1988 AIME Problems/Problem 13

Problem

Find $a$ if $a$ and $b$ are integers such that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1$ .

Solution

Solution 1

Let's work backwards! Let $F(x) = ax^{17} + bx^{16} + 1$ and let $P(x)$ be the polynomial such that $P(x)(x^2 - x - 1) = F(x)$ .

First, clearly the constant term of $P(x)$ must be $- 1$ . Now, we have $(x^2 - x - 1)(c_1x^{15} + c_2x^{14} + \cdots + c_{15}x - 1)$ , where $c_{15}$ is some coefficient. However, since $F(x)$ has no $x$ term, it must be true that $c_{15} = 1$ .

Let's find $c_{14}$ now. Notice that all we care about in finding $c_{14}$ is that $(x^2 - x - 1)(\cdots + c_{14}x^2 + x - 1) = \text{something} + 0x^2 + \text{something}$ . Therefore, $c_{14} = - 2$ . Undergoing a similar process, $c_{13} = 3$ , $c_{12} = - 5$ , $c_{11} = 8$ , and we see a nice pattern. The coefficients of $P(x)$ are just the Fibonacci sequence with alternating signs! Therefore, $a = c_1 = F_{16}$ , where $F_{16}$ denotes the 16th Fibonnaci number and $a = 987$ .

Solution 2

Let $F_n$ represent the $n$ th number in the Fibonacci sequence. Therefore,

$x^2 - x - 1 = 0\Longrightarrow x^n = F_n(x),\ n\in N\Longrightarrow x^{n + 2} = F_{n + 1}\cdot x + F_n,\ n\in N\ .$

The above uses the similarity between the Fibonacci recursive definition, $F_{n+2} - F_{n+1} - F_n = 0$ , and the polynomial $x^2 - x - 1 = 0$ .

$0 = ax^{17} + bx^{16} + 1 = a(F_{17}\cdot x + F_{16}) + b(F_{16}\cdot x + F_{15}) + 1\Longrightarrow$

$(aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q\Longrightarrow$

$aF_{17} + bF_{16} = 0$ and $aF_{16} + bF_{15} + 1 = 0\Longrightarrow$

$a = F_{16},\ b = - F_{17}\Longrightarrow \boxed {a = 987}\ .$

Solution 3

We can long divide and search for a pattern; then the remainder would be set to zero to solve for $a$ . Writing out a few examples quickly shows us that the remainders after each subtraction follow the Fibonacci sequence. Carrying out this pattern, we find that the remainder is $(F_{16} + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0$ . Since the coefficient of $x$ must be zero, this gives us two equations, $F_{16}b + F_{17}a = 0$ and $F_{15}b + F_{16}a + 1 = 0$ . Solving these two as above, we get that $a = 987$ .

There are various similar solutions which yield the same pattern, such as repeated substitution of $x^2 = x + 1$ into the larger polynomial.

Solution 4

The roots of $x^2-x-1$ are $\phi$ (the Golden Ratio) and $1-\phi$ . These two must also be roots of $ax^{17}+bx^{16}+1$ . Thus, we have two equations: $a\phi^{17}+b\phi^{16}+1=0$ and $a(1-\phi)^{17}+b(1-\phi)^{16}+1=0$ . Subtract these two and divide by $\sqrt{5}$ to get $\frac{a(\phi^{17}-(1-\phi)^{17})}{\sqrt{5}}+\frac{b(\phi^{16}-(1-\phi)^{16})}{\sqrt{5}}=0$ . Noting that the formula for the $n$ th Fibonacci number is $\frac{\phi^n-(1-\phi)^n}{\sqrt{5}}$ , we have $1597a+987b=0$ . Since $1597$ and $987$ are coprime, the solutions to this equation under the integers are of the form $a=987k$ and $b=-1597k$ , of which the only integral solutions for $a$ on $[0,999]$ are $0$ and $987$ . $(a,b)=(0,0)$ cannot work since $x^2-x-1$ does not divide $1$ , so the answer must be $\boxed{987}$ . (Note that this solution would not be valid on an Olympiad or any test that does not restrict answers as integers between $000$ and $999$ ).

1988 AIME (Problems • Answer Key • Resources)
Preceded by Problem 12	Followed by Problem 14
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15
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