álgebra

Giovana Martins escreveu:

Não sei se esta é a forma mais exata para resolver o problema, porém, eu pensei em algo assim.

[latex]\mathrm{f(a)=a^2-(bc)a+(b^2+c^2-1)=0, com\ a,b,c\in\mathbb{Z}}[/latex]

[latex]\mathrm{(a_1,a_2)=\left ( \frac{bc+\sqrt{b^2c^2-4(b^2+c^2-1)}}{2},\frac{bc-\sqrt{b^2c^2-4(b^2+c^2-1)}}{2} \right )\ (I)}[/latex]

[latex]\mathrm{b^2c^2-4(b^2+c^2-1)=0\to b=\pm \sqrt{\frac{4c^2-4}{c^2-4}}\geq 0\to \left\{\begin{matrix}
\mathrm{c<-2\ (N\tilde{a}o\ conv\acute{e}m!)}\\
\mathrm{-1\leq c\leq 1\to 0\leq c\leq 1}\\
\mathrm{c>2}
\end{matrix}\right.}[/latex]

[latex]\mathrm{Se\ c=0\to b=\pm 1. \ Por\ outro\ lado, se\ c=1\to b=0\ \therefore\ (b,c)=(\pm 1,0),(b,c)=(0,1) }[/latex]

[latex]\mathrm{Se\ c>2\to infinitas\ soluc\tilde{o}es!}[/latex]

[latex]\mathrm{De\ (I), para\ o\ par\ (b,c)=(\pm 1,0)\to (a_1,b,c)=(a_2,b,c)=(0,\pm1,0) }[/latex]

[latex]\mathrm{De\ (I), para\ o\ par\ (b,c)=(0,1)\to (a_1,b,c)=(a_2,b,c)=(0,0,1) }[/latex]

[latex]\mathrm{De\ forma\ an\acute{a}loga\ para\ f(b)=0\ e\ f(c)=0:}[/latex]

[latex]\mathrm{\left\{\begin{matrix}
\mathrm{(b_1,c,a)=(b_2,c,a)=(0,\pm 1,0)\ e\ (b_1,c,a)=(b_2,c,a)=(0,0,1)}\\
\mathrm{(c_1,a,b)=(c_2,a,b)=(0,\pm 1,0)\ e\ (c_1,a,b)=(c_2,a,b )=(0 ,0,1)}
\end{matrix}\right.}[/latex]

[latex]\mathrm{\therefore\ (a,b,c)=\left \{ (\pm 1,0,0),(0,\pm 1,0),(0,0,\pm 1) \right \}}[/latex]