Set theory or Number theory?

First recognize that $f(\mathbb{Z}) \cap g(\mathbb(Z) = \emptyset$ is equivalent to the statement that the Diophantine equation $x^2-y^2=a+b$ has no solution for $x,y \in \mathbb{Z}$ . Next we consider two cases.

Case 1 : Let $x \equiv y \mod 2$ . This gives us that $x^2-y^2 \equiv 0 \equiv a+b \mod 4$ . And in particular, $x=\frac{a+b}{4}+1,y=\frac{a+b}{4}-1$ is always a solution in this case. Thus, $a + b \not\equiv 0 \mod 4$ .

Case 2 : $x\not\equiv y \mod 2$ . This gives us that $x^2-y^2 \equiv 1 \equiv a+b \mod 2$ . And in particular, $x=\frac{a+b+1}{2},y=\frac{a+b-1}{2}$ is a solution in this case. Thus, $a+b \not\equiv 1,3 \mod 4$ . This leaves one possibility, $a+b \equiv 2 \mod 4$ , the smallest solution in non-negative numbers being $\boxed{a+b = 2}$ .

For $a+b = 0$ and $a+b=1$ it is easy to find instances showing that

$f(\mathbb{Z}) \cap g(\mathbb{Z}) \ne \emptyset$ .

The next smallest possibility $\boxed{a+b=2}$ , (with $a=b=1$ ) is the required minimum.

Proof

Suppose on the contrary that there are some integers x and y such that

$x^2-1=y^2+1 \implies x^2-y^2=2 \implies (x-y)(x+y)=2$

By the fundamental theorem of arithmetic we can say that one of $(x-y)$ and $(x+y)$ is $\pm 1$ and the other one is $\pm 2$ . Either way adding the two factors gives

$2x= \pm 3$

Since this is contrary to the assumption that $x \in \mathbb{Z}$ we can complete the proof by contradiction.