Let $C \subseteq F^n_2$ be a linear $[n,k,d]_2-$code.
Show that if $C$ is a perfect code and $d = 3$, then $n$ is an odd number.
($d$ represents the minimum distance)
A $[n,k,d]_q$ code is perfect if $q^k=\frac{q^n}{|B_t(x)|}$, where:
$|B_t(x)|= \sum_{i=0}^t {{n}\choose{i}}(q-1)^i$
$t=\lfloor \frac{d-1}{2} \rfloor$
In this particulary case we have $q=2$ and $d=3$. So $t=\lfloor \frac{3-1}{2} \rfloor=1$. Now we can calculate $|B_t(x)|$.
$|B_t(x)|= \sum_{i=0}^t {{n}\choose{i}}(q-1)^i={{n}\choose{0}}+{{n}\choose{1}}=n+1$
Since the code is perfect we have $q^k=\frac{q^n}{|B_t(x)|} \Leftrightarrow 2^k=\frac{2^n}{n+1}$
Singleton Limit says that the $d \leqslant n-dim(C)+1$, So $d \leqslant n-k+1 \Leftrightarrow 3 \leqslant n-k+1 \Leftrightarrow k \leqslant n-2$
Assuming $k=n-2$, we have $2^{n-2}=\frac{2^n}{n+1} \Leftrightarrow n=3$ (and its an odd number)
However this is assuming $k=n-2$, how do I do for the case where $k<n-2$?
EDIT:
I think I got it. From $2^k=\frac{2^n}{n+1}$ , we write it as $2^{n-k}=n+1$. Because $2^{n-k}$ is a power of $2$ it will always be even. So it can't be equal to $n+1$ if $n+1$ is odd, so $n+1$ must be even,therefore $n$ must be odd. Singleton Limit isn't necessary for this.