it is known that if $x$ is integer and $x=-2n$ with $n >0 $ ,$\zeta(x)=0$ and is integer , here My question is to ask if there is a fixed real number for which $\zeta(x)$ is integer ?
Note: If there is no real number how I disproof the titled question ?
First, notice that $\zeta(x)$ is continuous for all $x>1$. Second, $\zeta(x)\to+\infty$ as $x\to1^+$ and $\zeta(2)<2$.
Thus, it follows by the intermediate value theorem, $\zeta(x)$ attains all integers greater than or equal to $2$.
Similarly argument for $x\in[-2,1)$ shows $\zeta(x)$ will take all negative integers including zero.