AIME 1988 · 第 13 题

Solution 1 (Fibonacci Numbers)

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

$$\begin{aligned} x^2 - x - 1 = 0&\Longrightarrow x^n = F_n(x), \ n\in N \\ &\Longrightarrow x^{n + 2} = F_{n + 1}\cdot x + F_n,\ n\in N. \end{aligned}$$ The above uses the similarity between the Fibonacci recursion|recursive definition, $F_{n+2} - F_{n+1} - F_n = 0$, and the polynomial $x^2 - x - 1 = 0$.

$$\begin{aligned} 0 = ax^{17} + bx^{16} + 1 = a(F_{17}\cdot x + F_{16}) + b(F_{16}\cdot x + F_{15}) + 1 &\Longrightarrow (aF_{17} + bF_{16})\cdot x + (aF_{16} + bF_{15} + 1) = 0,\ x\not\in Q \\ &\Longrightarrow aF_{17} + bF_{16} = 0 \text{ and } aF_{16} + bF_{15} + 1 = 0 \\ &\Longrightarrow a = F_{16},\ b = - F_{17} \\ &\Longrightarrow a = \boxed {987}. \end{aligned}$$

Solution 2 (Fibonacci Numbers)

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

$$(F_{16}b + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0.$$ 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 $a = \boxed{987}$.

There are various similar solutions which yield the same pattern, such as repeated substitution of $x^2 = x + 1$ into the polynomial with a higher degree, as shown in Solution 6.

Solution 3 (Fibonacci Numbers: For Beginners, Less Technical)

Trying to divide $ax^{17} + bx^{16} + 1$ by $x^2-x-1$ would be very tough, so let's try to divide using smaller degrees of $x$. Doing $\frac{ax^3+bx^2+1}{x^2-x-1}$, we get the following systems of equations:

$$\begin{aligned} a+b &= -1, \\ 2a+b &= 0. \end{aligned}$$ Continuing with $\frac{ax^4+bx^3+1}{x^2-x-1}$:

$$\begin{aligned} 2a+b &= -1, \\ 3a+2b &= 0. \end{aligned}$$ There is somewhat of a pattern showing up, so let's try $\frac{ax^5+bx^4+1}{x^2-x-1}$ to verify. We get:

$$\begin{aligned} 3a+2b &= -1, \\ 5a+3b &= 0. \end{aligned}$$ Now we begin to see that our pattern is actually the Fibonacci Numbers! Using the previous equations, we can make a general statement about $\frac{ax^n+bx^{n-1}+1}{x^2-x-1}$:

$$\begin{aligned} af_{n-1}+bf_{n-2} &= -1, \\ af_n+bf_{n-1} &= 0. \end{aligned}$$ Also, noticing our solutions from the previous systems of equations, we can create the following statement:

If $ax^n+bx^{n-1}+1$ has $x^2-x-1$ as a factor, then $a=f_{n-1}$ and $b = f_n.$

Thus, if $ax^{17}+bx^{16}+1$ has $x^2-x-1$ as a factor, we get that $a = 987$ and $b = -1597,$ so $a = \boxed {987}$.

Solution 4 (Fibonacci Numbers: Not Rigorous)

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

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 $c_{i}$ 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 $a = \boxed{987}$.

Solution 5 (Fibonacci Numbers)

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:

$$\begin{aligned} a\phi^{17}+b\phi^{16}+1=0, \\ a(1-\phi)^{17}+b(1-\phi)^{16}+1=0. \end{aligned}$$ 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 $999$).

Solution 6 (Reduces the Powers)

We are given that $x^2 - x - 1$ is a factor of $ax^{17} + bx^{16} + 1,$ so the roots of $x^2 - x - 1$ must also be roots of $ax^{17} + bx^{16} + 1.$

Let $x=r$ be a root of $x^2 - x - 1$ so that $r^2 - r - 1 = 0,$ or $r^2 = r + 1.$ It follows that

$$ar^{17} + br^{16} + 1 = 0. \hspace{20mm} (\bigstar)$$ Note that

$$\begin{aligned} r^4 &= (r+1)^2 \\ &= r^2 + 2r + 1 \\ &= (r+1) + 2r + 1 \\ &= 3r + 2, \\ r^8 &= (3r+2)^2 \\ &= 9r^2 + 12r + 4 \\ &= 9(r+1) + 12r + 4 \\ &= 21r + 13, \\ r^{16} &= (21r + 13)^2 \\ &= 441r^2 + 546r + 169 \\ &= 441(r+1) +546r + 169 \\ &= 987r + 610. \end{aligned}$$ We rewrite the left side of $(\bigstar)$ as a linear expression of $r:$

$$\begin{aligned} (ar+b)r^{16} + 1 &= 0 \\ (ar+b)(987r + 610) + 1 &= 0 \\ 987ar^2 + (610a+987b)r + 610b + 1 &= 0 \\ 987a(r+1) + (610a+987b)r + 610b + 1 &= 0 \\ (1597a+987b)r + (987a + 610b + 1) &= 0. \end{aligned}$$ Since this linear equation has two solutions of $r,$ it must be an identity. Therefore, we have the following system of equations:

$$\begin{aligned} 1597a+987b &= 0, \\ 987a+610b &= -1. \end{aligned}$$ To eliminate $b,$ we multiply the first equation by $610$ and multiply the second equation by $987,$ then subtract the resulting equations:

$$\begin{aligned} 610(1597a)+610(987b) &= 0, \\ 987(987a)+987(610b) &= -987, \end{aligned}$$ from which

$$\begin{aligned} (610\cdot1597-987\cdot987)a&=987 \\ (974170-974169)a&=987 \\ a&=\boxed{987}. \end{aligned}$$ ~MRENTHUSIASM

Solution 7 (Uses the Roots)

For simplicity, let $f(x) =ax^{17} + bx^{16} + 1$ and $g(x) = x^2-x-1$. Notice that the roots of $g(x)$ are also roots of $f(x)$. Let these roots be $u,v$. We get the system

$$\begin{aligned} au^{17} + bu^{16} + 1 &= 0, \\ av^{17} + bv^{16} + 1 &= 0. \end{aligned}$$ If we multiply the first equation by $v^{16}$ and the second by $u^{16}$ we get

$$\begin{aligned} u^{17} v^{16} a + u^{16} v^{16} b + v^{16} &= 0, \\ u^{16} v^{17} a + u^{16} v^{16} b + u^{16} &= 0. \end{aligned}$$ Now subtracting, we get

$$a(u^{17}v^{16} -u^{16} v^{17}) = u^{16}-v^{16} \implies a = \frac{u^{16} - v^{16}}{u^{17}v^{16} -u^{16} v^{17}}.$$ By Vieta's, $uv=-1$ so the denominator becomes $u-v$. By difference of squares and dividing out $u-v$ we get

$$a= (u^8+v^8)(u^4+v^4)(u^2+v^2)(u+v).$$ A simple exercise of Vieta's gets us $a= \boxed{987}.$

~bobthegod78

Solution 8 (Engineer's Induction, only use if you don't have time left)

We see that $ax^{17} + bx^{16} + 1 = (x^2-x-1)P(x)$ for some polynomial $P(x)$. Working forwards, we notice that the constant term of $P(x)$ must equal $-1$, to multiply the constant term of $(x^2-x-1)$ into $1$. We can similarly continue to use this logic, by repeatedly cancelling out the middle term, and obtain the process: $(x^2-x-1)(-1) = -x^2 + x + 1$ $(x^2-x-1)(-1 + x) = x^3 - 2x^2 + 1$ $(x^2-x-1)(-1 + x - 2x^2) = -2x^4 + 3x^3 + 1$ $(x^2-x-1)(-1 + x - 2x^2 + 3x^3) = 3x^5 - 5x^4 + 1$. By this time, you can hopefully notice that the coefficient of the $x^n$ term in $P(x)$ is equal to $(-1)^{n-1} * F_{n}$, where $F_n$ equals the $n$th number in the Fibonacci sequence. From here, we just need to find the coefficient of the $x^{15}$ term in $P(x)$, which happens to be $F_{15} = \boxed{987}$. Again, try to only use Engineer's Induction when you have no other options. A rigorous proof is usually not needed, but when you have extra time, checking a solution with a rigorous method is better than worrying about your Engineer's Induction solution.

~slight edits in $\LaTeX$ by Yiyj1