Let the set C = {(x, y) mid x^2 - 2^y = 2023, | vedclass

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(A) Given the equation $x^2 - 2^y = 2023$ where $x, y \in \mathbb{N}$.Rearranging the equation,we get $x^2 - 2023 = 2^y$.If $y = 1$,then $x^2 - 2023 =

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$46$

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(A) Given the equation $x^2 - 2^y = 2023$ where $x, y \in \mathbb{N}$.Rearranging the equation,we get $x^2 - 2023 = 2^y$.If $y = 1$,then $x^2 - 2023 = 2^1 = 2$,which implies $x^2 = 2025$,so $x = 45$.If $y = 2$,then $x^2 - 2023 = 4$,which implies $x^2 = 2027$ (not a perfect square).If $y = 3$,then $x^2 - 2023 = 8$,which implies $x^2 = 2031$ (not a perfect square).If $y = 4$,then $x^2 - 2023 = 16$,which implies $x^2 = 2039$ (not a perfect square).If $y = 5$,then $x^2 - 2023 = 32$,which implies $x^2 = 2055$ (not a perfect square).If $y = 6$,then $x^2 - 2023 = 64$,which implies $x^2 = 2087$ (not a perfect square).If $y = 7$,then $x^2 - 2023 = 128$,which implies $x^2 = 2151$ (not a perfect square).If $y = 8$,then $x^2 - 2023 = 256$,which implies $x^2 = 2279$ (not a perfect square).If $y = 9$,then $x^2 - 2023 = 512$,which implies $x^2 = 2535$ (not a perfect square).If $y = 10$,then $x^2 - 2023 = 1024$,which implies $x^2 = 3047$ (not a perfect square).For $y > 1$,considering the equation modulo $3$: $x^2 - 2^y \equiv 2023 \pmod 3 \Rightarrow x^2 - (-1)^y \equiv 1 \pmod 3$.If $y$ is even,$x^2 - 1 \equiv 1 \Rightarrow x^2 \equiv 2 \pmod 3$,which is impossible.If $y$ is odd,$x^2 + 1 \equiv 1 \Rightarrow x^2 \equiv 0 \pmod 3$,so $x$ must be a multiple of $3$. Let $x = 3k$.Then $9k^2 - 2^y = 2023$. For $y > 1$,$2^y$ is a multiple of $4$. $9k^2 - 2023 = 2^y$. This leads to no further solutions.Thus,the only solution is $(45, 1)$.The sum $\sum_{(x, y) \in C} (x + y) = 45 + 1 = 46$.

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

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