Let $v\_q(n,r)=\sum\_{i=0}^r \binom{n}i (q-1)^i$ denote a number of points in a ball of radius $r$ in the Hamming metric on the cube $\Sigma^n$, where $|\Sigma|=q$. What is the maximal number of points in $\Sigma^n$ with mutual distances at least $d$ (later: $d$-distant point sets)? **Gilbert**'s bound says that we may find $M$ such points if $(M-1)\cdot v\_q(n,d-1)< q^n$: add points one by one, while we have less than $M$ points, the $(d-1)$-balls centered at them do not cover $\Sigma^n$ and we may add another point. If $q$ is a power of prime, **Varshamov**'s bound improves the previous result by claiming that we may find $q^k$ $d$-distant points whenever $q^{k-1}v\_q(n,d-1)<q^n$. For the proof, we identify $\Sigma$ with a finite field and consider the maximal subspace $X$ of $\Sigma^n$ into which all points are $d$-distant. If $\dim X\geqslant k$, we are done, else there exists a point $v$ not covered by $(d-1)$-balls centered at elements of $X$, and the span of $X$ and $v$ is larger $d$-distant subspace.


My question is: is Varshamov's bound true if we do not require that $q$ is a prime power (say, for $q=6$)? This often happens and always does astonish me when arithmetical properties like being prime powers are crucial for purely combinatorial or topological claims, and I wonder what is the situation in this particular case.



Is primality essential in Varshamov's bound?
The analogue of the Varshamov inequality need not hold for instance for $\lvert\Sigma\rvert=q=6$: For $n=4$ and $d=3$ it would give a code $C\subseteq\Sigma^4$ with $d(C)\ge3$ and $\lvert C\rvert=36$. But such a code gives a pair of orthogonal Latin squares of order $6$, which is known not to exist (by Tarry's negative answer to Euler's [thirty-six officers problem](http://Thirty-six%20officers%20problem)).


(The construction of the orthogonal pair $A,B$ is as follows: Since $d(C)\ge3$, the map $C\to\Sigma^2$, $(i,j,a,b)\mapsto(i,j)$ is bijective. For $(i,j,a,b)\in C$ let $A$ and $B$ have entry $a$ and $b$ in the $i$-th row and $j$-th column, respectively.)


