By Paula Ribenboim

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**Additional resources for Algebraic Numbers (Pure & Applied Mathematics Monograph)**

**Sample text**

N, has to be verified by substituting. (iii) 2xt - x2 -Xt + 2X2 - X3 = at , = 0, = 0, -Xn-2 + 2Xn - t - Xn = 0, - Xn- t + 2Xn = a2 . (41) 50 1. Algebraic Identities and Equations We rewrite the equations of the system ( 41) from the second to the second to-last one in the form x 1 - x2 = x2 - X3 , x2 - X3 = X3 - X4 , . · Xn - 2 Xn- 1 = Xn- 1 - Xn· The ''trick" of elimination relies on the introduction of the new variable t = x1 - x2 , and in view of the preceding equations we have . , t = X 1 - X2 = X2 X3 = X3 - X4 = · · = Xn- 1 - Xn· · - The first and the last equations in the system (41) can also be expressed in terms of the variable t: Thus we obtain the system a 1 - t, a 1 - 2t, a 1 - 3t, X1 = X2 = X 1 - t X3 = X2 - t (42) Xn- 1 = Xn-2 - t = Xn = Xn- 1 - t = Xn = a 1 - (n - 1)t, a 1 - nt, From the last two equations of ( 42) we then determine t = ( a 1 -a2 ) / ( n + 1).

The terms of F(x�, x2 ) of the form aii xix� will be change4 to aii o:� , and we collect the remaining terms in pairs of the form aijXi� + a3ix{x�, where i > j. The symmetry of F(x�, x2 ) implies the equation aij = a3i, and thus we can factor out ai3 (x 1 x2 )i = aij � · It therefore remains to find j expressions for the sums of powers Si -j = xi- + x;- j . 2, we obtain the following expressions for S k = xt + x�: S 1 = X 1 + X2 = 0'1 , S2 = X� + X� = (x 1 + X2 ) 2 - 2X 1 X2 = 0'� - 20'2 , s3 = x� + x� = (x 1 + x2 ) 3 - 3x 1 x2 (x 1 + x2 ) = 0'� - 30'10'2 , S4 = xi + x� = (x 1 + x2 ) 4 - 4x l x2 (x� + x�) - 6x�x� = O't - 40'2 S 2 - 6 0'� = O't - 40'2 (0'� - 20'2 ) - 60'� = O't - 40"� 0'2 + 20'� ' 40 1.

I) F(x 1 , X2 ) = X�X2 + xfx� + X� X� + X 1 X� + X�X2 + x 1 x�. SoLUTION . 4: F(x�, x2 ) = (x�x2 + x 1 x�) + (xfx� + x�x� ) + (x�x2 + x 1 x�) = x 1 x2 (xi + x� ) + (x 1 x2 ) 2 (x� + x�) + x 1 x2 (x1 + x2 ) = er2 8 4 + er�8 2 + er2 er1 = er2 (erf - 4er�er2 + 2er�) + er�(er� - 2er2 ) + er2 er1 o = erier2 - 3er�er� + er 1 er2 . SOLUTION . We have G(x� , x2 ) = (x 1 - x2 ) 2ooo = [(x 1 - x2 ) 2pooo = (x� + x� - 2x 1 x2 ) 1 ooo = [(x 1 + x2 ) 2 - 4x 1 x2 pooo = (er� 40"2 ) 1000 . _ SOLUTION . 0 By combining factors in an appropriate way, we obtain H(x�, x2 ) = (x 1 + 4x2 )(4x 1 + x 2 )(2x 1 + 3x2 )(3x 1 + 2x2 ) = [4(x� + x�) + 1 1x 1 x2] (6(x� + x�) + 13x 1 x2] = [4(er� - 2er2 ) + 1 1er2] [6(er� - 2er2 ) + 13a2] = (4er� + 9er2 )(6a� + er2 ).