$\textbf{Edit:}$
Thank you all so much for all your help! Superglad I chose to ask. All your different approaches were very insightful and taught me a lot. Thank you!
Same question: Found here.
Hi,
I'm a bit stuck with my question. I've read through the answers in the above link, but I cannot quite understand the answers fully. I don't want to hijack the other thread either, so I thought I'd start a separate thread.
$\textbf{The question is:}$
For which ${n \ge 0}$ is the number ${n \cdot 2^n + 1}$ divisible by 3?
I'v gotten this far in my understanding:
For even ${n(n=2k)}$:
\begin{equation}
\begin{aligned}
2^n&=2^{2k} \equiv_3 4^k \equiv_3 1^k \equiv_3 1\\
n\cdot 2^n+1 &\equiv_3 n\cdot 1 + 1 \equiv_3 n +1
\end{aligned}
\end{equation}
For odd ${n(n=2k+1)}$:
\begin{equation}
\begin{aligned}
2^{2k+1}&=2^{2k}\cdot 2 \equiv_3 1^k \cdot -1 \equiv_3 1 \cdot – 1 \equiv_3 -1\\
n\cdot 2^n+1 &\equiv_3 n\cdot -1 + 1 \equiv_3 -n +1
\end{aligned}
\end{equation}
One of the answers in the link above mentioned
\begin{equation}
n\equiv 0 (\text{mod 2})\qquad n\equiv 2(\text{mod 3})
\end{equation}
which could be "consolidated as ${n\equiv 2 (\text{mod 6})}$".
I don't understand what the "consolidate"-part means.
So, would someone please like to share some knowledge and insights with me? It feels like I'm almost there, but still a far way to go.
Thanks
Best Answer
Respecting the comments from the OP, all the steps used in the answer have been rigorously expanded.
Of course, $n≠3k,~ k\in\mathbb Z^{+}$.
Take $n=3k-1,~k\in\mathbb Z^{+}$, then you have
$$\begin{align}&\frac{(3k-1)\times 2^{3k-1}+1}{3}\in\mathbb Z^{+}\\ \implies &\frac{2^{3k-1}-1}{3}\in\mathbb Z^{+} \\ \implies &3k-1=2m,~m\in\mathbb Z^{+}\\ \implies &\begin{cases}n=3k-1\\ 3k-1=2m\end {cases}\\ \implies &k=2k'-1,~k'\in\mathbb Z^{+}\\ \implies &n=3(2k'-1)-1,~k'\in\mathbb Z^{+}\\ \implies &n=6k'-4,~k'\in\mathbb Z^{+}\\ \implies &n=6k-4,~k'\longmapsto k .\end{align}$$
Finally take $n=3k-2,~k\in\mathbb Z^{+}$, you have
$$\begin{align}&\frac{(3k-2)\times 2^{3k-2}+1}{3}\in\mathbb Z^{+}\\ \implies &\frac{2^{3k-1}-1}{3}\in\mathbb Z^{+} \\ \implies &3k-1=2m,~m\in\mathbb Z^{+}\\ \implies &\begin{cases}n=3k-2\\ 3k-1=2m\end {cases}\\ \implies &k=2k'-1,~k'\in\mathbb Z^{+}\\ \implies &n=3(2k'-1)-2,~k'\in\mathbb Z^{+}\\ \implies &n=6k'-5,~k'\in\mathbb Z^{+}\\ \implies &n=6k-5,~k'\longmapsto k .\end{align}$$
Thus, our exact answer is:
$$\color {gold}{\boxed {\color{black} {n=6k-4,~n=6k-5,~k\in\mathbb Z^{+}.}}} $$
Explanations:
$$\frac{3k\times 2^{3k}+1}{3}=k\times 2^{3k}+\frac 13\not\in\mathbb Z^{+}.$$
$$\frac{2^{2m}-1}{3}=\frac{4^m-1^m}{3}=\frac{(4-1)(4^{m-1}+4^{m-2}+\cdots+1)}{3}\in\mathbb Z^{+}, ~\forall m\in\mathbb Z^{+}.$$