TITLE: Does $(\mathbb{Z}/n\mathbb{Z})^2$ ever admit a difference set when $n$ is odd?
QUESTION [9 upvotes]: A difference set of a group $G$ is a subset $D\subseteq G$ with the property that there exists an integer $\lambda>0$ such that for every non-identity member $g$ of $G$, there exist exactly $\lambda$ ordered pairs $(a,b)\in D\times D$ such that $g=ab^{-1}$. Note that $D=G$ is a difference set with $\lambda=|G|$, and so we typically only consider nontrival difference sets.
Davis showed that $(\mathbb{Z}/n\mathbb{Z})^2$ admits a nontrivial difference set when $n$ is a power of $2$. Are there any known difference sets when $n$ is odd? Perhaps cyclotomic difference sets?
As far as I know, the Paley-type construction Douglas Zare suggests in the comments (letting $D$ be the set of nonzero perfect squares in $\mathrm{GF}(n^2)$ when $n$ is prime) is only guaranteed to work when $n^2$ is $3\bmod 4$ (which never happens). However, there are hopefully weaker sufficient conditions for $n$ to satisfy, and I think the literature discusses this in the context of "cyclotomic difference sets," but I am not familiar with these results.

REPLY [9 votes]: Such difference sets exist. There exist (nontrivial) difference sets with 
$|G|  =  q^{d+1}[1+(q^{d+1}-1)/(q-1)]$,
$|D|   =  q^d(q^{d+1}-1)/(q-1)$,
$\lambda  =  q^d(q^d-1)/(q-1)$,
whenever $q$ is a prime power (R. L. McFarland, A family of difference sets in non-cyclic groups, JCT A, 15 (1973), pp. 1-10). More precisely, such difference sets exist in any abelian group of order $v$ which contains an elementary abelian subgroup of order $q^{d+1}$. 
Take $q=7$ and $d=1$, for instance. This shows that a nontrivial difference set in $(\mathbb{Z}/21\mathbb{Z})^2$ exists.