2007 AMC 10A Problem 20

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 $\boxed{\textbf{(D)}\ 194}$.

Because $a \cdot a^{-1} = 1$, the cross term in the square of $a^k + a^{-k}$ is always the constant $2$. That makes the sums $a^{2^j} + a^{-2^j}$ reachable by repeated squaring without ever knowing $a$ (which is the irrational $2 \pm \sqrt3$). Symmetric expressions in $a$ and $1/a$ should always be attacked through $a + 1/a$.

Answer-choice check: $a^4 + a^{-4} = (a^2 + a^{-2})^2 - 2$, and $a^2 + a^{-2}$ is an integer, so the answer is $2$ less than a perfect square. Among the choices only $194 = 196 - 2$ qualifies.

• Forgetting the cross term and writing $(a + 1/a)^2 = a^2 + 1/a^2$, which gives $16$ then $256$ (or $254$).
• Solving the quadratic $a^2 - 4a + 1 = 0$ and expanding $(2 + \sqrt3)^4$ by hand, which is slow and invites arithmetic errors.