Trying to prove the Pythagorean theorem using Picks theorem. Picks Theorem Let A be the area of a simply closed lattice square. Let B denote the number of lattice points on the square edges and I the number of points in the interior of the square. Then
$\large A=I + \frac{B}{2}-1$
Define three squares with areas
$\large A_{a} = I_{a} + \frac{B_{a}}{2}-1 = a^2$
$\large A_{b} = I_{b} + \frac{B_{b}}{2}-1 = b^2$
$\large A_{c} = I_{c} + \frac{B_{c}}{2}-1 = c^2$

Theorem
For $B_{c}=4$, $A_{c} = A_{a} + A_{b}$.
Proof
Case $B_{c}=4$,
$\large A_{c} = I_{c} + \frac{4}{2}-1$
$\large A_{c} = I_{c} + 1$
Observe that
$\large I_{c} = A_{a} + A_{b} - 1$
Substituting $I_{c}$
$\large A_{c} = (A_{a} + A_{b} - 1) + 1$
$\large A_{c} = A_{a} + A_{b}$
$\therefore \large c^2 = a^2 + b^2 $
Questions
Is that a valid proof for specific case $B_{c}=4$?
For general cases, can Picks be applied to prove Pythagoras?
Thanks.

The following describe a possible general proof. You just need to calculate the number of grid points in a triangle and the square as function of the sides of the right angle triangles.

