QJEEM24Ma31b5: JEE Main 2024 Mathematics - Class 11 JEEM24Ma31b5 Solution: Let the set . Then ∑ ( x , y ) ∈ C ( x + y ) is equal to…

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 $C$ .

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.

Also, $2023$ can be factored into prime factors to further analyze the possible solutions:

$2023 = 7 \times 17 \times 17$

Thus, allowing us to rewrite the equation as:

$x^2 - 2^y = 7 \times 17^2$

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:

$x^2 = 7 \times 17^2 + 2^y$

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.

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:

1. For $y=1$ :

$x^2 = 2023 + 2^1 = 2023 + 2 = 2025$

Surprisingly, we find a perfect square since $45^2 = 2025$ . Therefore, $(x, y) = (45, 1)$ is one solution.

1. For $y=2$ or higher:

$2^y$ 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 $y$ .

Hence, it appears there is only one possible solution: $(x, y) = (45, 1)$ .

Therefore, the sum $$\sum _{(x,y)\in C}(x+y)$$ for this set will consist of only this one pair:

$$\sum _{(x,y)\in C}(x+y)=45+1=46$$

So the answer is $46$ .