Proof of 3-perfect codes I am reading a the proof of a theorem that says that $3$-perfect codes can only have length $7$ or $23$. I do not understand the following:
It follows from the definition that if $n$ is the length of a $3$-perfect then we must have that $(n+1)((n+1)^3-3(n+1)+8) = 3 \cdot 2^k$ for some $k \in \mathbb{Z}^+$. Now the proof claims that $16$ can not divide $(n+1)$.
I do not why. Any help or hints appreciated. 
 A: Suppose the highest power of $2$ which divides $n+1$ be $m \ge 4$, i.e., $2^4 = 16$ divides $n + 1$. Thus, you have that
$$n + 1 = 2^{m}j \tag{1}\label{eq1A}$$
where $j$ is odd. Since $n + 1$ divides evenly into $3 \cdot 2^k$, this means $j = 1$ or $j = 3$. You also have the equation
$$\begin{equation}\begin{aligned}
(n+1)((n+1)^3-3(n+1)+8) & = 3 \cdot 2^k \\
2^m(j)(2^{3m}j^3 - 3j2^{m} + 8) & = 3 \cdot 2^k \\
j(2^{m+3})(2^{3m-3}j^3 - 3j2^{m-3} + 1) & = 3 \cdot 2^k \\
j(2^{3m-3}j^3 - 3j2^{m-3} + 1) & = 3 \cdot 2^{k-m-3}
\end{aligned}\end{equation}\tag{2}\label{eq2A}$$
Note that both $j$ and $2^{3m-3}j^3 - 3j2^{m-3} + 1$ are odd, so their product is odd. Thus, you must have $k = m + 3$ on the right, so it's value is just $3$.
If $j = 1$, then you have
$$\begin{equation}\begin{aligned}
j(2^{3m-3}j^3 - 3j2^{m-3} + 1) & = 3 \\
2^{3m-3} - 3(2^{m-3}) + 1 & = 3 \\
2^{3m-3} - 3(2^{m-3}) & = 2 \\
2^{3m-4} - 3(2^{m-4}) & = 1
\end{aligned}\end{equation}\tag{3}\label{eq3A}$$
If $m = 4$, then $3m - 4 = 8$ so you have $2^{8} - 3 = 1$, which is not true. However, if $m \gt 4$, then the left side has at least one factor of $2$ but the right side doesn't have any factors, so that can't be true.
If $j = 3$ instead, this then means that you have
$$\begin{equation}\begin{aligned}
3(2^{3m-3}j^3 - 3j2^{m-3} + 1) & = 3 \\
2^{3m-3}(3^3) - 3(3)2^{m-3} + 1 & = 1 \\
27(2^{3m-3}) & = 9(2^{m-3}) \\
3(2^{3m-3}) & = 2^{m-3}
\end{aligned}\end{equation}\tag{4}\label{eq4A}$$
However, there's at least one factor of $3$ on the left but no factor of $3$ on the right, so this can't be true either.
Since all of the possibilities have been considered and shown to not be possible, this means the original assumption, i.e., that $16$ divides $n + 1$, can't be true.
Update: A somewhat shorter & simpler way would be to note that, in \eqref{eq2A}, due to the relatively large size of the first term, you have for $j \ge 1$ and $m \ge 4$ that $2^{3m-3}j^3 - 3j2^{m-3} + 1 \gt 3$, so it's not possible as $3$ is the odd factor on the right.
