Seeking a proof of a general result on central divisors Define the central divisors $\mathbf{CD}$(N) of a natural number N as paired
factors of N that are closer together than any of its other factor pairs.
More formally,
$$\mathbf{CD}(N) = (a,N/a),\text{ where }a = \text{max } \{x \in \mathbb{N}\ :\ x|N\text{ and }x \le \sqrt{N}\}$$
I wish to prove:
$$\text{If }\mathbf{CD}(N) = (a,b),\text{ then }\mathbf{CD}(mN) = (m,N)\text{ for }(a - 1)(b - 1)
\lt m \le N\tag{1}$$
This is a more general case of the proposition proved in a
previous post that
$$\mathbf{CD}(mN) = (m,N)\text{ for }(\sqrt{ab} - 1)^2 \lt m \le N$$
since
$$( (\sqrt{ab} - 1)^2, ab ] \subseteq ( (a - 1)(b - 1), ab ]$$
from
$$(\sqrt{ab} - 1)^2 \ge (a - 1)(b - 1)$$
In (1), if a is 1, then N is a prime p and it follows that (m,p) are central
divisors as long as m $\le$ p since it's impossible to create the product
mp from two factors both smaller than p.  Without loss of generality, we may
therefore take 2 $\le$ a $\le$ b.
The lower limit on m is minimal and is a requirement since if
m = (a - 1)(b - 1), it's easily seen that (m,N) are not central
divisors:
$$m = (a - 1)(b - 1) \lt (a - 1)b \le a(b - 1) \lt ab = N$$
$$\mathbf{CD}(mN) = ((a - 1)b,a(b - 1))$$
The condition $\mathbf{CD}$(N) = (a,b) is also a requirement, as can be
seen by the counterexample a = 2, b = 6, m = 6, in which
$$m > (a - 1)(b - 1) = 5,$$
but (m,N) = (6,12) are not central divisors.
There are some obvious results if a or b is a prime, but I would like to
see a proof in the general case.  I have not been able to make progress without
resorting to manipulating the prime factorization of mab, so I am hoping there's
a simpler or more direct approach.

Denote by \eqref{1} the inequality
$$(a-1)(b-1) \lt m \lt (a+1)(b+1)\tag{1}\label{1}$$
First, if the product ab is a prime p, then its central divisors are {1,p} and
we have
$$1 \le m \le 2p + 1$$
It's straightforward to show that {m,p} are central divisors by considering the
possible factors of mp.
Otherwise, note that if \eqref{1} holds for central divisors a and b, then it also
holds for any other factor pair of ab.  This is also straightforward to show.
Next, let w,x,y,z be natural numbers with
$$(w-1)(x-1) \lt yz \lt (w+1)(x+1)\tag{2}\label{2}$$
Then
$$(y-x)(z-w) \le 0\tag{3}\label{3}$$
This can be shown by a proof by contradiction.
For {m,ab} to be central divisors, we must have
$$(m-ab)^2 \le (k-mab/k)^2\tag{4}\label{4}$$
for all divisors k of the product mab.  Each such divisor can be written as
$k = ij$, where i is a divisor of m and j is a divisor of ab.  It follows
that \eqref{1} holds for i and j:
$$(j-1)(ab/j-1) \lt i(m/i) \lt (j+1)(ab/j+1)$$
Applying \eqref{2} with w = j, x = ab/j, y = i and z = m/i, \eqref{3} can be written
$$(i-ab/j)(m/i-j) \le 0,$$
which reduces to \eqref{4}, QED.
