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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?

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  • $\begingroup$ "Formalization" does not mean "writing things with symbols". It means "fully explaining everything, without leaving gaps". This is a common misconception - using a bunch of symbols doesn't necessarily make your statement any more formal, useful, or understandable. $\endgroup$ – Deusovi Dec 14 '18 at 14:25
  • $\begingroup$ I would be glad if you share your thoughts with me. $\endgroup$ – Doesbaddel Dec 14 '18 at 14:54
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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 $0 \cdot n + n \cdot 0 = 0$.

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  • $\begingroup$ That would mean my formalization is wrong, right? $\endgroup$ – Doesbaddel Dec 14 '18 at 9:13
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    $\begingroup$ It's not your formalization that is wrong – it's your observation that is wrong. $\endgroup$ – Yuval Filmus Dec 14 '18 at 9:13
  • $\begingroup$ Oh, I see. I'm not really sure what the observation would be like otherwise. Could you give me a hint, please? $\endgroup$ – Doesbaddel Dec 14 '18 at 9:17
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    $\begingroup$ Note that $1^2=1$. $\endgroup$ – Yuval Filmus Dec 14 '18 at 9:24
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    $\begingroup$ My answer formalizes and proves your observation. $\endgroup$ – Yuval Filmus Dec 15 '18 at 9:49

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