If $x,y$ are integers greater than $1$ and $n$ is a positive integer such that $2^n + 1=xy$ , $\exists 1< a<n$ such that $2^a|x-1$ or $2^a|y-1$?

Question

If $x,y$ are integers greater than $1$ and $n$ is a positive integer such that $2^n + 1=xy$ , then is it true that either $2^n|x-1$ or $2^n|y-1$ ? I have only been able to observe that both $x,y$ are odd . Please help

EDIT : As is seen from an answer : the original claim does not hold ; so I ask does there exist $a>1 , a<n$ such that $2^a|x-1$ or $2^a|y-1$ ?

Answer 1

Let $x=2^{a+p}c+1,y=2^ad+1$ where $p\ge0$ and $c,d$ are add

$2^n+1=xy=2^{2a+p}cd+2^{a+p}c+2^ad+1$

$\iff2^n=2^a(2^{a+p}cd+2^pc+d)$

$\iff2^{n-a}-[2^{a+p}cd+2^pc]=d$ which is odd

If $p>0,2^{n-a}-[2^{a+p}cd+2^pc]$ is even

$\implies p=0\implies$ the highest power of $2$ that divides $x-1$ and $y-1$ will be the same

Answer 2

$2^n=xy-1>x-1$ or $>y-1$. So it is impossible.

Answer of Edit: $9=3.3$ and $33=11.3$. Here $3-1$ and $11-1$ are not divisible by $4$. But for $65=13.5$, $4$ and $12$ are both divisible by 4.

Answer 3

Neither of your original claim nor edited claim holds.

For the original claim, consider the counter example: $x = 19, y = 27, n=9$.

So, $2^9 + 1 = 513 = 19 \times 27$, but $18$ or $26$ is not divisible by $512$.

And if you ask the other way round, i.e. "Is it true that either $(x−1)|2^n$ or $(y-1)|2^n$ ?", the claim still does not hold, as $512$ is not divisible by $18$ or $26$.

For the edited claim, using the same example, $18$ or $26$ is not divisible by $2^a$ either for $a > 1$.