Let the set $C=\left\{(x, y) \mid x^2-2^y=2023, x, y \in \mathbb{N}\right\}$. Then $\sum_\limits{(x, y) \in C}(x+y)$ is equal to _________.

<p>First, let's consider the equation $x^2 - 2^y = 2023$ where $x$ and $y$ are natural numbers. Our goal is to find all the pairs $(x, y)$ that satisfy this equation and then sum the values of $x+y$ for each pair in set $C$. <p>Since $2023$ is an odd number, and $x^2$, the square of any natural number, is even when $x$ is even and odd when $x$ is odd, we can determine that for the left-hand side of the equation to be odd (thus equal to $2023$), $x$ must be odd since the right-hand side of the equation ($2^y$) is always even as it represents a power of two.</p> <p>Also, $2023$ can be factored into prime factors to further analyze the possible solutions:</p> <p>$2023 = 7 \times 17 \times 17$</p> <p>Thus, allowing us to rewrite the equation as:</p> <p>$x^2 - 2^y = 7 \times 17^2$</p> <p>The next step is to check for potential values of $x$ that would fit the equation, keeping in mind that $x$ must be odd. We can try to express $x^2$ as $7 \times 17^2$ plus a power of $2$, recognizing that we are looking for the decomposition of the form:</p> <p>$x^2 = 7 \times 17^2 + 2^y$</p> <p>By examining the powers of $2$ and keeping in mind that they grow very quickly, we can reason that $y$ cannot be very large because $x^2$ must not exceed $2023$ by a large margin.</p> <p>Let's start by trying the lowest values for $y$ since that would make $2^y$ small and $x$ has a better chance of being a natural number:</p> <ol> <li>For $y=1$:</li> </ol> <p>$x^2 = 2023 + 2^1 = 2023 + 2 = 2025$</p> <p>Surprisingly, we find a perfect square since $45^2 = 2025$. Therefore, $(x, y) = (45, 1)$ is one solution.</p> <ol> <li>For $y=2$ or higher:</li> </ol> <p>$2^y$ becomes at least $4$ and increases exponentially, so $x^2$ must be at least $2027$ or higher in such cases. There's no natural number between $45$ and $46$, and $46^2$ far exceeds the target (2116), making it impossible for $x^2$ to be less than $2116$ for any larger $y$.</p> <p>Hence, it appears there is only one possible solution: $(x, y) = (45, 1)$.</p> <p>Therefore, the sum $\sum_\limits{(x, y) \in C}(x+y)$ for this set will consist of only this one pair:</p> <p>$\sum_\limits{(x, y) \in C}(x+y) = 45 + 1 = 46$</p> <p>So the answer is $46$. </p></p>