Problem 224: Almost Right-angled Triangles II

Mathematical Development

Write

We want the positive integer solutions of

Any common divisor of $a,b,c$ would also divide

so every solution is primitive.

The classical Berggren matrices for the ternary quadratic form $x^2+y^2-z^2$ are

Each satisfies

so they preserve the equation $Q=-1$.

The smallest positive solution is

and the Hall-Berggren theory for this form shows that every positive solution of $Q=-1$ is obtained by repeatedly applying these matrices.

Because $a$ and $b$ play symmetric roles, we sort them after each transformation. When $a=b$, the first and third matrices produce the same sorted child, so only two distinct descendants remain. That gives a duplicate-free tree.

Therefore the counting problem is reduced to a depth-first traversal of that tree, pruning as soon as

Editorial

The direct factorization

is valid, but it is not the most convenient way to count all solutions up to a perimeter of seventy-five million. The better observation is that the equation lives on the quadratic surface

and that this surface has the same Berggren-style tree structure that ordinary Pythagorean triples have.

So instead of factoring $4m^2+1$ for millions of values of $m$, we start from the root $(2,2,3)$ and generate every solution exactly once. Each matrix application gives a new valid triple, sorting the first two coordinates restores the convention $a \le b$, and the perimeter bound cuts off the recursion naturally. That turns the problem into a pure tree walk with almost no arithmetic overhead.

Pseudocode

Store the three Berggren matrices that preserve a^2 + b^2 - c^2.

Initialize a stack with the root triple (2, 2, 3).
Set the answer counter to 0.

While the stack is not empty:
    remove one triple (a, b, c)
    reorder the first two entries so that a <= b

    If a + b + c exceeds the perimeter limit:
        discard this branch
        continue

    Count the current triple.

    If a = b:
        only the first two matrices give distinct sorted children
    otherwise:
        use all three matrices

    For each chosen matrix:
        compute the transformed triple
        keep it only if all three coordinates stay positive
        sort the first two coordinates
        push the child onto the stack

Return the counter.

Complexity Analysis

• Time: Linear in the number of generated triples; every valid solution is visited once.
• Space: Proportional to the maximum size of the DFS stack.

Answer

#include <algorithm>
#include <array>
#include <iostream>
#include <tuple>
#include <vector>

using namespace std;

using Matrix = array<array<long long, 3>, 3>;

const Matrix matrices[3] = {
    {{{1, -2, 2}, {2, -1, 2}, {2, -2, 3}}},
    {{{1, 2, 2}, {2, 1, 2}, {2, 2, 3}}},
    {{{-1, 2, 2}, {-2, 1, 2}, {-2, 2, 3}}},
};

tuple<long long, long long, long long> transform(const Matrix& matrix,
                                                 long long a,
                                                 long long b,
                                                 long long c) {
    long long x = matrix[0][0] * a + matrix[0][1] * b + matrix[0][2] * c;
    long long y = matrix[1][0] * a + matrix[1][1] * b + matrix[1][2] * c;
    long long z = matrix[2][0] * a + matrix[2][1] * b + matrix[2][2] * c;
    return {x, y, z};
}

int main() {
    const long long limit = 75000000LL;
    long long count = 0;
    vector<tuple<long long, long long, long long>> stack = {{2, 2, 3}};

    while (!stack.empty()) {
        long long a0, b0, c;
        tie(a0, b0, c) = stack.back();
        stack.pop_back();

        long long a = min(a0, b0);
        long long b = max(a0, b0);

        if (a + b + c > limit) {
            continue;
        }

        count++;
        int childCount = (a == b ? 2 : 3);
        for (int i = 0; i < childCount; i++) {
            long long x0, y0, z;
            tie(x0, y0, z) = transform(matrices[i], a, b, c);
            if (x0 <= 0 || y0 <= 0 || z <= 0) {
                continue;
            }

            long long x = min(x0, y0);
            long long y = max(x0, y0);
            stack.push_back(make_tuple(x, y, z));
        }
    }

    cout << count << '\n';
    return 0;
}

"""
Problem 224: Almost Right-angled Triangles II

We need positive solutions of

    a^2 + b^2 = c^2 - 1

with a <= b and a + b + c <= 75,000,000.

The Berggren matrices preserve x^2 + y^2 - z^2 = -1. Starting from the root
(2, 2, 3), they generate every positive primitive solution. Because a and b
play symmetric roles, sorted triples only need two children when a = b and all
three otherwise.
"""

MATRICES = (
    ((1, -2, 2), (2, -1, 2), (2, -2, 3)),
    ((1, 2, 2), (2, 1, 2), (2, 2, 3)),
    ((-1, 2, 2), (-2, 1, 2), (-2, 2, 3)),
)


def transform(matrix, triple):
    a, b, c = triple
    return (
        matrix[0][0] * a + matrix[0][1] * b + matrix[0][2] * c,
        matrix[1][0] * a + matrix[1][1] * b + matrix[1][2] * c,
        matrix[2][0] * a + matrix[2][1] * b + matrix[2][2] * c,
    )


def solve():
    limit = 75_000_000
    count = 0
    stack = [(2, 2, 3)]

    while stack:
        a, b, c = stack.pop()
        if a > b:
            a, b = b, a

        if a + b + c > limit:
            continue

        count += 1
        children = MATRICES[:2] if a == b else MATRICES

        for matrix in children:
            x, y, z = transform(matrix, (a, b, c))
            if x <= 0 or y <= 0 or z <= 0:
                continue
            if x > y:
                x, y = y, x
            stack.append((x, y, z))

    print(count)


if __name__ == "__main__":
    solve()