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



the opposite vertex on the hypotenuse c, and such that d -\- s is a square. 

 Take the sides to be the products of the numbers (1) by uv. Then s = 

 abc/(ab + c 2 ) equals uv times a fractional function of m, n, whose denomi- 

 nator is taken as u. Since d = a -f- b c = 2n(m ri)uv, the condition 

 d + s = D is of the form Av = D and holds if v = A. 



S. Tebay 154 noted the existence of an infinitude of pairs of right triangles 

 with the same hypotenuse such that the differences between the hypotenuse 

 and the legs are a square and double a square. 



G. de Longchamps 1540 stated and Svechnikoff proved that x z = y* -f- 2 2 has 

 an infinitude of solutions for which x + y is a biquadrate. 



Several 155 found right triangles with the base 105, and two right triangles 

 with the same base which is a mean proportion between the two perpen- 

 diculars. 



To find any number of dissimilar rational triangles of equal perimeter, 

 R. W. D. Christie 156 multiplied the sides of special triangles by suitable 

 common factors, while A. Cunningham employed (1) and solved 



m(m + ri) = const. 



A. Ge"rardin 157 noted that, to find two right triangles having the same 

 sum of squares of the hypotenuse and one leg, we have to solve 



(z 2 + */ 2 ) 2 + (2xyY = (cv 2 + 2 ) 2 + (2/5) 2 , 



and gave a solution in which x, y, a, /3 are functions of the seventh degree 

 of two parameters. 



R. Janculescu 158 noted that the problem to find a right triangle with 

 integral values for the sides and perpendicular from the right-angle leads 

 to 1/x 2 + I/?/ 2 = 1/2 2 . Thus x 2 + if = t\ Let d be the g.c.d. of x = da, 

 y = d/3, t = dy. Then z = daftly, so that d must be a multiple of y. 



E. Turriere 159 discussed right triangles each of whose sides is a sum of 

 two squares, as 9 = 3 2 , 40 = 2 2 + 6 2 , 41 = 4 2 + 5 2 . 



E. Bahier, 62 pp. 122-148, investigated right triangles with a given 

 perimeter. 



RIGHT TRIANGLE WITH A RATIONAL ANGLE-BISECTOR. 



Diophantus, VI, 18, found a rational right triangle with the bisector 

 of one acute angle rational. Let the bisector be 5N, altitude 4N, so that 

 one segment of the base is 3N. The other segment is taken to be 3 37V. 

 Then (by proportion) the hypotenuse is 4 4N. Equating its square to 

 (4A0 2 + 3 2 , we get N = 7/32. Multiply all our numbers by 32. Then 

 the sides are 28, 96, 100, and the bisector is 35. 



C. G. Bachet 74a in his commentary on the preceding noted that no rational 

 right triangle has a rational bisector of the right angle. 



1M Math. Quest. Educ. Times, 55, 1891, 99-101. 

 1640 Jour, de math, elem., 1892, 282. 

 155 Amer. Math. Monthly, 5, 1898, 51-4, 277-9. 

 Math. Quest. Educ. Times, (2), 14, 1908, 19-21. 

 157 Sphinx-Oedipe, 5, 1910, 187. 



168 Mathesis, (4), 3, 1913, 119-20. 



169 L'enseignement math., 19, 1917, 247-252. 



