Let the set C=(x,y)|x^2-2^y=2023,x,y N.

Question

Let the set $C={(x,y)|{x}^{2}-{2}^{y}=2023,x,y\in N}$. Then $\sum _{(x,y)\in C}(x+y)$ is equal to \_\_\_\_\_\_\_\_\_.

Answer: 46

Accepted Answer

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:

$2023=7	imes 17	imes 17$

Thus, allowing us to rewrite the equation as:

${x}^{2}-{2}^{y}=7	imes {17}^{2}$

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 $7	imes {17}^{2}$ plus a power of $2$, recognizing that we are looking for the decomposition of the form:

${x}^{2}=7	imes {17}^{2}+{2}^{y}$

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:

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.

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 $46$.

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