I just rewatched some old math books for the fun of it and was able to observe the following which I want to formalize (could you help me with it):
Choose 2 numbers with distance 1 for each pair. Proceed with...
First Example $(i)$
(3,4) and (6,7)
$3\cdot 6+4\cdot 7=46$ and
$3\cdot 7 + 4\cdot 6 = 45$
EDIT: Second example $(ii)$
(distance between pairs don't need to be the same distance)
(5,6) and (9,10)
$5 \cdot 9 + 6 \cdot 10 = 105$ and
$5\cdot 10 + 6\cdot 9 = 104$
Therefore, the outcome of both equations in $(i)$ and $(ii)$ have a distance of $1^2$.
Observation
If you try this for different examples and for distances $n\in \mathbb{N}_0$, you'll most likely notice, that the observation is satisfied for all pairs $(a,b),(c,d)\mid a,b,c,d\in\mathbb{N}\setminus \{0\}$ with a specific distance $n\geq 0$. I want to formalize this fact, but I'm not really sure, if my following thoughts are correct.
Informal, we have: For all pairs $(a,b),(c,d) \in \mathbb{N}\setminus 0:$ exists one $n$ in $\mathbb{N}_0$ with $b$ is the successor of $a$ with distance $n$ and $d$ is the successor of $c$ with distance $n$ for which we obtain equivalent that the outcome of $(ac+bd)$ has distance $n^2$ to the outcome $(ad+bc)$.
I would formalize this as the following: $$\forall a,b,c,d \in \mathbb{N}\setminus 0 : \exists n \in \mathbb{N}_0:(b=a+n) \land (d=c+n) \iff \underbrace{(ac+bd)}_{=x+n^2}- \underbrace{(ad+bc)}_{=x}=n^2$$
Do you think this is correct? If not, could you share your thoughts about it? Could we use group-theory to discribe these equations easier?
If $b = a + n$ and $d = c + n$ then $$ (ac+bd)-(ad+bc) = (b-a)(d-c) = n^2. $$ For example, if the numbers are $(0,n)$ and $(0,n)$ then $0\cdot 0 + n \cdot n = n^2$ whereas $0 \cdot n + n \cdot 0 = 0$.