2007 AMC 10A Problem 20

We do not need to find $a$. Squaring $a + \frac1a$ lets us climb from the first power to the second, and squaring again reaches the fourth.

1. Square once. Multiply out $\left(a + \frac1a\right)^2$:
$$\left(a + \frac1a\right)^2 = a^2 + 2 \cdot a \cdot \frac1a + \frac{1}{a^2} = a^2 + 2 + \frac{1}{a^2}.$$
The middle term is just $2$, because $a \cdot \frac1a = 1$.
2. Use $a + \frac1a = 4$. The left side is $4^2 = 16$, so
$$a^2 + \frac{1}{a^2} = 16 - 2 = 14.$$
3. Square again, the same way, with $a^2$ in place of $a$:
$$\left(a^2 + \frac{1}{a^2}\right)^2 = a^4 + 2 + \frac{1}{a^4}.$$
4. Use the value $14$. The left side is $14^2 = 196$, so
$$a^4 + \frac{1}{a^4} = 196 - 2 = 194.$$

The answer is $\boxed{\textbf{(D)}\ 194}$.

A number times its reciprocal is $1$, so when you square "number plus reciprocal" the middle term is always the plain number $2$. That lets you go from powers $1$ to $2$ to $4$ without ever solving for $a$, which here is a messy number ($2 + \sqrt3$ or $2 - \sqrt3$). Whenever you see $a$ and $\frac1a$ together, try squaring.

A quick check with the choices: the last step shows the answer is $2$ less than a perfect square (the square of the whole number $a^2 + \frac{1}{a^2}$). Among the choices, only $194 = 196 - 2 = 14^2 - 2$ has that form: $164 + 2$, $172 + 2$, $192 + 2$ and $212 + 2$ are not perfect squares.

The key identity: for any nonzero $a$,
$$\left(a^k + a^{-k}\right)^2 = a^{2k} + 2 + a^{-2k}, \quad\text{so}\quad a^{2k} + a^{-2k} = \left(a^k + a^{-k}\right)^2 - 2.$$

Apply it with $k = 1$: $a^2 + a^{-2} = 4^2 - 2 = 14$.

Apply it with $k = 2$: $a^4 + a^{-4} = 14^2 - 2 = 196 - 2 = 194$.

The answer is $\boxed{\textbf{(D)}\ 194}$.