# Difference between revisions of "2007 AMC 10A Problems/Problem 20"

## Problem

Suppose that the number $a$ satisfies the equation $4 = a + a^{ - 1}$. What is the value of $a^{4} + a^{ - 4}$?

$\text{(A)}\ 164 \qquad \text{(B)}\ 172 \qquad \text{(C)}\ 192 \qquad \text{(D)}\ 194 \qquad \text{(E)}\ 212$

## Solutions

### Solution 1

Notice that $(a^{k} + a^{-k})^2 = a^{2k} + a^{-2k} + 2$. Thus $a^4 + a^{-4} = (a^2 + a^{-2})^2 - 2 = [(a + a^{-1})^2 - 2]^2 - 2 = 194\ \mathrm{(D)}$.

### Solution 2(LIFEHACK)

Notice that $(a^{4} + a^{-4}) = (a^{2} + a^{-2} + 2$. Since \ \mathrm{(D)}$is the only option 2 less than a perfect square, that is correct. PS: Because this is a multiple choice test, this works. === Solution 3 ===$4a = a^2 + 1$. We apply the [[quadratic formula]] to get$a = \frac{4 \pm \sqrt{12}}{2} = 2 \pm \sqrt{3}$. Thus$ (Error compiling LaTeX. ! Missing $inserted.)a^4 + a^{-4} = (2+\sqrt{3})^4 + \frac{1}{(2+\sqrt{3})^4} = (2+\sqrt{3})^4 + (2-\sqrt{3})^4$(so it doesn't matter which root of$a$we use). Using the [[binomial theorem]] we can expand this out and collect terms to get$194$.

=== Solution 3 === (similar to Solution 1) We know that$(Error compiling LaTeX. ! Missing$ inserted.)a+\frac{1}{a}=4$. We can square both sides to get$a^2+\frac{1}{a^2}+2=16$, so$a^2+\frac{1}{a^2}=14$. Squaring both sides again gives$a^4+\frac{1}{a^4}+2=14^2=196$, so$a^4+\frac{1}{a^4}=\boxed{194}$. === Solution 4 === We let$ (Error compiling LaTeX. ! Missing $inserted.)a$and$1/a$be roots of a certain quadratic. Specifically$x^2-4x+1=0$. We use [[Newton's Sums]] given the coefficients to find$S_4$.$S_4=\boxed{194}$=== Solution 5 === Let$a$=$\cos(x)$+$i\sin(x)$. Then$a + a^{-1} = 2\cos(x)$so$\cos(x) = 2$. Then by [[De Moivre's Theorem]],$a^4 + a^{-4}$=$2\cos(4x)$ and solving gets 194.