$n$ is a divider of $c$ if and only if $n = 2(c \mod (n-1)) - (c \mod(n-2)) + 2$ While working on Integer factorization problem I came to this conclusion:
If and only if $n$ is a divider of $c$
$$c\mod n = 0$$
Than
$$n = 2(c \mod (n-1)) - (c \mod(n-2)) + 2$$


*

*c,n are positive integers


And $c$ is a composite of two prime numbers with the same binary length, aka bit length, without leading zeroes
I found this while I been examining the mod function of $c$
$$f(n) = c \mod n$$
While I have no mathematical proof for this behavior, it seems to be working.  
In this question I am not looking for such proof by simply help with this equation.
How can I extract $n$? Or how can I simplify it?
Working Example
$$437 = 19*23$$
$$23 = 2(437 \mod 22) -(437 \mod 21) + 2$$
Not working example
$$319 = 11 * 29$$
$$29 \not= 2(319 \mod 28) -(319 \mod 27) + 2$$


*

*11 and 29 does not have the same bit length.

 A: I found a mathematical proof to why this equation is true. How ever is not answering the question...
Here is the proof:
Lets assume that $n$ is the bigger divider such that:
$$\frac{c}{n}=a$$
$$a<n$$


*

*a is a positive integer


We also know that $a$ and $n$ have the same bit count, so that means:
$$a<n<2a$$
In order to continue I will use real life example.
Lets assume that we have $c$ liters of wine and want to split it between $n$ bottles of wine.
We made it, we felt $c$ bottles with $a$ litters of wine!
Now we been asked to give one bottle away, so we used it to fill the remain $n - 1$ bottles, since $n-1$ is bigger than $a$, we left with $a$
$$c\mod n-1 = a$$
And again we been asked to give one bottle away, so we used it to fill the remain
$n-2$ bottles, since $2a$ is bigger than $n$, we successfully increased the total litter amount in the bottles to $a+1$ litters, and we left with $2a-(n-2)$ litters of wine.
$$c\mod n-2=2a-(n-2)$$
So lets get back to our equation. This is what we are getting:
$$n = 2a - (2a-(n-2)) + 2$$
$$0=0$$
So this equation is true when:
$$a<n-2<2a$$
