# Difference between revisions of "1990 AIME Problems/Problem 15"

## Problem

Find $a_{}^{}x^5 + b_{}y^5$ if the real numbers $a_{}^{}$, $b_{}^{}$, $x_{}^{}$, and $y_{}^{}$ satisfy the equations $$ax + by = 3^{}_{},$$ $$ax^2 + by^2 = 7^{}_{},$$ $$ax^3 + by^3 = 16^{}_{},$$ $$ax^4 + by^4 = 42^{}_{}.$$

## Solution

Set $S = (x + y)$ and $P = xy$. Then the relationship

$$(ax^n + by^n)(x + y) = (ax^{n + 1} + by^{n + 1}) + (xy)(ax^{n - 1} + by^{n - 1})$$

can be exploited:

$\begin{eqnarray*}(ax^2 + by^2)(x + y) & = & (ax^3 + by^3) + (xy)(ax + by) \\ (ax^3 + by^3)(x + y) & = & (ax^4 + by^4) + (xy)(ax^2 + by^2)\end{eqnarray*}$

Therefore:

$\begin{eqnarray*}7S & = & 16 + 3P \\ 16S & = & 42 + 7P\end{eqnarray*}$

Consequently, $S = - 14$ and $P = - 38$. Finally:

$\begin{eqnarray*}(ax^4 + by^4)(x + y) & = & (ax^5 + by^5) + (xy)(ax^3 + by^3) \\ (42)(S) & = & (ax^5 + by^5) + (P)(16) \\ (42)( - 14) & = & (ax^5 + by^5) + ( - 38)(16) \\ ax^5 + by^5 & = & \boxed{20}\end{eqnarray*}$