I was wondering if the following were true. It makes sense but I'm having trouble concocting any formal reasoning.
Let $2^n+1=xy$ for some integers $x,y>1$ and $n>0$. For $a\in\mathbb{Z}^+$, does $2^a\mid (x-1)$ $\iff$ $2^a\mid (y-1)$?
Without loss of generality, one only needs to prove one direction. However, I'm not sure how to approach the problem. Here's my attempt: Suppose $2^a\mid (x-1)$. Then $x\equiv 1$ mod $2^a$, so $y\equiv xy=2^n+1\equiv 2^r+1 \hspace{5mm}(\text{mod }2^a),\hspace{5mm}\text{ where $0\leq r I'm having trouble continuing from here, since I don't know any extra information to determine $2^r\equiv 0$ mod $2^a$. It doesn't seem like that could be true for arbitrary $a\in\mathbb{Z}$, so I must have veered horribly off-track. I appreciate any help!