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

Notice that $(a^{4} + a^{-4}) = (a^{2} + a^{-2})^{2} - 2$ . Since 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 $a^4 + a^{-4} = (2+\sqrt{3})^4 + \frac{1}{(2+\sqrt{3})^4} = (2+\sqrt{3})^4 + (2-\sqrt{3})^4$ (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 4

(similar to Solution 1) We know that $a+\frac{1}{a}=4$ . 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 5

We let $a$ 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 6

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.

2007 AMC 10A (Problems • Answer Key • Resources)
Preceded by Problem 19	Followed by Problem 21
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25
All AMC 10 Problems and Solutions

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

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