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%I #3 Mar 30 2012 17:34:21
%S 4,8,12,16,18,20,22,24,28,30,32,36,40,42,44,48,52,54,56,60,64,66,72
%N Possible sides in the Euler V=E-F+2 as roots in a cubic polynomial of the form: P(x)=(x-V)*(x-F)*(x+E) =x^3+(E-V-F)*x^2+(V*F-E(V+F))*x=E*F*V Solves here for F ( Face, Edge, Vertex).
%C Polynomial cubic of Euler's V,F,E: V=E-F+2 P(x)=(x-V)*(x-F)*(x+E) =x^3+(E-V-F)*x^2+(V*F-E(V+F))*x=E*F*V letting (E-V-F)=-2 V+F=E+2 and product: p=V*F I got P(x)=x3-2*x2+(p-E*(E+2))*x+E*p Setting that polynomial equal to zero gives roots that agree with Euler's equation. In the exceptional groups: ( down to two integer variables) p=16*m ; m-> {1,3,15} E=6*n ; n->{1,2,5} The program works to produce the right roots for {-E,V,F}
%F F roots such that:x^3+(E-V-F)*x^2+(V*F-E(V+F))*x=E*F*V and that are exceptional like ( tetrahedron, cube, octahedron, dodecahedron, icosahedron)
%e Program to get roots for tetrahedron, (cube, octahedron),
%e (dodecahedron,
%e icosahedron):
%e a = {1, 2, 5}
%e b = {1, 3, 15}
%e g[n_, m_] := x /. Solve[e [a[[m]]]*p[b[[m]]] - e [a[[m]]]*(e[a[[
%e m]]] + 2)*x + p[b[[m]]]* x - 2* x^2 + x^3 == 0, x][[n]]
%e Table[g[n, m], {n, 1, 3}, {m, 1, 3}]
%e {{-6, -12, -30}, {4, 6, 12}, {4, 8, 20}}
%t ExpandAll[(x - v)*(x - f)*(x + e)]; e[n_] := 6*n; p[m_] := 16*m; a0 = Table[If[IntegerQ[x /. Solve[e [m]*p[p0] - e [m]*(e[m] + 2)*x +p[p0]* x - 2* x^2 + x^3 == 0, x][[1]]] && IntegerQ[x /. Solve[e [m]*p[p0] - e [m]*(e[m] + 2)*x + p[p0]* x - 2* x^2 + x^3 == 0,x][[2]]] && IntegerQ[x /. Solve[e [m]*p[p0] - e [m]*(e[m] + 2)*x +p[p0]* x - 2* x^2 + x^3 == 0, x][[3]]], {Abs[x] /. Solve[e [m]*p[p0] - e [m]*(e[m] + 2)*x + p[p0]* x - 2* x^2 + x^3 == 0, x][[3]]}, {}], {m, 1, 12}, {p0, 1, 33}]
%Y Cf. edge: A008458; vertex: A118081.
%K nonn,uned
%O 1,1
%A _Roger L. Bagula_, Jul 06 2007
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