Separable Integers Reloaded

I am an amateur number theory enthusiast, therefore, my problem(and/or)solution may have errors in them. Please bear with me and notify me if any errors are present. Thanks in advance.

Also look at the original problem . This problem is an attempt at generalizing the original one. On the request of Calvin Lin.

Solution:

We have to find the sum of all positive integers $A$ , where

$A=n\times(n+9)$ and $A=m\times(m+18)$ where $n$ , $m$ are both(unequal) positive integers.

$\Rightarrow A=\quad { n }^{ 2 }+9n\quad ={ \quad m }^{ 2 }+18m$ .............. $(1)$

It is easy to observe that $n>m$ hence we can write:

$n= m + d$ where $d$ is also a natural number.

from $(1)$

$\Rightarrow { (m+d) }^{ 2 }+9(m+d)={ m }^{ 2 }+18m$

on simplification, the above equation yields:

${ d }^{ 2 }+(2m+9)d-9m=0$ ................................................ $(2)$

extracting the roots of the above quadratic in $d$ and neglecting the negative root, we get:

$d\quad =\quad \frac { -(2m+9)+\sqrt { { (2m+9) }^{ 2 }+36m } }{ 2 }$ ........ $(3)$

Since $d$ is a positive integer, $R.H.S.$ of equation $(3)$ should also be an integer. This is possible if and only if the expression under the radical sign is a perfect square. Note that the parity of $m$ doesn't affect this.

$\Rightarrow { (2m+9) }^{ 2 }+36m=\quad { 4m }^{ 2 }+72m+81=\quad { (2m+9+a) }^{ 2 }$ where $a$ is a positive integer. Reason for this is left as an exercise.

solving the above equation we get:

$m\quad =\quad \frac { a\quad \times \quad (18+a) }{ 4\quad \times \quad (9-a) }$ ............................... $(4)$

From equation $(4)$ it is easy to see that $a$ has to be an even number and that $a<9$

We can now easily find value(s) of $m$ by plugging in values for $a$ which yield integer result for $m$ . In this particular case, $m$ can take values $12$ and $52$ .

Hence, the required separable number(s) $A$ are:

$A1\quad =\quad 12\times (12+18)\quad =\quad 360$

$A2\quad =\quad 52\times (52+18)\quad =\quad 3640$

THEREFORE, THE REQUIRED ANSWER IS: $\boxed { A1+A2\quad =\quad 360+3640\quad =\quad 4000 }$

This solution can be generalized by taking variables in place of $9$ and $18$ , although not satisfactorily. Please suggest improvements. Thanks again!