Define the following equivalance relation over $\mathbb{N} \times \mathbb{N}:$ $(m,n) \sim (m', n')$ if and only if $m + n' = m' + n'.$ Equivalently, $(m,n) \sim (m', n')$ if and only if $m - m' = n - n'.$
Define the following multiplication operation $\cdot: \mathbb{N} \times \mathbb{N} \rightarrow \mathbb{N} \times \mathbb{N}$ as follows:
$(m,n)\cdot(p,q) = (mp + nq, mq + pn).$
So the first term in the result is a "dot product", the second is a "cross product".
It is required to prove that this multiplication operation can be extended to the equivlance classes of $\sim.$
In other words show that $[(m,n)] \cdot [(p,q)] = [(m,n) \cdot (p,q)]$ is well define. So, given $(m,n) \sim (m', n')$ and $(p,q) \sim (p',q')$, show that $(m,n)\cdot (p,q) \sim (m',n')\cdot (p',q').$
So we have to show that $mp + nq + m'q' + n'p = m'p' + n'q' + mq + np$ given that $m + n' = m' + n'$ and $p + q' = p' + q.$
Any suggestions would be appreciated.
The equivalence classes of $\sim$ are straight 45 degree lines on the lattice of numbers in $\mathbb{N} \times \mathbb{N}$
Thus the problem is to show that for any choice of $ (m,n)$, the mapping $(p,q) \rightarrow (m,n)\cdot(p,q)$ maps straight lines onto other such straight lines.
In addition, show the result line only depends on which line $(m,n)$ was on, so if you used another point on the same line as (m,n), say $(m', n'),$ you would get the same result
If you interpret the ordered pairs as complex numbers, then the multiplication can be represented as follows: let $z =(z_1, z_2)$ and $w = (w_1, w_2)$ be two such pairs. Then, by interpreting them as complex numbers $z\cdot w \equiv zw + 2\Re(z)\Re(w).$
This suggests the problem could be tackled geometrically, but the transformations are very nonlinear and don't have an easy interpretation. For example there is no reason to expect that the transformation would be isometric or preserve angles or similar.
Alternatively, the operator $\cdot$ is distributive with respect to the usual vector addition, where $(m,n) + (p,q) = (m + p, n + q)$. That is $(m,n) \cdot ( (a,b) + (p,q)) = (m,n)\cdot (a,b) + (m,n) \cdot (p,q).$
We have a multiplicative identity since $(m,n) \cdot (1,0) = (m,n)$. It is also associative, though the proof is a little monotonous.
Thus, I think it might be possible to use abstract algebra, since $(\mathbb{N} \times \mathbb{N}, +, \cdot)$ is a semiring.
It is easy to see that addition plays nice with the equivalence relation so that $[(m,n)] + [(p,q)] = [(m,n) + (p,q)].$
Could $\sim$ induce an ideal or quotient semiring of some sort? I am not sufficiently acquainted with semirings to pursue this further though.
The equivalence relation you've described is used to construct integers out of natural numbers as your building material, nothing else.
You start with the naturals. How can you build integers only having naturals? The anticipating key observation is that an object you want to construct, $-3$, e.g., is $1-4=7-10=997-1000$ etc. But you only can use naturals to describe the phenomenon. So the desired $-3$ may be viewed as a collection (equivalence class) of pairs of naturals $\{(1,4),(7,10),(997,1000)\dots\}$.
Now what two pairs have in common (the equivalence of two pairs expressed in term of naturals) is that the sum of the outermost coordinates equals the sum of the innermost ones. In that way you never leave the naturals to describe what you have in mind.
From here you declare $$-3=\{(m,n)\in\mathbb{N}^2|m+3=n\}.$$ This equivalence class is a $45^\circ$ line in $\mathbb{N}^2$. It crosses the first coordinate axes in (no surprise now) $(0,-3)$.
Observe that the defined multiplication is the usual multiplication of integers, here viewed as pairs of naturals.