# 1961 AHSME Problems/Problem 33

## Problem

The number of solutions of $2^{2x}-3^{2y}=55$, in which $x$ and $y$ are integers, is: $$\textbf{(A)} \ 0 \qquad\textbf{(B)} \ 1 \qquad \textbf{(C)} \ 2 \qquad\textbf{(D)} \ 3\qquad \textbf{(E)} \ \text{More than three, but finite}$$

## Solution

Let $a = 2^x$ and $b = 3^y$. Substituting these values results in $$a^2 - b^2 = 55$$ Factor the difference of squares to get $$(a + b)(a - b) = 55$$ If $y < 0$, then $55 + 3^{2y} < 64$, so $y$ can not be negative. If $x < 0$, then $2^{2x} < 1$. Since $3^{2y}$ is always positive, the result would be way less than $55$, so $x$ can not be negative. Thus, $x$ and $y$ have to be nonnegative, so $a$ and $b$ are integers. Thus, $$a+b=55 \text{ and } a-b=1$$ $$\text{or}$$ $$a+b=11 \text{ and } a-b=5$$ From the first case, $a = 28$ and $b = 27$. Since $2^x = 28$ does not have an integral solution, the first case does not work. From the second case, $a = 8$ and $b = 3$. Thus, $x = 3$ and $y = 1$. Thus, there is only one solution, which is answer choice $\boxed{\textbf{(B)}}$.