Problem. Let $d=(a,b)$, $b=\beta d$ and $n>1$. If $\beta$ is odd number, prove that $(n^a+1,n^b-1)\le 2$.
Solution (from the book). Each common divisor of numbers $n^a+1$ and $n^b-1$ has to be divisor of their sum $n^a+n^b=n^b(n^{a-b}+1)$, that is, it has to be a common divisor of $n^b-1$ and $n^{a-b}+1$ (we assume a>b). Keeping up, we conclude that the common divisor of numbers $n^a+1$ and $n^b-1$ has to be a divisor of number $n^d+1$. Let $x=n^d$, such that $n^b-1=x^\beta-1$ and $\beta$ is odd. To divide $n^b-1$ with $n^d-1$ means to divide $x%\beta-1$ with $x+1$. Remainder of this division is $-2$, and this means that numbers $n^a+1$ and $n^b-1$ cannot have common divisor greater than 2. $\square$
I'm horrible at number theory, because it always seems to abstract for me, and solutions always seem kind of forced, as if first the solution was written, and then the problem.
So I've decided to start from the beginning of my Introduction to number theory book, and this problem was supposed to be an easy example of previously learned.
There is a bunch of things I don't understand in this proof, beginning with "it has to be a common divisor of $n^b-1$ and $n^{a-b}+1$". After that, I understand literally nothing. I wish someone could write a more detailed explanation, but less formal, so that it can include many words and explanations, and examples, too.
Also, are there any hints/directions you could give me that I could follow when encountering number theory problems. What am I supposed to see, what am I supposed to look for, how much "deep" do I have to go to prove something (sometimes, some things sound pretty obvious to me, yet they are explained pretty long, and some things, as stuff I mention I don't understand, are simply gone though without a brief of explanation.
I hope I'm not asking too much.