1) $a^2b^2 = 1$, $a^2 + b^2 = 11$. Therefore a qualifying equation is $\boxed{x^2-11x+1=0}$ 2) This can be solved in several different ways. These different approaches are discussed in the book %%HREF%%Competition Algebra%%http://www.mathallstar.org/Home/Home/books/110-competition-algebra%%. Because $(a+1)$ and $(b+1)$ are the roots of $$(x-1)^2 -3(x-1)-1=0\Leftrightarrow x^2-5x+3=0$$ Thus, $\frac{1}{a+1}$ and $\frac{1}{b+1}$ are the roots of $$\Big(\frac{1}{x}\Big)^2 -5\Big(\frac{1}{x}\Big)+3 = 0 \Leftrightarrow 1-5x+3x^2=0$$ Hence, the answer is $\boxed{\frac{5}{3}}$. 3) The answer is $\boxed{x_{n+2}^2 - 3x_{n+1}-x_n=0}$ where $x_0=2$ and $x_1=3$. Because $a$ and $b$ are the roots of $x^2-3x-1=0$, therefore \begin{align} a^2 - 3a -1 = 0 & \implies & a^{n+2} -3 a^{n+1} - a^n = 0&\qquad\scriptsize{(multiplying\ a^n})\\ b^2 - 3b - 1 = 0 & \implies & b^{n+2} - 3b^{n+1}-b^n = 0&\qquad\scriptsize{(multiplying\ b^n}) \end{align} Adding these two relations leads to the result. More details can be found at These different approaches are discussed in the book  Competition Algebra .