4 + ?/ 4 =j= D . 617 



cerning negative theorems, and cited in the same letter the theorem under 

 discussion. 



Frenicle de Bessy 8 (f!765) gave a proof, published posthumously, the 

 principle of which is doubtless due to Fermat in view of the letters just 

 cited. It suffices to prove it for a primitive right triangle. Denote the 

 sides by 2mn, ra 2 n 2 . If the area is a square, the odd leg ra 2 n 2 is a square 

 I 2 and the even leg 2mn the double of a square. Thus we have a second 

 primitive triangle whose hypotenuse is m, odd leg I and even leg n. Since 

 mn is a square and m is relatively prime to n, m and n are both squares. 

 Denote the sides of the second triangle by 2ef, e 2 db/ 2 , where e and / are rela- 

 tively prime. Since n = 2ef is a square, one of the numbers e, f is an odd 

 square and the other the double of a square. Let e = r 2 , /=2s 2 . Also 

 e 2_|_p = m j s a sq uare a z t Thus a, e, f are the sides of a third primitive right 

 triangle whose area is the square r 2 s 2 . Its sides are less than the corre- 

 sponding sides of the second triangle : 



a<a 2 = m, f<2ef=n, e<(e+f)(e-f) = l. 



The sides of the second are less than the corresponding sides of the first: 

 m<m?+ri 2 ,n< 2mn, l<P = m 2 n-. Hence from the first primitive triangle 

 with a square area we have derived another primitive triangle (the third 9 ) 

 with a square area and with smaller sides. 



G. Wertheim 10 reproduced the last proof in slightly modified form. 



Frenicle proved in like manner (p. 175) that no right triangle has each 

 leg a square and hence the area of a right triangle is never the double of a 

 square. He concluded (p. 178) that no square is the sum of two biquadrates 

 and that x 4 4z 4 = y 2 is impossible in integers. 



Fermat had proposed to St. Croix Sept., 1636 that he find two bi- 

 quadrates whose sum is a biquadrate (Oeuvres, II, 65; III, 287), to Frenicle, 

 May (?), 1640 (II, 195). 



G. W. Leibniz 11 proved, in a manuscript dated Dec. 29, 1678, that the 

 area of a primitive right triangle with integral sides is not a square. The 

 sides are x 2 y 2 , 2xy, one being even. Then if x 2 y 2 and xy are both squares, 

 x and y are both squares; also x-\-y and x y (since a common factor 2 

 would make x 2 y 2 even, contrary to the above). But y, x y, x, x+y are 

 not all squares. For, if so, the last three give squares in arithmetical 

 progression whose common difference is a square, " which is absurd." Fur- 

 ther, iix z y 2 = xy, then (y-{-x}y= D, (xy)x= D, and each of the four fac- 

 tors would be a square, just disproved. He noted several corollaries. In 

 view of the triangle formed from x and 1, (x l)x(x+l) is not a square. 

 The difference of two biquadrates is not a square. For, if v 4 w 4 = D, the 



8 Traite des Triangles Rectangles en Nombres, Paris, 1676, 101-6; Mem. Acad. Sc. Paris, 



5, 1666-1699; ed. Paris, 5, 1729, 174; Recuil de plusieurs traitez de mathematique de 



1'Acad. Roy. Sc. Paris, 1676. 

 B Identical with Format's second triangle. 

 10 Zeitschrift Math. Phys., 44, 1899, Hist. Lit. Abt., 4-7. 

 u Math. Schriften (ed., C. I. Gerhardt), 7, 1863, 120-5. In a fragment, dated July, 1679, 



Leibniz merely stated that the problem is impossible; see L. Couturat, Opuscules et 



fragments in&iitis de Leibniz, Paris, 1903, 578. 



