Suppose that $2^n + 1 =xy$, where $x$ and $y$ are integers > $1$ and $n>0$.
Show that $2^a|(x-1)$ iff $2^a|(y-1)$.
Here $a|b$ implies $a$ divides $b$.
Show that $2^a|(x-1)$ iff $2^a|(y-1)$.
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number-theory
2 Answers
1
$$(x-1)(y-1) = 1 + xy - x - y = 2^n - (x - 1) - (y - 1)$$
Then if $2^a$ divides $(x-1)$ then it definetile divides $(y-1)$, because
$$(y - 1) = 2^n - (x - 1) - (x-1)(y-1).$$
0
WLOG let $x-1=p2^a,y-1=q2^b$ where $p,q$ are odd and $a\ge b>0$
$$1+2^n=xy=(p2^a+1)(q2^b+1)=pq2^{a+b}+p2^a+q2^b+1$$
$$q=2^{n-b}-pq2^a-p2^{a-b}$$
As $n-b>0$
$q$ will be even if $a>b$
and $q$ will remain odd if $a=b$