2020 AMC 10A Problem 14

Put the two fractions over the common denominator $x^2 y^2$:
$$x + y + \frac{x^5 + y^5}{x^2 y^2}.$$
Here $x + y = 4$ and $x^2 y^2 = (xy)^2 = 4$, so we need $x^5 + y^5$.

Let $s_n = x^n + y^n$. Then $s_1 = 4$ and
$$s_2 = (x+y)^2 - 2xy = 16 + 4 = 20.$$
Since $x$ and $y$ are roots of $t^2 - 4t - 2 = 0$, each power sum satisfies $s_{n} = 4 s_{n-1} + 2 s_{n-2}$:
$$s_3 = 4 \cdot 20 + 2 \cdot 4 = 88,\qquad s_4 = 4 \cdot 88 + 2 \cdot 20 = 392,\qquad s_5 = 4 \cdot 392 + 2 \cdot 88 = 1744.$$

Therefore the expression equals
$$4 + \frac{1744}{4} = 4 + 436 = 440.$$

The answer is $\boxed{\textbf{(D)}\ 440}$.

Any symmetric expression in $x$ and $y$ is a polynomial in $x + y$ and $xy$, so it can be evaluated without finding $x$ and $y$ (which are irrational here). The recurrence $s_n = (x+y)s_{n-1} - xy\,s_{n-2}$ generates all power sums mechanically; alternatively $s_5 = s_2 s_3 - (xy)^2 s_1 = 20 \cdot 88 - 4 \cdot 4 = 1744$.

• A sign slip in x^3 + y^3 = (x+y)^3 - 3xy(x+y) = 64 + 24 = 88 (writing 64 - 24 = 40) derails everything downstream.
• Forgetting the outer $x + y = 4$ and answering $436$ (not offered, but a warning sign), or multiplying the fractions instead of adding them.