JEE Main · 2024 · 29 Jan 2024 · Shift 2 · Sets Relations and Functions

$x^2-2^y=2023\Rightarrow x^2=2023+2^y$, with $x,y$ natural numbers.

For $y\ge 2$, $2^y\equiv 0(\mathrm{mod}4)$, so $x^2\equiv 2023(\mathrm{mod}4)$. Since $2023=4(505)+3$, this requires $x^2\equiv 3(\mathrm{mod}4)$ — impossible, as a perfect square is always $\equiv 0$ or $1(\mathrm{mod}4)$. So no solutions exist for $y\ge 2$.

Check $y=1$: $x^2=2023+2=2025=45^2$. Valid: $(x,y)=(45,1)$.

(If $y=0$ were allowed, $x^2=2024$ is not a perfect square, since $44^2=1936$ and $45^2=2025$.)

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