JEE Mathematics 2024 Question with Solution

Given: $x^2 - 2^y = 2023$ with $x, y \in N$.

Find: $\sum_{(x,y) \in C}(x+y)$.

From the solution working, test natural values of $y$ so that $2023 + 2^y$ becomes a perfect square.

For $y = 1$,

So,

Thus, the solution in $C$ is $(x,y) = (45,1)$.

Now,

Therefore, the required numerical value is $46$.

Given: $x^2 - 2^y = 2023$.

Find: the value of $\sum_{(x,y) \in C}(x+y)$.

Rearrange the equation as

We need the right-hand side to be a perfect square.

Trying the smallest natural value shown in the extracted solution,

Hence $x=45$.

So the pair is $(45,1)$ and

Hence the sum over the set is $46$.