Convolution that is equiprobable to a uniform distribution I am struggling with finding two distributions a and b on the non
non-negative integers (both not concentrated at 0, this is t trivial)
such that the convolution of a and b is the equiprobable  distribution on the set 0,1,...n-1
n is not a prime number.
Any ideas?
 A: This "solution" could be considered as a "comment" on... the excellent "solution" given by @Did as a "comment", 
$$\tag{*}\frac1{k\ell}(1+s+\cdots+s^{k\ell-1})=\frac1k(1+s^\ell+\cdots+s^{(k-1)\ell})\frac1\ell(1+s+\cdots+s^{\ell-1})$$
but I am close to believe it helps to understand it:
A) through an example:
Consider the uniform distribution $d_1$ on the set $\{0,1\}$ and the uniform distribution $d_2$ on the set $\{0,2,4,6,8\}$.
Let $X,Y$ be independent Random Variables (R.V.) that follow $d_1,d_2$ resp.
Then R.V. $Z:=X+Y$ follows a uniform distribution on $\{0,1,2,3,4,5,6,7,8,9\}$.
(we have thus answered the question because the distribution of the sum of 2 independent R.V. is the convolution of their distributions).
Formal proof: Taking into account that $P(X=a)=\frac12$ (for $ \ b=0,1$) and $P(Y=b)=\frac15$, (for $ \ b=0,2,4,6,8$), the two cases to be considered are:
$P(Z=2k)=P(X=0).P(Y=2k)=\frac12.\frac15=\frac{1}{10}$
$P(Z=2k+1)=P(X=1).P(Y=2k)=\frac12.\frac15=\frac{1}{10}$
b) by giving it a direct proof: Using the formula for the sum of a finite geometric progression (the first one with ratio $s^{\ell}$, the second one with ratio $\ell$, (*) is transformed into the evident relationship :
$$\dfrac{1-(s^{\ell})^k}{1-s^{\ell}}\dfrac{1-s^\ell}{1-s}=\dfrac{1-s^{\ell k}}{1-s}$$
