By E. W. HOBSON, A. E. H. LOVE (Editors)
Read or Download Proceedings of the Fifth International Congress of Mathematicians (Cambridge, 22-28 August 1912) - Volume II PDF
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21, 72, 75; 60, 80, 100; 56, 105, 119; 72, 135, 153; 72, 96, 120; 35, 120, 125. 57, 176, 185. 65, 156, 169. 64, 120, 136; 60, 87 ; 60, 144, 156; 60, 252, 273; 160, 384, 416; 460, 483, 667 ; 55, 300, 305. 243, 324, 425. 440, 525, 685. 240, 320, 400; 480, 504, 696; 387, 516, 645; 264, 495, 561; 455, 528, 697; 560, 588, 812; 111, 680, 689. 473, 864, 985. 315, 988, 1037. 40, 96. 104; 63, Such groups of right-angled triangles could be obtained by solving the following simultaneous equations, but it is easier to find them from the table of right-angled triangles, with the aid of Barlow's Table I, as I have done.
10 ; also, Miscellaneous Notes and Queries, Vol. x, No. , 1892), p. 225. XVI. As stated elsewhere, a prime number of the form 4m -f 1 is the sum of two squares in one way only, and therefore is the hypotenuse of only one rational right-angled triangle ; but the square of such a prime is the sum of two squares in two ways, and therefore is the common hypotenuse of two different right-angled triangles; the cube of such a prime is the sum of two squares in three ways, and so is the common hypotenuse of three different right-angled triangles ; and, generally, the nth.
Coast and Geodetic Survey, now of the Carnegie Institution of Washington. Another method of the same proof. Let CD = CF = m, AE =AF=n, OD = OE = OF = r, in Fig. 1 ; then AC = a = m + n, AB = b = n + r, BC = c = m + r. n + r) (n + r) = r (2m + 2n + 2r) (1), since each expression is double the area of the triangle ABC. Therefore and mn = mr -j- nr + r2, 2mn = 2mr + 2nr 4- 2r- (2). '< + rf, a2 = b2 + c2. or This method was communicated by Lucius Brown, Hudson, Mass. The following is not new, but is of interest because of its simplicity.
