CHAP. IV] PROBLEMS ON AREAS OF RIGHT TRIANGLES. 179 



a = d + l,b = d 1 and equate the quartic in d to the square of 3 + 

 22d/3 - 43d 2 /27; thus d = 202752/179200, which gives the last answer. 

 A longer analogous discussion led to the new value r = 50929/46200, which 

 yields an answer involving numbers of ten digits. 



T. Leybourn 94 took x/(x + y) and y/(x + y) as the legs, since each 

 exceeds double the area by a square. Take x = m 2 n 2 , y = 2mn. Then 

 the hypotenuse exceeds double the area by a square if m 4 + 4ran 3 n 4 = D . 

 Take m = 1 + v, n = 4, and equate the quartic in v to the square of 

 v 2 - 130y + 1, whence v = 4223/66. Or take m = y + 5, n = v 3, and 

 equate the quartic in v to (2v 2 - 23Qv - 2) 2 , whence v = 7619/176. 



Malezieux 95 proposed to find two right triangles the sum or difference of 

 whose perimeters is a square; the difference of the areas a square; the 

 difference of the least side of the first and the least side of the second equals 

 the difference of the two largest sides of the first or of the two largest sides 

 of the second, the difference being a cube; the difference of the largest leg 

 of the first and the least leg of the second is a square; and the sum of the 

 least side of the first and the medium side of the second is a square. 



L. Euler 96 discussed the problem proposed by Fermat (on the margin 

 of his Diophantus VI, 14) : Find a right triangle such that each leg exceeds 

 the area by a square. Euler denoted the legs by 2x/z, y/z, where x = ab, 

 y = a 2 b 2 . Subtract the area xy/z 2 . Hence 2xz xy and yz xy are 

 to be squares. Let their product be the square of xy yzpfq. Hence 



z x = x 2 y(p q) 2 /k, 2z y = x(2qx py) 2 /k, k = 2q 2 x 2 p 2 yx. 



It remains only to make k a square, say r 2 ^ 2 . Thus x : y = p 2 : 2q 2 r 2 . 

 Taking the proportionality factor with z, we may set x = p 2 , y = 2q 2 r 2 . 

 Then z = p 2 + (p - q) 2 (2q 2 - r 2 )/r 2 . The condition 4x 2 + y 2 = D be- 

 comes E = 4p 4 + (2q 2 r 2 } 2 = D. Special solutions are obtained by set- 

 ting V# = 2p 2 =F r 2 , 2p 2 2q 2 or r 2 + 2q 2 db 2p 2 . Returning in 20 to 

 the general case, Euler expressed k = r 2 x 2 in the form 



ab(a 2 - 6 2 ) = (2q 2 - r 2 ) = 2t 2 - u\ 



Every product of primes 2, 8m 1 and a square is of the form 2t 2 u 2 

 and only such products. Moreover, if a product of two numbers whose 

 g.c.d. is 1 or 2 is of the form 2t 2 u 2 , each factor is. Hence a, b, a + b, 

 a b must each be of the form 2t 2 u 2 . Conversely, when this is the 

 case, solutions of the initial problem can be readily found. Euler tabulated 

 the permissible values a < 200 for each permissible b < 100, and gave 

 formulas for p, q, r, z. 



To find a right triangle whose area increased by the square of the 

 hypotenuse is a square, J. Whitley 97 wrote rs(r 2 s 2 ) + (r 2 + s 2 ) 2 = a 2 



94 Math. Quest, from Ladies' Diary, 1, 1817, 173-5. 



95 Elements de Geometric de M. le Due de Bourgogne, par de Malezieux, 1722. Solved by E. 



Fauquembergue, Sphinx-Oedipe, 2, 1907-8, 15-16. 



96 Novi Comm. Acad. Petrop., 2, 1749, 49; Comm. Arith., I, 62. 



97 The Gentleman's Math. Companion, 2, No. 10, 1807, 69. 



