Can you find a function that follows a rule while rational and it's differentiable everywhere? (relates to the Pythagorean theorem)

Question

can you find a function that follows a rule while rational and it's differentiable? let's call this function $\alpha(x)$

$x$'s simplest form is $\frac{a}{b}$ when $x$ is a fraction.

when $a^2+b^2=c^2$ and $c$ is an integer then $\alpha(x)$ a rational number

when $a^2+b^2≠c^2$ then $\alpha(x)$ is a irrational number

and when x isn't a fraction $\alpha(x)$ can be either.

$\alpha(x)$ must be differentiable everywhere

Answer 1

$$\alpha(x)=\sqrt{x^2+1}$$ satisfies the conditions above:
when $x=\frac ab$ then $\sqrt{x^2+1}=\frac{\sqrt{a^2+b^2}}{|b|}$ and it's rational when there is such integer $c$ that $a^2+b^2=c^2$ and it's irrational when no such $c$ exist (however, I'm not sure if I should prove it too).
$$\left(\sqrt{x^2+1}\right)'=\frac{x}{\sqrt{x^2+1}}$$ is defined everywhere on $\mathbb{R}$ as $x^2+1>0$.