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

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 ??

• $2^{17}-1>10^5$, so you don't have very many more to check – J. W. Tanner Nov 7 at 23:44
• $2^p-1$ with $p\in\{2,3,5,7,13\}$ is (Mersenne) prime – J. W. Tanner Nov 7 at 23:53
• $2^{2k}-1=(2^k-1)(2^k+1)$ – J. W. Tanner Nov 7 at 23:54
• @J.W.Tanner what about $n=$ odd – Ellen Ellen Nov 8 at 9:20
• I'm sure there are tables of factorizations of Mersenne numbers somewhere on the web. There are also programs to do factorizations. And there's the Online Encyclopedia of Integer Sequences. – Gerry Myerson Nov 8 at 11:24

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.
• Thank you very much @J.W.Tanner , but if $n=$ odd then how I prove it ?? – Ellen Ellen Nov 8 at 18:25
• How I prove $n=$ only odd – Ellen Ellen Nov 8 at 21:49
• He never said that. he said that for odd $n$ you can check a table. In general if $n=2^jk$ then it applies his factorization of $2^n-1$ ; $j$ times and has at least $j$ factors. Similar if you found a factorization of $2^n-1$ when it's divisible by 3, etc. – Roddy MacPhee Nov 9 at 13:21