Animated Solution for Mathematics - Basic Mathematics: Let the set $C=\{(x,y)\mid x^2-2^y=2023,x,y\in N\}$. Then ${\sum }_{(x,y)\in C}(x+y)$ is equal to

The given equation is a Diophantine equation of the form:

We are tasked with finding all integer solutions $(x,y)$ that satisfy this relationship.

We begin by testing the simplest case where $y=1$. Substituting this into the equation, we get:

Taking the square root, we find $x=45$. Thus, we have identified our first valid solution pair: $(45,1)$.

Now, we consider the case where $y\ge 2$. To analyze this, we employ modular arithmetic, specifically looking at the equation modulo $4$.

For any $y\ge 2$, the term $2^y$ is a multiple of $4$, which implies:

Substituting this into our original equation, we obtain:

Since $2023=2020+3$, we can simplify the congruence to:

However, we know that for any integer $x$, a perfect square can only be congruent to $0$ or $1$ modulo $4$. Specifically:

If $x$ is even, $x=2k\Rightarrow x^2=4k^2\equiv 0(\mathrm{mod}4)$.

If $x$ is odd, $x=2k+1\Rightarrow x^2=4k^2+4k+1\equiv 1(\mathrm{mod}4)$.

Because $x^2\equiv 3(\mathrm{mod}4)$ is impossible, there are no integer solutions for $y\ge 2$.

The only valid pair is $(45,1)$. Consequently, the sum of these values is: