Good Triangles

Suppose the three points have the same color. Let the vector from $A$ to $B$ be $(a,b)$ and the vector from $A$ to $C$ be $(c,d)$. By the same-color condition, both $a+b$ and $c+d$ are even. Hence $a,b$ have the same parity, and $c,d$ also have the same parity.

The parallelogram area is

$$W=|ad-bc|.$$

Because of the parity condition above, $ad-bc$ is always even. Therefore, for $W$ to be prime, we must have $W=2$.

We count ordered triples $(A,B,C)$ first, then divide the answer by $6$. Fix a vector $u=(a,b)$ and count vectors $v=(c,d)$ satisfying

$$ad-bc=\pm 2.$$

Let $g=\gcd(a,b)$. The linear Diophantine equation $ad-bc=\pm 2$ has an integer solution only if $g$ divides $2$. Using the extended Euclidean algorithm, find one solution $v_0=(c_0,d_0)$. Then all solutions are of the form

$$(c,d)=\left(c_0+\frac{a}{g}t,\ d_0+\frac{b}{g}t\right).$$

For each $t$, the following conditions must hold.

• $c$ and $d$ must lie in the possible vector range inside the grid.
• $c+d$ must be even.

For each valid vector pair $(u,v)$, count how many positions of $A$ make all three points $A,A+u,A+v$ lie inside the grid. This is obtained by computing the required span in the $x$-coordinates and $y$-coordinates.

After summing all ordered triples, divide by $6$ to obtain the answer.

Solution written by GPT5.5