Find all $n$ such that $x+y$ is a power of $2$ whenever $xy=n$. 
Let $n$ be a positive integer. $n$ is called a golden integer if $n$
  is composite, and if we write $n$ as $n=xy$, where $x$ and $y$ are
  positive integers, then $x+y$ is a power of two ($x+y=2^{r}$). Find
  all golden integers.

It is obvious by definition and the fact that $n=1\times n$ that there is an integer $a$, such that $n=2^{a}-1$. $n$ is composite, hence there are two integers, say $x$ and $y$, such that $1<x\leq y<n$ and $n=xy$. Thus we have the following system,
$$x+y=2^{b}$$
$$xy=2^{a}-1$$
Consequently,
$$(x+1)(y+1)=2^{b}(2^{a-b}+1)$$
$$(x-1)(y-1)=2^{b}(2^{a-b}-1)$$
Does anyone know how to continue this solution?
 A: If $n$ is a golden number and $n=xy$ is a decomposition with $x,y>1$, then indeed
$$x+y=2^b\qquad\text{ and }\qquad xy=2^a-1,$$
for positive integers $a$ and $b$. Note that $b<a$ because
$$2^b=x+y<xy=2^a-1.$$
Suppose $a$ is prime, say $a=p$. Then $x$ and $y$ divide $2^p-1$ and hence${}^1$ $x\equiv y\equiv1\pmod{p}$. Then because $b<a=p$ and
$$2^b=x+y\equiv2\pmod{p},$$
we see that $b-1$ strictly divides $p-1$, so $2(b-1)\leq p-1$. But also
$$2^p-1=xy=x(2^b-x)\leq 2^{2b-2},$$
which shows that $p\leq 2b-2$. This means $2b-2\leq p-1<p\leq 2b-2$, a contradiction.
Hence $a$ is composite, say $a=uv$ with $u,v>1$. Then
$$n=2^a-1=(2^u-1)\sum_{k=0}^{v-1}2^{ku},$$
and because $n$ is golden we have
$$(2^u-1)+\sum_{k=0}^{v-1}2^{ku}=2^c,$$
for some integer $c$. This implies $v=2$, and by symmetry also $u=2$ and so $n=15$.



*

*Note that $2^m-1$ and $2^n-1$ are coprime if and only if $m$ and $n$ are coprime. So if $q$ is a prime dividing $2^p-1$ then for all $m<p$ the prime $q$ does not divide $2^m-1$. So modulo $q$ the number $2$ has order $p$, which implies that $p\mid q-1$ and so $q\equiv1\pmod{p}$. It follows that all divisors of $2^p-1$ are congruent to $1$ modulo $p$.

