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 $x+y$
 A: HINT: $$2^{28}-1=\left(2^{14}-1\right)\left(2^{14}+1\right)=\left(2^7-1\right)\left(2^7+1\right)\left(2^{14}+1\right)$$
A: 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 $2^7-1=127$...
A: Here is a method using general techniques that do not require use of special factorization formulas (so it may work more widely). Suppose $\rm\:f,g\in \Bbb Z[x]\:$ are polynomials with integer coefficients and suppose that we are searching for divisors of $\rm\:f(g(a))\:$ near $\rm\:g(a),\:$ say $\rm\:g(a)-n\:$ for small $\rm\:n.\:$ Testing such candidate divisors is especially simple by modular arithmetic since
$$\begin{eqnarray}\rm\:mod\ g(a)-n\!:\,\ g(a)\equiv n\:\Rightarrow\:f(g(a))\equiv f(n)\\  \\
\rm\Rightarrow\ \ g(a)-n\mid f(g(a))\iff g(a)-n\mid f(n)\end{eqnarray}$$
We seek divisors of $\rm\:(2^7)^4\!-\!1\:$ near $\,2^7\!=128,\:$ say $\rm\:2^7\!-n \in [120,130].\,$ Thus, as above
$$\rm\begin{eqnarray} mod\ 2^7\!-n\!:\,\ 2^7\equiv n\:\Rightarrow\: (2^7)^4\!-1\equiv n^4-1\\ 
\\
\rm \Rightarrow\ \ 2^7\!-n\mid 2^{28}\!-1\iff \color{#C00}{2^7\!-n\mid n^4\!-1}\end{eqnarray}$$ 
Applying this $\rm\color{#C00}{test}$ to the necessarily odd divisors in the interval $[120,130]$ yields
$$\begin{eqnarray} 
\rm \color{#C00}{128\!-n}&\: \mid\,&\rm \color{#C00}{n^4\!-1} \\
\hline
11^2=121&\: \nmid\,&7^4\!-1 = 49\cdot 51\rm\ \ by\ \ 11\nmid 49,51\\
3\cdot 41=123&\: \nmid\,&5^4\!-1 = 24\cdot 25\rm \ \ by\ \ 41\nmid 24,25\\
5^3=125&\:\nmid\,&3^4\!-1 = \ \ 8\cdot 9\ \ \rm \ \ by\ \ \ \ 5\nmid 8,9\\
    \color{#0A0}{127}&\mid&1^4\!-1 =\ 0\\
    \color{#0A0}{129}&\mid&\! (-1)^4\!-1 =\ 0\\
\end{eqnarray}$$
Thus the only divisors of $\,2^{28}\!-1\,$ in the interval $\,[120,130]\,$ are $\,\color{#0A0}{127,129}.$ Once one knows the above technique, testing the divisors as above is very fast, about a  half minute of mental arithmetic. As such, it is often more efficient to do this test than to search for special factorizations or other innate structure (different from than the compositional structure that this method employs).
