Key Concepts and Formulas

• Diophantine Equations: Equations where only integer solutions are sought.
• Modular Arithmetic: Analyzing remainders after division by a modulus. Notation: $a \equiv b \pmod{m}$ means $a$ and $b$ have the same remainder when divided by $m$.
• Properties of Squares Modulo 4: Any perfect square is congruent to either 0 or 1 modulo 4.

Step-by-Step Solution

Step 1: Analyze Parity (Modulo 2)

We are given the equation $x^2 - 2^y = 2023$, where $x, y \in \mathbb{N}$. We want to find all possible pairs $(x, y)$ that satisfy this equation. First, we rewrite the equation as $x^2 = 2023 + 2^y$. We analyze this equation modulo 2 to determine the parity of $x$.

Since 2023 is odd, $2023 \equiv 1 \pmod{2}$. For $y \geq 1$, $2^y$ is always even, so $2^y \equiv 0 \pmod{2}$. Therefore, $x^2 \equiv 1 + 0 \pmod{2}$ $x^2 \equiv 1 \pmod{2}$ This implies that $x^2$ is odd, and thus $x$ must be odd.

Step 2: Case Analysis on the Value of $y$

Now, we consider different cases based on the value of $y$.

Case 1: $y = 1$

Substitute $y = 1$ into the equation $x^2 = 2023 + 2^y$: $x^2 = 2023 + 2^1$ $x^2 = 2023 + 2$ $x^2 = 2025$ Taking the square root of both sides, we get $x = \sqrt{2025} = 45$. Since $x = 45$ is a natural number and is odd, $(x, y) = (45, 1)$ is a valid solution.

Case 2: $y \geq 2$

If $y \geq 2$, then $2^y$ is divisible by 4, so $2^y \equiv 0 \pmod{4}$. We analyze the original equation $x^2 - 2^y = 2023$ modulo 4.

First, we find the remainder of 2023 when divided by 4: $2023 = 4 \times 505 + 3$ So, $2023 \equiv 3 \pmod{4}$.

Now, substitute the congruences into the original equation modulo 4: $x^2 - 2^y \equiv 2023 \pmod{4}$ $x^2 - 0 \equiv 3 \pmod{4}$ $x^2 \equiv 3 \pmod{4}$

We know that any perfect square is congruent to either 0 or 1 modulo 4. This is because if $x$ is even, $x = 2k$, so $x^2 = 4k^2 \equiv 0 \pmod{4}$. If $x$ is odd, $x = 2k + 1$, so $x^2 = (2k + 1)^2 = 4k^2 + 4k + 1 \equiv 1 \pmod{4}$.

Since $x^2 \equiv 3 \pmod{4}$ contradicts the property that perfect squares are only congruent to 0 or 1 modulo 4, there are no integer solutions for $x$ when $y \geq 2$.

Step 3: Identify the Set $C$ and Calculate the Sum

The only solution we found is $(x, y) = (45, 1)$. Therefore, the set $C = \{(45, 1)\}$.

We are asked to calculate $\sum_{(x, y) \in C} (x + y)$. Since $C$ only contains one element, $\sum_{(x, y) \in C} (x + y) = 45 + 1 = 46$

Common Mistakes & Tips

• Always consider modular arithmetic to simplify Diophantine equations.
• Remember that squares can only be 0 or 1 mod 4.
• Don't forget to check small values of $y$ as they can sometimes lead to easy solutions.

Summary

We analyzed the equation $x^2 - 2^y = 2023$ using modular arithmetic. By considering the equation modulo 2, we determined that $x$ must be odd. Then, we performed a case analysis on $y$. For $y=1$, we found the solution $(45, 1)$. For $y \geq 2$, we analyzed the equation modulo 4 and found a contradiction, meaning there were no solutions in this case. Therefore, the only solution is $(45, 1)$, and the sum $x+y$ is $45+1 = 46$.

The final answer is \boxed{46}.