Difference between revisions of "1988 AIME Problems/Problem 13"

(Solution 6 (Decreases the Powers))
m (Solution 6 (Decreases the Powers))
(19 intermediate revisions by 2 users not shown)
Line 2: Line 2:
 
Find <math>a</math> if <math>a</math> and <math>b</math> are integers such that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1</math>.
 
Find <math>a</math> if <math>a</math> and <math>b</math> are integers such that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1</math>.
  
== Solution 1 (Fibonacci Version 1) ==
+
== Solution 1 (Fibonacci Numbers) ==
 
Let <math>F_n</math> represent the <math>n</math>th number in the Fibonacci sequence. Therefore,
 
Let <math>F_n</math> represent the <math>n</math>th number in the Fibonacci sequence. Therefore,
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 16: Line 16:
 
\end{align*}</cmath>
 
\end{align*}</cmath>
  
== Solution 2 (Fibonacci Version 2) ==
+
== Solution 2 (Fibonacci Numbers) ==
 
We can long divide and search for a pattern; then the remainder would be set to zero to solve for <math>a</math>. 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 <cmath>(F_{16}b + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0.</cmath> Since the coefficient of <math>x</math> must be zero, this gives us two equations, <math>F_{16}b + F_{17}a = 0</math> and <math>F_{15}b + F_{16}a + 1 = 0</math>. Solving these two as above, we get that <math>a = \boxed{987}</math>.
 
We can long divide and search for a pattern; then the remainder would be set to zero to solve for <math>a</math>. 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 <cmath>(F_{16}b + F_{17}a)x + F_{15}b + F_{16}a + 1 = 0.</cmath> Since the coefficient of <math>x</math> must be zero, this gives us two equations, <math>F_{16}b + F_{17}a = 0</math> and <math>F_{15}b + F_{16}a + 1 = 0</math>. Solving these two as above, we get that <math>a = \boxed{987}</math>.
  
 
There are various similar solutions which yield the same pattern, such as repeated substitution of <math>x^2 = x + 1</math> into the polynomial with a higher degree, as shown in Solution 6.
 
There are various similar solutions which yield the same pattern, such as repeated substitution of <math>x^2 = x + 1</math> into the polynomial with a higher degree, as shown in Solution 6.
  
== Solution 3 (Fibonacci Version 3: For Beginners, Less Technical) ==
+
== Solution 3 (Fibonacci Numbers: For Beginners, Less Technical) ==
 
Trying to divide <math>ax^{17} + bx^{16} + 1</math> by <math>x^2-x-1</math> would be very tough, so let's try to divide using smaller degrees of <math>x</math>. Doing <math>\frac{ax^3+bx^2+1}{x^2-x-1}</math>, we get the following systems of equations:
 
Trying to divide <math>ax^{17} + bx^{16} + 1</math> by <math>x^2-x-1</math> would be very tough, so let's try to divide using smaller degrees of <math>x</math>. Doing <math>\frac{ax^3+bx^2+1}{x^2-x-1}</math>, we get the following systems of equations:
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 48: Line 48:
 
Thus, if <math>ax^{17}+bx^{16}+1</math> has <math>x^2-x-1</math> as a factor, we get that <math>a = 987</math> and <math>b = -1597,</math> so <math>a = \boxed {987}</math>.
 
Thus, if <math>ax^{17}+bx^{16}+1</math> has <math>x^2-x-1</math> as a factor, we get that <math>a = 987</math> and <math>b = -1597,</math> so <math>a = \boxed {987}</math>.
  
== Solution 4 (Fibonacci Version 4: Not Rigorous)==
+
== Solution 4 (Fibonacci Numbers: Not Rigorous)==
 
Let's work backwards! Let <math>F(x) = ax^{17} + bx^{16} + 1</math> and let <math>P(x)</math> be the polynomial such that <math>P(x)(x^2 - x - 1) = F(x)</math>.
 
Let's work backwards! Let <math>F(x) = ax^{17} + bx^{16} + 1</math> and let <math>P(x)</math> be the polynomial such that <math>P(x)(x^2 - x - 1) = F(x)</math>.
  
Line 55: Line 55:
 
Let's find <math>c_{14}</math> now. Notice that all we care about in finding <math>c_{14}</math> is that <math>(x^2 - x - 1)(\cdots + c_{14}x^2 + x - 1) = \text{something} + 0x^2 + \text{something}</math>. Therefore, <math>c_{14} = - 2</math>. Undergoing a similar process, <math>c_{13} = 3</math>, <math>c_{12} = - 5</math>, <math>c_{11} = 8</math>, and we see a nice pattern. The coefficients of <math>P(x)</math> are just the Fibonacci sequence with alternating signs! Therefore, <math>a = c_1 = F_{16}</math>, where <math>F_{16}</math> denotes the 16th Fibonnaci number and <math>a = \boxed{987}</math>.
 
Let's find <math>c_{14}</math> now. Notice that all we care about in finding <math>c_{14}</math> is that <math>(x^2 - x - 1)(\cdots + c_{14}x^2 + x - 1) = \text{something} + 0x^2 + \text{something}</math>. Therefore, <math>c_{14} = - 2</math>. Undergoing a similar process, <math>c_{13} = 3</math>, <math>c_{12} = - 5</math>, <math>c_{11} = 8</math>, and we see a nice pattern. The coefficients of <math>P(x)</math> are just the Fibonacci sequence with alternating signs! Therefore, <math>a = c_1 = F_{16}</math>, where <math>F_{16}</math> denotes the 16th Fibonnaci number and <math>a = \boxed{987}</math>.
  
== Solution 5 (Fibonacci Version 5) ==
+
== Solution 5 (Fibonacci Numbers) ==
 
The roots of <math>x^2-x-1</math> are <math>\phi</math> (the Golden Ratio) and <math>1-\phi</math>. These two must also be roots of <math>ax^{17}+bx^{16}+1</math>. Thus, we have two equations:  
 
The roots of <math>x^2-x-1</math> are <math>\phi</math> (the Golden Ratio) and <math>1-\phi</math>. These two must also be roots of <math>ax^{17}+bx^{16}+1</math>. Thus, we have two equations:  
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
Line 64: Line 64:
  
 
== Solution 6 (Decreases the Powers) ==
 
== Solution 6 (Decreases the Powers) ==
We are given that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1,</math> so the roots of <math>x^2 - x - 1</math> are roots of <math>ax^{17} + bx^{16} + 1.</math>
+
We are given that <math>x^2 - x - 1</math> is a factor of <math>ax^{17} + bx^{16} + 1,</math> so the roots of <math>x^2 - x - 1</math> must also be roots of <math>ax^{17} + bx^{16} + 1.</math>
  
Let <math>x=t</math> be a root of <math>x^2 - x - 1</math> so that <math>t^2 - t - 1 = 0,</math> or <math>t^2 = t + 1.</math> It follows that <cmath>at^{17} + bt^{16} + 1 = 0. \hspace{48.75mm} (\bigstar)</cmath>
+
Let <math>x=r</math> be a root of <math>x^2 - x - 1</math> so that <math>r^2 - r - 1 = 0,</math> or <math>r^2 = r + 1.</math> It follows that <cmath>ar^{17} + br^{16} + 1 = 0. \hspace{20mm} (\bigstar)</cmath>
 +
Note that
 +
<cmath>\begin{align*}
 +
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{align*}</cmath>
 +
We rewrite the left side of <math>(\bigstar)</math> as a linear expression of <math>r:</math>
 +
<cmath>\begin{align*}
 +
(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{align*}</cmath>
 +
Since this linear equation has two solutions of <math>r,</math> it must be an identity. Therefore, we have the following system of equations:
 +
<cmath>\begin{align*}
 +
1597a+987b &= 0, \\
 +
987a+610b &= -1.
 +
\end{align*}</cmath>
 +
To eliminate <math>b,</math> we multiply the first equation by <math>610</math> and multiply the second equation by <math>987,</math> then subtract the resulting equations:
 +
<cmath>\begin{align*}
 +
610(1597a)+610(987b) &= 0, \\
 +
987(987a)+987(610b) &= -987,
 +
\end{align*}</cmath>
 +
from which
 +
<cmath>\begin{align*}
 +
(610\cdot1597-987\cdot987)a&=987 \\
 +
(974170-974169)a&=987 \\
 +
a&=\boxed{987}.
 +
\end{align*}</cmath>
 +
~MRENTHUSIASM
  
Note that
+
== Solution 7 (Uses the Roots) ==
 +
 
 +
For simplicity, let <math>f(x) =ax^{17} + bx^{16} + 1</math> and <math>g(x) = x^2-x-1</math>. Notice that the roots of <math>g(x)</math> are also roots of <math>f(x)</math>. Let these roots be <math>u,v</math>. We get the system
 
<cmath>\begin{align*}
 
<cmath>\begin{align*}
t^4 &= (t+1)^2 \\
+
au^{17} + bu^{16} + 1 &= 0, \\
&= t^2 + 2t + 1 \\
+
av^{17} + bv^{16} + 1 &= 0.
&= (t+1) + 2t + 1 \\
 
&= 3t + 2, \\
 
t^8 &= (3t+2)^2 \\
 
&= 9t^2 + 12t + 4 \\
 
&= 9(t+1) + 12t + 4 \\
 
&= 21t + 13, \\
 
t^{16} &= (21t + 13)^2 \\
 
&= 441t^2 + 546t + 169 \\
 
&= 441(t+1) +546t + 169 \\
 
&= 987t + 610,
 
 
\end{align*}</cmath>
 
\end{align*}</cmath>
from which we rewrite the left side of <math>(\bigstar)</math> as a linear expression of <math>t:</math>
+
If we multiply the first equation by <math>v^{16}</math> and the second by <math>u^{16}</math> we get <cmath>\begin{align*}
 +
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{align*}</cmath>
 +
Now subtracting, we get <cmath>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}}.</cmath>
 +
By Vieta's, <math>uv=-1</math> so the denominator becomes <math>u-v</math>. By difference of squares and dividing out <math>u-v</math> we get <cmath>a= (u^8+v^8)(u^4+v^4)(u^2+v^2)(u+v).</cmath> A simple exercise of Vieta's gets us <math>a= \boxed{987}.</math>
 +
 
 +
~bobthegod78
  
 
== See also ==
 
== See also ==

Revision as of 14:13, 17 September 2021

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 1 (Fibonacci Numbers)

Let $F_n$ represent the $n$th number in the Fibonacci sequence. Therefore, \begin{align*} 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{align*} 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{align*} 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{align*}

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{align*} a+b &= -1, \\ 2a+b &= 0. \end{align*} Continuing with $\frac{ax^4+bx^3+1}{x^2-x-1}$: \begin{align*} 2a+b &= -1, \\ 3a+2b &= 0. \end{align*} 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{align*} 3a+2b &= -1, \\ 5a+3b &= 0. \end{align*} 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{align*} af_{n-1}+bf_{n-2} &= -1, \\ af_n+bf_{n-1} &= 0. \end{align*} 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{align*} a\phi^{17}+b\phi^{16}+1=0, \\ a(1-\phi)^{17}+b(1-\phi)^{16}+1=0. \end{align*} 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 (Decreases 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{align*} 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{align*} We rewrite the left side of $(\bigstar)$ as a linear expression of $r:$ \begin{align*} (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{align*} Since this linear equation has two solutions of $r,$ it must be an identity. Therefore, we have the following system of equations: \begin{align*} 1597a+987b &= 0, \\ 987a+610b &= -1. \end{align*} To eliminate $b,$ we multiply the first equation by $610$ and multiply the second equation by $987,$ then subtract the resulting equations: \begin{align*} 610(1597a)+610(987b) &= 0, \\ 987(987a)+987(610b) &= -987, \end{align*} from which \begin{align*} (610\cdot1597-987\cdot987)a&=987 \\ (974170-974169)a&=987 \\ a&=\boxed{987}. \end{align*} ~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{align*} au^{17} + bu^{16} + 1 &= 0, \\ av^{17} + bv^{16} + 1 &= 0. \end{align*} If we multiply the first equation by $v^{16}$ and the second by $u^{16}$ we get \begin{align*} 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{align*} 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

See also

1988 AIME (ProblemsAnswer KeyResources)
Preceded by
Problem 12
Followed by
Problem 14
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15
All AIME Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png