What is GCD of $7^{3001}-1$ and $7^{3000}+1$? 
Find GCD of $7^{3001}-1$  and $7^{3000}+1$.

My work. I noted that $(1)(7^{3001}-1) -(7)(7^{3000}+1)=-8$.
 A: Your remark that $7^{3001}-1=7\cdot (7^{3000}+1)-8$ is quite useful here. In fact, it implies that
$$\gcd(7^{3001}-1,7^{3000}+1)=\gcd(8,7^{3000}+1).$$
Now note that $7^{3000}+1$ is even, but 
$$7^{3000}+1=(8-1)^{3000}+1\equiv (-1)^{3000}+1=2 \pmod{4}.$$
Can you take it from here?
A: $7^n=(6+1)^n=6m+1$
$\\$ hence $7^n+1 $. Is when divided by 6 will give remainder as 2 $\\$ $7^n-1$ is multiple of 6. $\\$
Hence HCF will be 2.
A: Let $\, n = 3000,\ d = $ gcd. $\bmod d\!:\,\ \color{#0a0}{7^n\equiv -1},\,\ 1\equiv 7(\color{#0a0}{7^n})\equiv -7\,\Rightarrow\, 8\equiv 0\,\Rightarrow\, \color{#c00}{7\equiv -1}\ $ thus$\, -1\equiv \color{#c00}7^n\equiv (\color{#c00}{-1})^n\equiv 1\,\Rightarrow\, 2\equiv 0,\,$ so $\,d\mid 2.\,$ But $\,2\mid 7^n\!+\!1,7^{n+1}\!-1\,\Rightarrow\,2\mid d,\,$ so $\,d=2$. 

Another example: compute $\ \gcd(5^{300} - 1, 5^{200} + 6)\ $ from this question.
$\!\bmod \overbrace{(\color{#c00}{a^{\large 3}\!-\!1},\color{#0a0}{a^{\large 2}\!+\!6})}^{\textstyle\small a = 5^{\large 100}\  \text{in OP}}\!:\,\ \color{#c00}{1 \equiv (a^{\large 3})^{\large 2}}\equiv \color{#0a0}{(a^{\large 2})^{\large 3}\equiv (-6)^{\large 3}}\,\Rightarrow\,0\equiv \color{#c00}1\!-\!\color{#0a0}{(-6)^{\large 3}}\equiv\!217\equiv  7\cdot 31$ 
also $\, \color{#c00}1\equiv a(a^{\large 2})\equiv \color{#0a0}{-6a}\,$ so $\,a\equiv -1/6\ $ so  $\,\begin{align} a\equiv&\ \  \ 6/6\equiv 1\!\!\!\pmod{\!7}\\ a\equiv& \ 30/6\equiv 5\!\!\!\pmod{\!31}\end{align},\,$  indeed roots, so
$$\bbox[9px,border:1px solid #c00]{(a^{\large 3}\!-\!1,a^{\large 2}\!+\!6)\, =\, (a\!-\!1,7)\,(a\!-\!5,31)}$$
In OP $\!\bmod 7\!:\ a \equiv (-2)^{100}\!\equiv -2(-2^3)^{33}\equiv 2\,$ so $\ (a\!-\!1,7)=1$
and $\bmod 31\!:\ \ a\  =\  5^{100}\ \ \equiv\ \ \  5\ (5^3)^{33}\ \equiv\  5\ \ $  so $\ (a\!-\!5,31) = 31,\,$ so  OP $\,\bbox[3px,border:1px solid #c00]{\rm gcd = 31}$
