For both parts we will consider the cases $ n=3k, 3k+1, 3k+2$. Working in $ mod 7$ we have $ 2^{3} \equiv 1 (mod 7)$ hence the following: $ 2^{3k} \equiv 1 (mod7)$ $ 2^{3k+1} \equiv 2 (mod7)$ $ 2^{3k+2} \equiv 4(mod 7)$ Now the rest is easy. a) $ 2^{3k}-1 \equiv 0 (mod7)$ $ 2^{3k+1}-1 \equiv 1 (mod7)$ $ 2^{3k+2}-1 \equiv 3(mod 7)$ So only for $ n=3k$ where $ k\in\mathbb{Z}^{+}$ does $ 7|2^{n}-1$. b) $ 2^{3k}+1 \equiv 2 (mod7)$ $ 2^{3k+1}+1 \equiv 3 (mod7)$ $ 2^{3k+2}+1 \equiv 5(mod 7)$ So there is no value of $ n\in\mathbb{Z}^{+}$ such that $ 7|2^{n}+1$.

Here is my solution for part (a) (wow this thread is 12 years old!) Let $k$ be any number.\begin{align} \nonumber7\equiv 0\pmod 7\\ \nonumber (7+1)\equiv 1\pmod 7\\ \nonumber(7+1)^k\equiv 1\pmod 7 \\ \nonumber8^k\equiv1\pmod7\\ \nonumber2^{3k}\equiv 1\pmod7\\ \nonumber2^{3k}-1\equiv 0\pmod7\end{align}We also know that $2^n-1\equiv 0\pmod 7$. Thus, $n=3k$, which means that $n$ must be a multiple of $3$.

There are $3$ possible cases: $\implies n\equiv0\pmod3$ $\implies n\equiv1\pmod3$ $\implies n\equiv2\pmod3$ If $n\equiv0\pmod3$, we can express $n$ as $3y$, where $y\in \mathbb{Z}^+$. If $n\equiv1\pmod3$, we can express $n$ as $3y + 1$, where $y\in \mathbb{Z}^+$. If $n\equiv2\pmod3$, we can express $n$ as $3y + 2$, where $y\in \mathbb{Z}^+$. (Note that $y$ in these 3 lines must not be the same constant.) Note that: $\implies 2^{3y}\equiv1\pmod7$ $\implies 2^{3y + 1}\equiv2\pmod7$ $\implies 2^{3y + 2}\equiv4\pmod7$ (a): $7|(2^n - 1)$ iff (if and only if) $2^n \equiv 1\pmod7$. We check our cases and see that this is only true when $n$ can be expressed as $3y$. This implies that $n = 3y$ where $y \in \mathbb{Z}^+$. (b): $7|(2^n + 1)$ iff (if and only if) $2^n \equiv 1\pmod7\equiv6\pmod7$. We check our cases to see that no such $n$ satisfies this. Credits to my blog

By Euler's Theorem, $2^6\equiv1\pmod7$, so we only need to consider $n\in[0,6]$, since it cycles. If $n\pmod7\in\{0,3,6\}$, then $2^n-1\equiv0\pmod7$ and $2^n+1\equiv0\pmod7$. If $n\pmod7\in\{1,4\}$, then $2^n-1\equiv1\pmod7$ and $2^n+1\equiv3\pmod7$. If $n\pmod7\in\{2,5\}$, then $2^n-1\equiv3\pmod7$ and $2^n+1\equiv5\pmod7$. So $\textbf{(A)}~\boxed{n\in3\mathbb N}$ and $\textbf{(B)}~\square$.

Wikipedian1337 wrote: (a) Find all positive integers $ n$ for which $ 2^n-1$ is divisible by $ 7$. (b) Prove that there is no positive integer $ n$ for which $ 2^n+1$ is divisible by $ 7$. For part (a) Note that: $$2^n-1 \equiv 0 \pmod 7 \implies 2^n \equiv 1 \pmod 7 \implies n=3k \; \forall k \in \mathbb N$$That becuase $2^3 \equiv 1 \pmod 7$ and $1^k=1$ for all $k \in \mathbb N$. For part (b) use case work: We know that $n$ can be $3i$ , $3i+1$ and $3i+2$ so: Case 1.- $n=3i$ $$2^{3i} \equiv -1 \pmod 7 \implies 1 \equiv -1 \pmod 7 \implies \; \text{contradiction!!}$$Case 2.- $n=3i+1$ $$2^{3i+1} \equiv -1 \pmod 7 \implies 2 \equiv -1 \pmod 7 \implies \; \text{contradiction!!}$$Case 3.- $n=3i+2$ $$2^{3i+2} \equiv -1 \pmod 7 \implies 4 \equiv -1 \pmod 7 \implies \; \text{contradiction!!}$$So we proved that no such $n$ for part (b). Thus we are done

(a) We have $2^3\equiv 1\text{(mod 7)}$, but $2^1\equiv 2\text{(mod 7)}$, so $\text{ord}_72=3$, and hence $3|n$. (b) Notice that $2^n$ is a QR mod $7$, but $-1$ isn't, contradiction.

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Let's bash it a bit

a) Fermat's Little Theorem gives $2^6\equiv 1 \pmod 7.$ But since $7|(2^3-1)(2^3+1)$ we have $2^3\equiv 1\pmod 7.$ So all $n$ divisible by $3$ satisfy. b) Let $n=3r+s,$ where $s\in \{1,2\}.$ Then $$2^n\equiv (2^3)^r\cdot 2^s \equiv 2\text{ or } 4 \not\equiv 1 \pmod 7.$$$\text{Q.E.D.}$

Observe that since $2^3=8\equiv1\pmod{7}$, $2^n\pmod{7}$ is cyclic with period $3$: $1,2,4,1,2,4,\dots$. Thus, for (a), the residues of $2^n-1$ modulo $7$ are $0$, $1$, and $3$ when $n\equiv0,1,2\pmod{3}$, in that order, so $\boxed{\text{all }n\text{ s.t. }3|n}$ is the requested answer. Similarly, for (b), the residues of $2^n+1$ modulo $7$ are $2$, $3$, and $5$, and in particular, $0$ is not one such residue. Hence proven. $\blacksquare$

We split into cases. Case 1: $n \equiv 0 \pmod 3$ \[2^{3k}-1 \equiv 8^k-1 \equiv 1-1 \equiv 0 \pmod 3\]\[2^{3k}+1 \equiv 8^k+1\equiv 1+1 \equiv 2 \pmod 3\] Case 2: $n \equiv 1 \pmod 3$ \[2^{3k+1}-1 \equiv 2\cdot8^k-1 \equiv 2-1 \equiv 1 \pmod 3\]\[2^{3k+1}+1 \equiv 2\cdot8^k+1\equiv 2+1 \equiv 3 \pmod 3\] Case 3: $n \equiv 2 \pmod 3$ \[2^{3k+2}-1 \equiv 4\cdot8^k-1 \equiv 4-1 \equiv 3 \pmod 3\]\[2^{3k+2}+1 \equiv 4\cdot8^k+1\equiv 4+1 \equiv 5 \pmod 3\] Thus, the only case that satisfies $2^n-1$ is divisible by $7$ is when $n \equiv 0\mod3$ or $n$ is divisible by $3$ and no cases satisfy $7$ dividing $2^n+1$.

(a)$7 | 2^n-1 \implies 2^n \equiv 1 \pmod 7 \implies n=3k \forall k \in \mathbb N$ since $\text{ord}_7(2) = 3.$ (b) $7 | 2^n+1 \implies 2^n \equiv 6 \pmod 7.$ However, the residues of $2^n \pmod 7 \forall n \in \mathbb N$ is $1, 2,$ or $4.$ Hence, there is no positive integer $ n$ for which $ 2^n+1$ is divisible by $ 7$.

(a) We claim that the answer is when $n$ is a multiple of 3. We will use proof by induction. The base case is when $n=3$, which works, now $n=3k$ is 1 mod 7, $n=3k+3$, is 2^3 mod 7 which is 1 mod 7, hence, proved. (b) $2^n$ needs to be 6 mod 7, but it can only be 1 mod 7, 2 mod 7, 4 mod 7, hence, proved.

Notice that $2^3\equiv 1\pmod 7$. Now, we have $$2^{n}\equiv 1\pmod 7$$if $n\equiv 0\pmod 3$, $$2^n\equiv 2\pmod 7$$if $n\equiv 1\pmod 3$, and $$2^n\equiv 4\pmod 7$$otherwise. (a) We need $2^n\equiv 1\pmod 7$, so $\boxed{n\equiv 0\pmod 3}$. (b) There are no $6\pmod 7$ residues, done.

Quick solution : a)$7\mid 2^n-1\iff 2^n\equiv 1[7]$ Thus : $o_7(2)\mid n \iff 3|n$ Hence : $n\equiv 0[3]$ b) $2^n\equiv -1[7]\equiv 3\nmid n$ So : $n\equiv 1[3]$ Or $n\equiv 2[3]$ Hence : $\exists k\in \mathbb{N}:n=3k+1$ or $\exists k\in \mathbb{N}:n=3k+2$ Thus : $2^{3k+1}\equiv 6[7]$ Or $2^{3k+2}\equiv 6[7]$ $\iff (2^3)^k×2\equiv 6[7]$ Or $(2^3)^k×4\equiv 6[7]$ Because : $o_7(2)=3$ , we get that : $2\equiv 6[7]$ Or $4\equiv 6[7]$ Absurde, so there exists no such positive integers

We can investigate $2^k \pmod 7$. $$ \begin{tabular}{c|c} $k$ & $2^k \pmod 7$ \\[5pt] \hline 1 & 2 \\[5pt] \hline 2 & 4 \\[5pt] \hline 3 & 1 \\[5pt] \hline 4 & 2 \\[5pt] \hline 5 & 4 \\[5pt] \hline 6 & 1 \\[5pt] \hline 7 & 2 \\[5pt] \hline \end{tabular} $$It seems to be a pattern of repeating $2, 4, 1$. (a) Subtracting $1$, we have a pattern of repeating $1, 3, 0$, so the answer is $\boxed{x \equiv 0\pmod 3}$. $\blacksquare$ (b) Adding $1$, we have a pattern of repeating $2, 4, 1$, so there are no values of $x$. $\blacksquare$

a) 2^n-1 is divisible by 7 iff 2^n is 1 mod 7 iff n is 0 mod 3 or 3k for any natural number k. b) For the sake of contradiction, let's suppose that 2^n+1 is 0 mod 7, so 2^n is 6 mod 7. But, since 2^n is only 2,4 or 1 mod seven, then here we have contradiction.