Find all natural from the format: $2^{n}-1,n\in\Bbb N^{*}$ such that: $\begin{cases}2^{n}-1<10^{5}\\2^{n}-1=pq \:(\text{distinct primes})\end{cases}$

Question

Problem :

Find all natural from the format : $2^{n}-1,n\in \mathbb N^{*}$ they are less then $10^{5}$ and written as product two different $p,q$ prime numbers

My try :

$n=1$ so $1<10^{5}$ $×$

$n=2$ so $3<10^{5}$ but $3≠p.q$ , $p,q=$ prime number $×$

$n=3$ so $7<10^{5}$ but $7≠p.q$ $×$

$n=4$ so $15<10^{5}$ and $15=3.5$ $√$

But I need method to find all this number ??

Answer 1

For $n=2k$, $2^n-1=(2^k+1)(2^k-1)$.

Factorizations of $2^n-1$ for $n=2k+1$ can be found in the Cunningham tables.

For $n=2, 3, 5, 7$, or $13$, $2^n-1$ is prime.

With this information, you should be able to answer your question.