Croatian Mathematical Competitions

A line segment of length $\sqrt{5}$ between two integer points must have a vector $\vec{v} = (a, b)$ with $a^2 + b^2 = 5$. The integer solutions are $(\pm 1, \pm 2)$ and $(\pm 2, \pm 1)$.

So, the possible displacements are $(1,2)$, $(1,-2)$, $(-1,2)$, $(-1,-2)$, $(2,1)$, $(2,-1)$, $(-2,1)$, $(-2,-1)$.

For each such vector, we count the number of segments whose endpoints are both in the grid $1 \le x \le 200$, $1 \le y \le 100$.

Let us consider $(a, b)$ with $a = 1, b = 2$.

The starting point $(x, y)$ must satisfy $1 \le x \le 200 - 1 = 199$ and $1 \le y \le 100 - 2 = 98$.

So, for $(1,2)$, the number of such segments is $199 \times 98$.

Similarly, for $(1,-2)$, $1 \le x \le 199$, $3 \le y \le 100$ (since $y + 2 \le 100 \implies y \le 98$, $y \ge 1$; for $y - 2 \ge 1 \implies y \ge 3$), so $199 \times 98$.

For $(2,1)$: $1 \le x \le 198$, $1 \le y \le 99$, so $198 \times 99$.

For $(2,-1)$: $1 \le x \le 198$, $2 \le y \le 100$, so $198 \times 99$.

Similarly for the negatives, but each segment is counted twice (once from each endpoint), so we only count each vector once and multiply by 2.

So, the total number is:

$2 \times [(199 \times 98) + (198 \times 99)] = 2 \times (199 \times 98 + 198 \times 99)$

Calculate:

$199 \times 98 = (200 - 1) \times 98 = 200 \times 98 - 98 = 19600 - 98 = 19502$

$198 \times 99 = (200 - 2) \times 99 = 200 \times 99 - 2 \times 99 = 19800 - 198 = 19602$

Sum: $19502 + 19602 = 39104$

Total: $2 \times 39104 = 78208$

Answer: $78\,208$