Đặt $x=\sqrt[8]{2207-\frac{1}{2207-\frac{1}{2207-...}}}⇒x=\sqrt[8]{2207-\frac{1}{x^8}}$
$⇔x^8=2207-\frac{1}{x^8}$
$⇔x^{16}-2207x^8+1=0$
$⇔(x^8+47x^4+1)(x^4+7x^2+1)(x^2+3x+1)(x^2-3x+1)=0$
$⇔x^2-3x+1=0(*)$
$⇒Δ=9-4=5$
Khi đó, $PT(*)$ có 2 nghiệm: $x_{1,2}=\frac{3±\sqrt{5}}2$
Vậy $(a;b;c;d)=(3;±1;5;2)$
*Tham khảo: Putnam Exam | 1995:B4*