222 HISTORY OF THE THEORY OF NUMBERS. [CHAP, v 



of the half face angles. Then the cosine of the dihedral angle (6, c) is 

 [6 2 + c 2 - a 2 (l + 6 2 c 2 )]/{26c(l + a 2 ) }. Adding and subtracting 1, we ob- 

 tain as factors of the numerators 



Z) = a + 6 + c abc, A = a + 6 + c + abc, 

 B = a 6 + c + abc, C = a + b c + abc. 



Hence s = sin (6, c) = ^ABCD/{2bc(l + a 2 )}. If /, g, h are the tangents 

 of the half dihedral angles, then 



f\ A f / -t * J*O\ /< I 7 0\ /^ I 7 O\ /7 7 / -t I O\ / ~i I 0\ / 



s = 2//(l +/ 2 ), (1 + 6 2 )(1 + h 2 )/bh = (1 + c 2 )(l + g*)/cg, 



etc. If the latter equation has rational solutions, we obtain 32 distinct 

 rational trihedrals, since we may replace 6 by its reciprocal, etc. 



To obtain a rational tetrahedron, we may take two rational trihedrals 

 having a common dihedral angle and subject to the condition that the 

 edges converge (in the earlier notation, 66' < 1, cc' <!,/=/'). While 

 the tetrahedron now has a rational volume, it remains to make the sixth 

 edge rational. The condition is that 61 + bl + cl + c\ 2 26i6 2 CiC 2 2m 

 be a square, where 



1 + 66' 6-6' 1 + cc' c - c' 



^2 7 7/ > ^1 , ., C 2 



66" 6 + 6" 1-cc" c + c" 



166c6'c'(l - / 2 ) 



6') (c 



K. Schwering 150 discussed rational tetrahedra by use of the formula 

 36F 2 = f*g*WF, F = (1 - cos 2 )(! - cos 2 0) - (cos 7 - cos a cos 0) 2 , 



where /, g, h are the edges from the vertex D, and a, /5, 7 are the face 

 angles at D, while a, 6, c are the sides of the base of the tetrahedron. 

 The first problem is to choose rational values of the cosines such that F 

 shall be the square of a rational number. The first term of F must be the 

 sum of two squares. Give 1 cos 2 a the form of a fraction whose de- 

 nominator is a perfect square. Then its numerator is a divisor of a sum 

 of the squares of two integers and hence is itself the sum of two squares. 

 Thus 1 cos 2 a. equals the sum of two rational squares. Hence cos 2 a is 

 one of three rational squares whose sum is unity; likewise for cos 2 /?. Con- 

 sider the integral squares equal to the sums of the squares of three inte- 

 gers; for instance 



(m 2 -J- n 2 + p 2 + g 2 ) 2 = (m? + n? -- p* g 2 ) 2 + (2mp -f- 2?ig) 2 + (2mq 2np) 2 . 



If 



Q 2 = M 2 + N 2 + P\ Ql = MI + NI + PJ, 

 we take 



M M l MM l - NP l + PN, 



cosa = -, COS/3 = ^, cos y = 



and find that F is the square of (NN l + 

 180 Jour, fur Math., 115, 1895, 301-7. 



