Từ giả thiết suy ra
$(a^{11}+b^{11})^2=(a^{10}+b^{10})(a^{12}+b^{12})\Leftrightarrow 2a^{11}b^{11} =a^{10}b^{12}+a^{12}b^{10}$
$\Leftrightarrow a^{10}b^{10}(a^2+b^2-2ab)=0\Leftrightarrow a^{10}b^{10}(a-b)^2=0.\Leftrightarrow \left[ {\begin{matrix} a =0\\ b=0\\a=b \end{matrix}} \right.$
+ Nếu $a=0\Rightarrow b^{10}=b^{11}=b^{12}\Rightarrow\left[ {\begin{matrix} b=0\\b =1\end{matrix}} \right.\Rightarrow P=a^{2014}+b^{2014}= \left[ {\begin{matrix} 0\\1\end{matrix}} \right.$.
+ Nếu $b=0$, tương tự $P =\left[ {\begin{matrix} 0\\1\end{matrix}} \right..$
+ Nếu $a=b\Rightarrow 2b^{10}=2b^{11}=2b^{12}\Rightarrow\left[ {\begin{matrix} b=0\\b =1\end{matrix}} \right.\Rightarrow P=a^{2014}+b^{2014}= \left[ {\begin{matrix} 0\\1\end{matrix}} \right.$.