How about this...

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

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

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 $\mathbb P[X>1]=0$. 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.

How about this...

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

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

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.

How about this...

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

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

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 $\mathbb P[X>1]=0$. 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.

How about this...

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

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

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 $\mathbb P[X>1]=0$. 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.