What is the smallest positive integer $n$ such that $(a+1)^{7^n} = (a+1)$? In $\mathbb{F} = \mathbb{F}_7[X]/(X^2+1)$ let $a$ be the class of $X$. What is the smallest positive integer $n$ such that $(a+1)^{7^n} = (a+1)$ ?
I know that the answer is $2$ but I don't see how it works. Can someone explain the logic to me? Because if I get it correctly, $a = X$ in this case and we're reducing $mod$ $X^2 +1$ so how does $ a^{49} +1$ equal to $a+1$ ?
 A: Since this is in characteristic $7$, $(x+1)^7 = x^7 + 1$ for any $x$, and
$(a+1)^{49} = (a^7+1)^7 = a^{49} + 1$.  Since $a^2 = -1$, $a^{48} = 1$ so 
$a^{49} + 1 = a + 1$.
A: [Note: this answer applies to the original more general question with exponent $7n$ vs. $7^{\large n}$]
$ \mathbb{F}_{\large 7}[x]/(x^2\!+\!1)\cong \Bbb F_{\large 7}[\,i\,]\,$ where $\,\overline{a\!+\!bi} = (a\!+\!bi)^{\large 7}\! = a\!-\!bi\,$ is the  conjugation automorphism.
If $\,w = 1\!+\!i\,$ then $\,(1\!+\!w)^{\large 7n}\! = 1\!+\!i \!\iff\! \bar w^{\large n} = w.\,$  Taking norm $\,N(x) = x\bar x\, $ of both sides yields $2^{\large n}\! = 2\,$ in $\,\Bbb F_{\large 7}\!\!\iff n\equiv 1\!\pmod{\!3},\,$ by $\,2\,$ has order $3$. $\,n = 7\,$ works: $\ \bar w^7\! = \bar {\bar w} = w,\, $ but no smaller $n$ works: $\,\bar w\neq w\,$ so $\,n\neq 1,\,$ and $\,n \neq  4\,$ else $\, \bar w^4\! = w = \bar w^7\Rightarrow\, 1 = \bar w^3 \Rightarrow\, \bar w = \bar w^4\! = w$.
A: The question can be rephrased as

What is the smallest positive integer $n$ such that $(a+1)^{7n-1} = 1$?

Let $m$ be the order of $a+1$ in the multiplicative group $\Bbb{F}^\times$, which has order $48$.
Then $m$ divides $\gcd(7n-1,48)$. Let's see the possibilities:
$$
\begin{array}{r|rrrrrrr}
n & 1 & 2 & 3 & 4 & 5 & 6 & 7 \\
7n-1 & 6 & 13 & 20 & 27 & 34 & 41 & 48 \\
\gcd(7n-1,48) & 6 & 1 & 4 & 3 & 2 & 1 & 48
\end{array}
$$
By checking that $(a+1)^k \ne 1$ for $k=1,2,3,4,6$, we arrive at $n=7$.
