Proof validity: pick's theorem

[harsh tundra]

Using the following dissection and the fact that the number of lattice points on a line between (0,0) and (a,b) for integer a,b is gcd(a,b) (https://math.stackexchange.com/questions/628117/how-to-count-lattice-points-on-a-line if you need to see more), it is fairly simple to see that the number of interior points in the internal square is I = (a+b+1)^2 - 2ab - 2(a+b+gcd(a,b)) and that the number of boundary points is B=4gcd(a,b). This formula for I comes from subtracting the total number of lattice points in the small triangles from the number of lattice points in the bigger square.

But then using Pick's theorem again with our formulas for I and B and doing basic algebra gives A=a^2 + b^2. Or, c^2 = a^2 + b^2.
This only works for integer values of a and b but is this valid? Also does it only work for half/integer or integer values of c^2 since pick's theorem shows that area can only be a half integer or integer (and the resulting endpoints of the c square side will not be at integer coordinates) ?

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[winter mantle]

I = (a+b+1)^2 - 2ab - 2(a+b+gcd(a,b))
Can you explain your reasoning?
B = 4 gcd(a,b)
I believe you're double-counting the corners

[harsh tundra]

@winter mantle
The number of lattice points in a shape can be calculated using Pick's theorem. Let us first use this to find the number of lattice points in one of the triangles. Rearranging Pick's theorem for $I$, we get the equation $I=A - \frac{B}{2} + 1$. The number of lattice points on a line between points $(a,b)$ and $(c,d)$ is $gcd(a-c, b-d) + 1$. Substituting into Pick's theorem and rearranging, we get
\begin{equation}
I=\frac{ab - (a+b+gcd(a+b))}{2}+1.
\end{equation}
\hspace{\parindent} This is equal to the number of interior lattice points per outer triangle. Multiplying by 4 for 4 triangles and simplifying, we get
\begin{equation}
2ab-2\left(a+b+\gcd\left(a,b\right)\right)+4
\end{equation}
which is the total number of lattice points in the triangles surrounding the square. We can subtract this from the number of lattice points in the entire outer square to get just the number of lattice points inside the square. The number of lattice points in the larger square is given by simply (a+b+1)^2 (the + 1 comes from the extra endpoint the side length doesn't "account for". draw up one example and this should make sense)

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[harsh tundra]

Also can you elaborate on the double corner counting?

[winter mantle]

Right, that makes sense. You're not double-counting the corners (though "the number of lattice points on a line between (0,0) and (a,b) for integer a,b is gcd(a,b)" should be gcd(a, b) + 1).

c^2 = a^2 + b^2 is just the Pythagorean theorem; it's true for any real a, b, c arranged around a right triangle like that, but if a and b are integers than c is never a half-integer

(that's actually a pretty neat proof of the Pythagorean theorem for integers)

[harsh tundra]

I lowkey kinda realized this after thinking about it. So yea it does work for only integer values of a,b,c. But as long as the rest of the reasoning works that's good

[harsh tundra]

I'd kinda like to post this somewhere but idk what to do with it (mainly for college apps and for getting internships). I'm also still 17 so I especially don't know what to do with it. I did this at 16 and I've kinda sat on it cause I didn't know what to do and everyone I emailed just ghosted me or couldn't help

Like, this by itself obviously isn't enough to post to a site like arXiv so is there anything I can do to get it out there

[winter mantle]

a blog post maybe?

or something similar

[harsh tundra]

good idea

[hybrid steppe]

@harsh tundra

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[visual pumice]

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