Is it possible to obtain the Uniform distribution as the difference of two independent random variables? Is it possible to have two independent random variables X,Y with identical distribution, such that  $X-Y \sim \text{Uniform}[a,b]$? I am almost certain that is not, but maybe I am overlooking something?

Assume that $X$ and $Y$ are i.i.d. and that $Z=X-Y$ is uniformly distributed on $(a,b)$. 
Then $(Y,X)$ is distributed like $(X,Y)$ hence $Y-X$ is distributed like $X-Y$. Since $Y-X=-Z$ is uniformly distributed on $(-b,-a)$, this proves that $(a,b)=(-b,-a)$. Thus $a=-b$ and the characteristic function $\varphi_Z$ of $Z$ is such that $\varphi_Z(t)=\sin(bt)/(bt)$ for every $t$. 
By independence of $X$ and $Y$, $E(\mathrm e^{\mathrm itZ})=E(\mathrm e^{\mathrm itX})E(\mathrm e^{-\mathrm itY})$. Furthermore, since $Y$ is distributed like $X$, $\varphi_Z(t)=\varphi_X(t)\varphi_X(-t)=|\varphi_X(t)|^2\geqslant0$ for every $t$. But obviously, $\varphi_Z(t)<0$ for some well chosen $t$, say $t=3\pi/(2b)$, which is absurd.
