Đáp án:
Ta có a^4+4b^4=(a^2+2b^2)^2−4a^2b^2
=(a^2−2ab+2b^2)(a^2+2ab+2b^2)a^4+4b^4
=(a^2+2b^2)^2−4a^2b^2=(a^2−2ab+2b^2)(a^2+2ab+2b^2)
Để a^4+4b^4∈P ⇔ $\left \{ {{a^2+2b^2-2ab=1} \atop {a^2+2b^2+2ab=p(p∈P)}} \right.$
⇒a^2−2ab+2b^2=1
⇔(a−b)^2+b^2=1⇒\(\left[ \begin{array}{l}a-b=0; b=1\\a-b=1; b=0\end{array} \right.\)