conjecture posed here in one of the answers.
The sum of $ n \geq 3 $ independent r.v.'s distributed on $ [-\alpha, \alpha] $ cannot be uniform on $ [-n\alpha, n\alpha] $.
Is this conjecture true if the random variables are allowed to be distributed on $[-n\alpha, n\alpha]$?
The modifed conjecture is false.
Let:
$U$ be $\text{Unif}(-\frac{3\alpha}4,\frac{3\alpha}4)$.
$X$ take the values $-\frac{3\alpha}4$ and $\frac{3\alpha}4$ each with probability 0.5.
$Y$ take the values $-\frac{6\alpha}4$ and $\frac{6\alpha}4$ each with probability 0.5.
Then,
$U+X\text~\text{Unif}(-\frac{6\alpha}4,\frac{6\alpha}4)$ and $U+X+Y\text~\text{Unif}(-3\alpha,3\alpha)$.
In general, it can be shown by induction that if:
$U$ is $\text{Unif}(-\beta,\beta)$
$X_1$ takes the values $-\beta$ and $\beta$ each with probability 0.5.
$X_2$ takes the values $-2 \beta$ and $2\beta$ each with probability 0.5.
$X_3$ takes the values $-2^2 \beta$ and $2^2 \beta$ each with probability 0.5.
...
$X_{n-1}$ takes the values $-2^{n-2}\beta$ and $2^{n-2}\beta$ each with probability 0.5.
Then,
$$U+\sum_{i=1}^{n-1} X_i\text{~Unif}\left(-2^{n-1}\beta,2^{n-1}\beta\right)$$
Therefore, if we take $\beta=n 2^{1-n}\alpha$, then the sum of the $n$ random variables will be uniform on $ [-n\alpha, n\alpha]$.