Sum of Independent Variables = Uniform? Is there a probability density (or measure), such that the sum of two such independent random variables is distributed uniform?  In other words, what is the Inverse Fourier Transform of the square-root of the Fourier Transform of a uniform density?
What about the Bernoulli distribution with two mass points at +1 and -1?  Is there a measure such that the convolution of this measure with itself gives the Bernoulli distribution?  I presume this measure will have negative mass points.
 A: Assume that $X$ and $Y$ are i.i.d. and that $X+Y$ is uniform on $(-1,1)$, then the characteristic function of $X$, defined by $\varphi:t\mapsto E[\mathrm e^{\mathrm i tX}]$, is such that
$$
\varphi(t)^2=E[\mathrm e^{\mathrm i t(X+Y)}]=\int_{-1}^1\mathrm e^{\mathrm i tu}\frac12\mathrm du=\frac{\sin t}t.
$$
Expanding the sine function at $0$ and using the fact that $\varphi(0)=1$ and that $\varphi$ is continuous yields
$$
\varphi(t)=1-\tfrac1{12}t^2+\tfrac1{1440}t^4-\tfrac1{24192}t^6+o(t^6).
$$
Since $\varphi$ is a characteristic function, the determinant
$$
\left|\begin{matrix}\varphi(0)&\varphi(t)&\varphi(2t)\\
\varphi(-t)&\varphi(0)&\varphi(t)\\ \varphi(-2t)&\varphi(-t)&\varphi(0)\end{matrix}\right|
$$
is a nonnegative real number, for every real number $t$. This determinant is
$$
D(t)=1-2|\varphi(t)|^2-|\varphi(2t)|^2+2\Re(\varphi(t)^2\varphi(-2t)).
$$
Plugging the expansion of $\varphi(t)$ in $D(t)$ yields
$$
D(t)=-\tfrac1{540}t^6+o(t^6),
$$
which is impossible.
The question for the Bernoulli distributions, say on $\{-1,1\}$, is simpler. Assume that $X+Y$ is a solution. Then $|X|\leqslant\frac12$ with full probability (why?), $P[X=\frac12]\ne0$ and $P[X=-\frac12]\ne0$ (why?) hence $P[X+Y=0]\ne0$ (why?), a contradiction.
