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 $2^{1277}-1$ ?
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.