Finding $x+y$ given that $(2^{28}-1)$ is divisible by $x,y$ and between $120 ,130$ The number $(2^{28}-1)$ is divisible by $x,y$.If each of  $x,y$ is between $120 ,130$ .How to find $x+y$

Using the decomposition $a^2-b^2=(a+b)(a-b)$ you can factorize $(2^{28}-1)$ as
$$
  2^{28}-1
=
  (2^{14}+1)(2^{14}-1)
=
  (2^{14}+1)(2^7+1)(2^7-1)
$$
Now, $2^7+1=129$ and $2^7-1=127$...
