How about this...  

Uniform distribution on $[-1,1]$ has [characteristic function][1]

$$
\varphi(t) = \frac{\sin t}{t}
$$

Suppose $X,Y$ are IID and $Z:=X+Y$ is uniformly distributed on $[-1,1]$.
Of course $X,Y$ are bounded:
$$
\mathbb P[X>1]^2 = \mathbb P[X>1, Y>1] \le \mathbb P[Z>2] = 0
$$
so $\mathbb P[X>2]=0$.  Similarly $\mathbb P[X<-2]=0$.  The characteristic function of a bounded random variable is an entire function.  But the characteristic function $\psi(t)$ of $X$ and $Y$ satisfies
$$
\psi(t)^2 = \varphi(t)
$$
so it cannot be differentiable at $t=\pi$, where $\varphi(t)$ changes sign.

>!More... $\varphi(\pi)=0$ so $\psi(\pi)=0$. But if $\psi'(\pi)$ exists, we get $-1/\pi = \varphi'(\pi) = 2\psi(\pi)\psi'(\pi) = 0$, contradiction.  


  [1]: https://en.wikipedia.org/wiki/Characteristic_function_(probability_theory)