Any odd number can be written as: $$ n = \left(\frac{n + 1}{2}\right)^2 - \left(\frac{n - 1}{2}\right)^2 $$ If I take: $$ a^2=\left(\frac{n + 1}{2}\right)^2 $$ and: $$ b^2=\left(\frac{n - 1}{2}\right)^2 $$ If I subtract $1$ from $a$ or $b$, then $3$ to the previous, then $5$ and so on, I will always obtain $n$ for $a$ and $n - 2$ for $b$. What I am doing is subtracting a perfect square number from $a$ or $b$, because the sum of first odd number is always a perfect square number.
For example, take: $$ n = 35, a^2 = 324, b^2 = 289 $$
I get the following sequences:
$$ 324, 323, 320, 315, 308, 299, (288), 275, 260, 243, 224, 203, 180, 155, 128, 99, 68, 35 $$ and: $$ 289, (288), 285, 280, 273, 264, 253, 240, 225, 208, 189, 168, 145, 120, 93, 64, 33 $$
Observe that the number between parentheses is repeated in both sequences (if $n$ is prime this will not happen): $$ c=288 $$
If I do the following: $$ a^2-c=6^2 $$ and: $$ b^2-c=1^2 $$ and then: $$ 35=6^2-1^2=(6-1)\times (6+1) $$ finally: $$ 35=5\times 7 $$
Moreover, for any natural number $k$, I have found that if: $$ n=6k\pm 1, n>24 $$ then: $$ a^2 \equiv (a^2 - c)\mod{144} $$ and: $$ b^2 \equiv (b^2 - c)\mod{144} $$
Let's get to the point:
If $n=6k\pm 1$, how to obtain $(a^2 - c)$ and $(b^2 - c)$?
Is this possible?
The statement that if $n=6k \pm 1$ then $c$ exists is false, and that it would be divisible by $144$ is even more false.
Setup
First realize that the sum of the first few odd numbers gives perfect squares. $$ \begin{align} 1^2 &= 1 \\ 2^2 &= 1+3 \\ 3^2 &= 1+3+5 \\ 4^2 &= 1+3+5+7 \\ n^2 &= \sum_{k=0}^n 2k+1 \end{align} $$ It's easy to find lots of ways to prove this fact, I leave this to you.
Next, notice that whatever the common sequence number $c$ is, we have that $$ n = (a^2 - c) - (b^2 - c) = a^2 - b^2 $$ And since we reach the common number by subtracting odd numbers starting from $1$, this means that the expressions $a^2-c$ and $b^2-c$ must be perfect squares.
Since they are perfect squares, we can write $$ x^2 = a^2 -c \\ y^2 = b^2 -c \\ n = x^2-y^2 = (x+y)(x-y) $$
Notice that since we choose $n$ to be an odd number, only an odd number times another odd number can give us an odd number. Therefore, $x+y$ and $x-y$ must both be odd, which guarantees $x$ and $y$ both are integers.
Working Backwards
The trick to understanding this problem is actually working our way backwards. Once we have $a$, $b$, $x$, and $y$, we know that $$ a^2-x^2 = b^2-y^2 = c $$
Pick any two odd numbers to multiply together, for instance:
$$ \begin{align} 5 \times 3 = 15 &= 8^2 - 7^2 \\ &= (4+1)(4-1) \end{align} $$ And therefore, $8^2 - 4^2 = 7^2 - 1^2 = c = 48$.
This is why $c$ only exists if $n$ is not prime. We already showed that as long as $n$ can be factored, it can be factored into odd factors, and we can then run this process to find $c$. (Technically, we still need to prove the converse, that there is no other way to obtain a common value $c$, but all the statements made so far can easily be tweaked to be biconditional.)
This also means that some $n$ values will give us multiple values for $c$, as long as the number can be represented as products in more than one way. See: $$ \begin{align} 45 = 23^2 - 22^2 &= 15 \times 3 &= (9+6)(9-6) \\ &= 9 \times 5 &= (7+2)(7-2) \end{align} $$ You end up with both: $$ 23^2-9^2 = 22^2-6^2 = 448 \\ 23^2-7^2 = 22^2-2^2 = 480 $$
Confirm with the following sequences: $$ 529, 528, 525, 520, 513, 504, 493, \color{red}{480}, 465, \color{blue}{448}, 429, 408, 385, 360, 333, 304, 273, 240, 205, 168, 129, 88, 45 \\ 484, 483, \color{red}{480}, 475, 468, 459, \color{blue}{448}, 435, 420, 403, 384, 363, 340, 315, 288, 259, 228, 195, 160, 123, 84, 43 $$
Conclusion
The common value $c$ only exists if $n$ is not prime. In fact, if $n$ has $k$ factors, then there are $\frac{k}{2}-1$ values for $c$ (rounded up if $n$ is a perfect square).
We have not made any statements about the divisibility of $c$, but it is trivial to at least show it must be even.