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 $\boxed{\textbf{(B)}}$ .

1961 AHSC (Problems • Answer Key • Resources)
Preceded by Problem 32	Followed by Problem 34
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1961 AHSME Problems/Problem 33

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