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Last edited by hbghlyj 2020-2-1 10:331.如果ax+cy+bz=X,cx+by+az=Y,bx+ay+cz=Z,证明$\left( {{{\rm{a}}^2} + {b^2} + {c^2} - bc - ca - ab} \right)\left( {{x^2} + {y^2} + {z^2} - yz - zx - xy} \right) = {X^2} + {Y^2} + {Z^2} - YZ - XZ - XY$
2.(1)$k\in\mathbf N_+$,求证：$x^2+x+1|x^k+1+(x+1)^k\Leftrightarrow k\equiv 2,4\pmod6$
(2)$(x^2+x+1)^2|x^k+1+(x+1)^k\Leftrightarrow k\equiv4\pmod6$
(3)不存在$k\in\mathbf N_+$,$(x^2+x+1)^3|x^k+1+(x+1)^k$
来自这贴的第六组原12题
3.${{\rm{f}}_1} = {{\rm{x}}^2},{g_1} = xy,{h_1} = {y^2},{f_{n + 1}} = {\left( {{f_n} + {g_n}} \right)^2},{g_{n + 1}} = g_n^2,{h_{n + 1}} = {\left( {{g_n} + {h_n}} \right)^2},$求证：${{\rm{x}}^2} + {\rm{xy}} + {{\rm{y}}^2}|{f_n} + {g_n} + {h_n}$
\[{x^2}{\left( {x + y} \right)^2} + {x^2}{y^2} + {y^2}{\left( {x + y} \right)^2}={\left( {{x^2} + xy + {y^2}} \right)^2}\]\[{x^4}{({x^2} + 2xy + 2{y^2})^2} + {x^4}{y^4} + {y^4}{(2{x^2} + 2xy + {y^2})^2} = ({x^2} - xy + {y^2})({x^2} + xy + {y^2})({x^4} + 4{x^3}y + 7{x^2}{y^2} + 4x{y^3} + {y^4})\]\[{\left( {{{\rm{x}}^{\rm{8}}}{\rm{ + 4}}{{\rm{x}}^{\rm{7}}}{\rm{y + 8}}{{\rm{x}}^{\rm{6}}}{{\rm{y}}^{\rm{2}}}{\rm{ + 8}}{{\rm{x}}^{\rm{5}}}{{\rm{y}}^{\rm{3}}}{\rm{ + 5}}{{\rm{x}}^{\rm{4}}}{{\rm{y}}^{\rm{4}}}} \right)^{\rm{2}}} + {{\rm{x}}^{\rm{8}}}{{\rm{y}}^{\rm{8}}} + {\left( {{\rm{5}}{{\rm{x}}^{\rm{4}}}{{\rm{y}}^{\rm{4}}}{\rm{ + 8}}{{\rm{x}}^{\rm{3}}}{{\rm{y}}^{\rm{5}}}{\rm{ + 8}}{{\rm{x}}^{\rm{2}}}{{\rm{y}}^{\rm{6}}}{\rm{ + 4x}}{{\rm{y}}^{\rm{7}}}{\rm{ + }}{{\rm{y}}^{\rm{8}}}} \right)^{\rm{2}}} = ({x^2} + xy + {y^2})({x^{14}} + 7{x^{13}}y + 24{x^{12}}{y^2} + 49{x^{11}}{y^3} + 65{x^{10}}{y^4} + 54{x^9}{y^5} + 25{x^8}{y^6} + {x^7}{y^7} + 25{x^6}{y^8} + 54{x^5}{y^9} + 65{x^4}{y^{10}} + 49{x^3}{y^{11}} + 24{x^2}{y^{12}} + 7x{y^{13}} + {y^{14}})\] |
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