Система уравнения (вдъхновена от руска олимпиада)

[ins-]

Да се реши системата:
[tex]\begin{cases} x+y^2=z^3 \\ x^4+y^5=z^6 \\ x^7+y^8=z^9 \end{cases}[/tex]

[Alex98]

[tex]\begin{array}{|l} x + y^2 =z^3\\x^4+y^5=z^6 \\ x^7 +y^8 =z^9 \end{array}[/tex]

1. [tex]z=0 \Rightarrow \begin{array}{|l} x+y^2=0 \\ x^4+y^5=0\\x^7 +y^8= 0 \end{array} \Rightarrow \begin{array}{|l}y^8 + y^5 = 0 \\ -y^{14}+ y^8 = 0 \end{array}\Rightarrow \begin{array}{|l}y^5(y^3+1) = 0 \\ y^8(1-y^3)(y^3+1) = 0 \end{array} \Rightarrow y_1=0 ; x_1=0 ; y_2=-1 ; x_2=-1[/tex]
2.[tex]z\ne 0 \Rightarrow \begin{array}{|l} (x+y^2)(x^7+y^8)=z^{12} \\ (x^4+ y^5)^2 = z^{12} \end{array} \Rightarrow xy^2(x^6+y^6)=2x^4y^5[/tex]

a) [tex]x=0 ; y\ne 0 \Rightarrow y^2=z^3 \Rightarrow y^8=z^{12} \Rightarrow z^9=z^{12}\Rightarrow z^9(z-1)(z^2+z+1)=0 \Rightarrow z=1 \Rightarrow y=1[/tex]
b)[tex]x\ne 0 ; y=0 \Rightarrow x=z^3 \Rightarrow x^4=z^{12} \Rightarrow z^{12}=z^6 \Rightarrow z^6(z+1)(z-1)(z^2+z+1)(z^2-z+1)=0 \Rightarrow z_4=1 ; x_4=1 ; z_5=-1;x_5=-1[/tex]
v)[tex]x\ne 0; y\ne 0 \Rightarrow x^6+y^6=2x^3y^3 \Rightarrow (x-y)^2 (x^2+xy+y^2)^2=0 \Rightarrow x=y[/tex]
[tex]\Rightarrow \begin{array}{|l} x(x+1) = z^3\\x^4(x+1)=z^6 \\ x^7(x+1) = z^9 \end{array}[/tex]
разделяме второто на първото [tex]x^3=z^3 \Rightarrow (x-z)(x^2+xz+z^2)=0 \Rightarrow x=z[/tex] това е възможно защото [tex]z\ne 0 \Rightarrow x+1\ne 0 ; x\ne 0[/tex]
[tex]\Rightarrow z(z^2-z-1)=0 \Rightarrow x_6=y_6=z_6=\frac{1+\sqrt{5}}{2} ; x_7=y_7=z_7=\frac{1-\sqrt{5}}{2}[/tex]

[tex]x,y,z=(0,0,0);(-1,-1,0);(0,1,1);(1,0,1);(-1,0,-1);(\frac{1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2},\frac{1+\sqrt{5}}{2});(\frac{1-\sqrt{5}}{2},\frac{1-\sqrt{5}}{2},\frac{1-\sqrt{5}}{2})[/tex]