We will show that if two points have the same "Long John" coordinates they are equal, and we will show this by contradiction.
Suppose there are four numbers a,b, x, and y, all integers, such that a²+b=x²+y and a+b²=x+y², with a²+b <> a+b² and suppose further that y != b (since everything is symmetric, there is no loss of generality).
Rewriting the two equations we get:
We replace a-x in the first equation by its value in the second equation. Since y<>b we can divide by y-b both sides, and we have:
Since all numbers are integers, this means that y+b=1 and a+x=1, or that they are both equal to -1. Again, since the two things are symmetric, we take that y+b=a+x=1, so that a-b=y-x. Replacing the values of y+b and a+x in equations (A) and (B) we get a+b=x+y. Therefore a=y and b=x. But this implies that a²+b=x²+y=a+b², which is wrong (since the two numbers are distinct). Hence there is a contradiction, and so the assumption is wrong: there is no two different pairs of numbers having the same "Long John" coordinates.