Let the set $C = \{(x, y) \mid x^2 - 2^y = 2023, x, y \in \mathbb{N}\}$. Then $\sum_{(x, y) \in C} (x + y)$…
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 _______.
Solution
Given $x^{2}-2^{y}=2023$
$\Rightarrow x^{2}=2^{y}+2023$
Now, if $y=1$ we get,
$\Rightarrow x^{2}=2025$
$\Rightarrow x=45$
And there is no other value which will satisfy the above equation,
Hence, $x=45$ and $y=1$
So, $\sum_{(x,y) \in C}(x+y)=45+1=46$
Alternative Solution:
Sure, here's how to solve this problem step by step.
The given equation is $x^2 - 2^y = 2023$, where $x$ and $y$ are natural numbers. We need to find the sum of all $(x+y)$ for which this equation holds true.
Firstly, we can see that $x^2$ must be slightly greater than 2023 and $2^y$ slightly less. So let's find a square number near to 2023. The square of 44 is 1936 and the square of 45 is 2025, which is very close to 2023. So, possible values for $x$ are 44 and 45.
Now, let's substitute these values into the equation:
1. For $x=44$, we have $44^2 - 2^y = 2023$. Solving for $y$ we get $2^y = 7$, which does not have a natural number solution, so $x$ cannot be 44.
2. For $x=45$, we have $45^2 - 2^y = 2023$. Solving for $y$ we get $2^y = 2$, which gives $y=1$.
So the only solution to the equation is $x=45$ and $y=1$.
Hence, the sum of all $(x+y)$ for the solutions is