Find the closest/next $x\in\mathbb{N}$ that satisfies the equation $Ⲭ_ℕ(f(x)) = 1$, where $Ⲭ_\mathbb{N}(x): \mathbb{R} \rightarrow \mathbb{N}$ is defined by $$ \begin{cases} 1 &\text{if } f(x) \in \mathbb{N}\\ 0 &\text{if } f(x) \notin \mathbb{N}, \end{cases} $$
So $Ⲭ_ℕ(x)$ is the indicator function
$f(x)$ can be choosen freely, but is known.
I am trying to ask if there is a method or an approach known, to solve - for any function (lets call it $f(x)$) that maps a $x \in ℕ$ into the real numbers - to find the next $x' \in ℕ$ so that $f(x')$ in the indicator function $Ⲭ_ℕ$ results in 1?
Some examples to show what I mean (if I find more, I will edit it):
Square root of a linear function
Let $f'(x) = \sqrt{k}$ and $x_1 = 3$ (e.g.)
$x_{test} = f'(x_1) = \sqrt{3} \approx 1.73205081$
$x_{solution} = \lceil{x_{test}}\rceil^2 = 2^2 = 4 \in ℕ \rightarrow Ⲭ_ℕ(f'(x_{solution})) = Ⲭ_ℕ(\sqrt{x_{solution}}) = Ⲭ_ℕ(2) = 1$
Fractions
Let $f''(x) = x * \frac{a}{b}$ and $x_2 = 2, a = 6, b = 7$ (free to choose, just an example)
$x_{test'} = f''(x_2) = 2 * \frac{a}{b} = \frac{12}{7}$
$x_{solution'} = \lceil(\frac{12}{7}) / {a}\rceil * b = (\lceil\frac{12}{42}\rceil * 7) = 7 \in ℕ \rightarrow Ⲭ_ℕ(f''(x_{solution'})) = Ⲭ_ℕ(f''(7)) = Ⲭ_ℕ(6) = 1$
So is there any approach or method to solve this for any real function (e.g. square root with multiple variables: $\sqrt{(a+x)-b}$ or $\sqrt {x^2-q}$) that maybe requires more than rounding?
Thank you in advance!
R