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

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 $\boxed{1}$ solution.
1
Level 5
Number Theory
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:<|START_OF_LATENT|><|PRIOR_PREFIX|>To determine the number of integer solutions for the equation \(2^{2x} - 3^{2y} = 55\), we start by rewriting the equation in a more manageable form. We can express \(2^{2x}\) as \((2^x)^2\) and \(3^{2y}\) as \((3^y)^2\). Thus, the equation becomes:

\[(2^x)^2 - (3^y)^2 = 55.\]

This is a difference of squares, which can be factored:

\[(2^x - 3^y)(2^x + 3^y) = 55.\]

Next, we need to find pairs of integers \((a, b)\) such that \(a \cdot b = 55\), where \(a = 2^x - 3^y\) and \(b = 2^x + 3^y\). The integer factor pairs of 55 are:

\[(1, 55), (5, 11), (-1, -55), (-5, -11).\]

We will analyze each factor pair to find possible values for \(2^x\) and \(3^y\).

1. For the pair \((1, 55)\):
   - We have \(2^x - 3^y = 1\) and \(2^x + 3^y = 55\).
   - Adding these equations gives \(2 \cdot 2^x = 56\) or \(2^x = 28\), which is not a power of 2.
   
2. For the pair \((5, 11)\):
   - We have \(2^x - 3^y = 5\) and \(2^x + 3^y = 11\).
   - Adding these gives \(2 \cdot 2^x = 16\) or \(2^x = 8\), thus \(x = 3\).
   - Substituting \(x = 3\) into \(2^x - 3^y = 5\) gives \(8 - 3^y = 5\) or \(3^y = 3\), leading to \(y = 1\).
   - This gives us one solution: \((x, y) = (3, 1)\).

3. For the pair \((-1, -55)\):
   - We have \(2^x - 3^y = -1\) and \(2^x + 3^y = -55\).
   - This leads to contradictions since \(2^x\) and \(3^y\) are both positive.

4. For the pair \((-5, -11)\):
   - We have \(2^x - 3^y = -5\) and \(2^x + 3^y = -11\).
   - This also leads to contradictions for the same reason as above.

After analyzing all factor pairs, we find that the only valid solution occurs for the pair \((5, 11)\), yielding the single solution \((x, y) = (3, 1)\). Therefore, the total number of integer solutions to the equation \(2^{2x} - 3^{2y} = 55\) is \(1\).<|END_OF_LATENT|> The final answer is $1$.