Question

Let p,q be integers and let a, ẞ be the roots of the equation, x²-x-1=0 where a ≠ β. For n = 0,1,2, let a = pa" + qß". Fact: If a and b are rational numbers and a+b√5 = 0, then a=0=b. If a = 28, then q + p/2 =

Answer 1

6

Answer 2

The roots of $x^2-x-1=0$ are $\alpha = \frac{1+\sqrt{5}}{2}$ and $\beta = \frac{1-\sqrt{5}}{2}$. The sequence $a_n = p\alpha^n + q\beta^n$ satisfies $a_{n+2} = a_{n+1} + a_n$. $a_0 = p+q$. $a_1 = p\alpha + q\beta = \frac{p+q}{2} + \frac{p-q}{2}\sqrt{5}$. $a_4 = \frac{7(p+q)}{2} + \frac{3(p-q)}{2}\sqrt{5}$. Given $a_4 = 28$. Equating rational and irrational parts: $\frac{7(p+q)}{2} = 28 \implies p+q = 8$. $\frac{3(p-q)}{2} = 0 \implies p=q$. Solving $p+q=8$ and $p=q$ gives $p=4, q=4$. Therefore, $q + p/2 = 4 + 4/2 = 6$.