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Mersenne numbers are numbers of the form $2^p-1$ where $p$ is a prime number. Some of them are prime for exemple $2^5-1$ or $2^7-1$ and some of them are composite like $2^{11}-1$ or $2^{23}-1$.

I'm interested about the composite for example $2^{11}-1 = 2047 = 23 \cdot 89 = (2 \cdot 11+1)(8 \cdot 11+1)$.

Prime factors of Mersenne numbers are of the form $2kp+1$.

I have noticed intersting about how to find the $k$ by solving an equation :

$(2^p-2)/(2 \cdot p) = 2 \cdot p \cdot k_1 \cdot k_2 + k_1 + k_2$ seems to give the $k$ for a factor of composite Mersenne numbers but I don't know how to prove it.

My question is : if we know all factors of $(2^p-2)/(2 \cdot p)$, can we found easily $k_1$ and $k_2$ ?

I'm asking this question because factor of $2^{1277}-1$ are unknown but we know all factors of $(2^{1277}-2)/(2 \cdot 1277).$

So can we use this equation to know factors of $2^{1277}-1$ ?

Aurel-BG
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1 Answers1

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First of all the explanation how the given equation gives the desired factors assuming $k_1$ and $k_2$ are known :

We have $$M:=2^p-1=(2k_1p+1)(2k_2p+1)=4k_1k_2p^2+2(k_1+k_2)p+1$$

hence $$2^p-2=2p(2k_1k_2p+k_1+k_2)$$ hence $$\frac{2^p-2}{2p}=2k_1k_2p+k_1+k_2$$

So knowing $k_1$ and $k_2$ would indeed give us the factors. But this diophantine equation cannot be solved without the factorization of $M$ , even if we know the factors of $\frac{2^p-2}{2p}$ which is equivalent to have the factors of $M-1$.

In the comment it is mentioned that having the factors of $N$ does not allow to factor $N-1$ or $N+1$. If this were possible , we could easily refine this technique to factor every number efficiently which is to our knowledge impossible.

Peter
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  • In some cases, it is possible to factor $N$ knowing the factors of $N-1$. For example $N=713=91$. $N=(6k_{1} +1)(6k_{2}+1)=36k_{1}k_{2} + 6(k_{1} + k_{2}) + 1$. So $N=(61+1)(62+1)=36*2 +6(2+1) +1$. So if we can find a number $M=36k_{1}k_{2}=72$, we will be able to use it to factor $N$. It turns out we do have such a number. In the multiplication table, $91$ is surrounded by $4$ numbers ($72$, $96$, $112$ and $84$) at a corner of a square. The number $L=90$ is also surrounded by ($72$, $90$, $110$, and $88$). (cont) – user25406 Nov 16 '23 at 12:47
  • (cont)- All we need is to express $72=89$ from $L=90=910$ to $72=612$ of $N=91$. So knowing $72=89=612$, we can find $91=(6+1)(12+1)$ of the number we wanted to factor. We cannot always find a number $M$ with the same $36k_{1}k_{2}$ but we can find numbers closer to $N$. For example $133$ and $130$ share the same $108=912=618$. I am pretty sure that we can find many examples of numbers $M$ differing from $N$ by $1$, $2$, $3$...that can be used to factor $N$. We should not stick to $N-1$ and $N+1$ and should consider differences larger than $1$ to get better chances at factoring $N$. – user25406 Nov 16 '23 at 13:00
  • It is clear that if starting with the factors of $N-1$, we can factor $N$ but the opposite is also true in the case of $91$ and $90$, that is knowing the factors of $91$, we can get the factors of $90$ in this particular case. The same thing is expected to apply to $N$ and $N+3$. No one is saying that this is a general rule. – user25406 Nov 16 '23 at 20:57