extremal problem from AOPS thread (BOMC-2)

[Niloy Da Fermat]

Find all positive integers $ x,y $ such that $ 2^x-1=xy $

[sourav das]

Hint

Use---
i)Smallest prime idea;
ii)Order

Solution

Note (x,y)=(1,1) is a solution.

Note (x,y) both odd.
Let $x>1$ and let $p$ be the smallest prime such that $p|x$.
Let $ord_p(2)=s$, then $s|x$. but $s<p$ which implies a smaller prime other than $p$ divides $x$. Contradiction .(As $s$ cannot be 1)

[Tahmid Hasan]

$x \mid 2^x-1$....(1)
now $x=1$ is trivial here.(in this case $y=1$)
$2^x-1$ is odd,so $x$ must be odd too.
now
CASE 1: $x$ is a prime(other that $2$)
by FLT we get $x \mid 2^x-2$....(2)
by (1),(2) $x \mid 1$
a contradiction!
CASE 2: $x$ is composite.we write $x=ap$ for a prime $p$ and some numerical $a$,so $p \mid x$
by factor theorem we get $2^p-1 \mid 2^x-1$
hence $p \mid 2^p-1$
which is the same as CASE 1.
so $(x,y)=(1,1)$ is the only solution.

[*Mahi*]

CASE 2: $x$ is composite.we write $x=ap$ for a prime $p$ and some numerical $a$,so $p \mid x$
by factor theorem we get $2^p-1 \mid 2^x-1$
hence $p \mid 2^p-1$

Why? What if $p \mid \frac{2^x-1}{2^p-1}$?

[Sazid Akhter Turzo]

I agree with Mahi. Before he posted the reply, I'd been confused to see your solution.
Turzo