Problem 251: Cardano Triplets

Formal Mathematical Development

Theorem 1 (Fundamental Equation). A triple $(a,b,c)$ of positive integers satisfies $\sqrt[3]{a + b\sqrt{c}} + \sqrt[3]{a - b\sqrt{c}} = 1$ if and only if

Proof. Set $u = \sqrt[3]{a + b\sqrt{c}}$ and $v = \sqrt[3]{a - b\sqrt{c}}$. Then:

Cubing the identity $u + v = 1$:

Therefore $u^3 v^3 = (uv)^3 = \frac{(1-2a)^3}{27}$, which yields:

Multiplying both sides by 27 and rearranging:

It remains to factor the left side. Observe that $a = -1$ is a double root of $8a^3 + 15a^2 + 6a - 1$:

and synthetic division yields $8a^3 + 15a^2 + 6a - 1 = (a+1)(8a^2 + 7a - 1)$. Factoring the quadratic: $8a^2 + 7a - 1 = (a+1)(8a-1)$. Hence:

Lemma 1 (Congruence Constraint). Every Cardano triplet satisfies $a \equiv 2 \pmod{3}$.

Proof. Since $(a+1)^2(8a-1) = 27b^2 c$, the left side must be divisible by $3^3 = 27$. We analyze the $3$-adic valuation.

Write $v_3(n)$ for the largest power of $3$ dividing $n$. We need $2\,v_3(a+1) + v_3(8a-1) \ge 3$.

• Case $3 \nmid (a+1)$: Then $v_3(a+1) = 0$, so we require $v_3(8a-1) \ge 3$. But $8a - 1 \equiv 2a - 1 \pmod{3}$, and $3 \nmid (a+1)$ means $a \not\equiv 2 \pmod{3}$. If $a \equiv 0 \pmod{3}$, then $8a - 1 \equiv -1 \pmod{3}$, so $v_3(8a-1) = 0 < 3$. If $a \equiv 1 \pmod{3}$, then $8a - 1 \equiv 1 \pmod{3}$, so $v_3(8a-1) = 0 < 3$. Both cases fail.

• Case $3 \mid (a+1)$: Then $a \equiv 2 \pmod{3}$ and $v_3(a+1) \ge 1$. Write $a + 1 = 3t$, so $a = 3t - 1$. Then $8a - 1 = 24t - 9 = 3(8t - 3)$, giving $v_3(8a-1) \ge 1$. The total is $2\,v_3(3t) + v_3(3(8t-3)) \ge 2 \cdot 1 + 1 = 3$, which suffices.

Therefore $a \equiv 2 \pmod{3}$ is both necessary and sufficient for the divisibility by 27. $\blacksquare$

Theorem 2 (Parametrization). Setting $a = 3t - 1$ for $t \ge 1$, the Cardano equation reduces to

For each $t$, decompose $8t - 3 = s^2 m$ where $m$ is squarefree and $s \ge 1$. Then every solution arising from this decomposition has the form $a = 3t - 1$, $b = ts$, $c = m$.

Proof. Substituting $a = 3t - 1$ into $(a+1)^2(8a-1) = 27b^2 c$:

Now write $8t - 3 = s^2 m$ with $m$ squarefree. Then $t^2 s^2 m = b^2 c$. Setting $b = ts$ and $c = m$ gives a valid solution with $c$ squarefree. Moreover, since $m$ is squarefree, the only $d$ with $d^2 \mid m$ is $d = 1$, so this is the unique factorization yielding a squarefree $c$ from the given decomposition.

Note that different square-factor decompositions of $8t - 3$ (when it has multiple square factors) yield distinct valid triplets. For each divisor $s^2$ of $8t - 3$, we get a triplet $(3t-1, ts, (8t-3)/s^2)$. $\blacksquare$

Remark. The constraint $a + b + c \le N$ translates to $3t - 1 + ts + m \le N$, where $N = 110{,}000{,}000$.

Editorial

A Cardano Triplet (a, b, c) satisfies: cbrt(a + bsqrt(c)) + cbrt(a - bsqrt(c)) = 1 This reduces to (a+1)^2 * (8a-1) = 27 * b^2 * c. Setting a = 3m-1 yields m^2 * (8m-3) = b^2 * c. Parametrize: b = ef, m = er with gcd(f,r) = 1 and f^2 | (8m-3). Then c = r^2*(8m-3)/f^2, and a+b+c = (3m-1) + ef + r^2(8m-3)/f^2 <= N. For each (r, f) pair, valid e values form an arithmetic progression mod f^2.

Pseudocode

f = 1 case: closed-form count
f >= 3, odd, gcd(f,r) = 1

Complexity Analysis

Time complexity: The outer loop runs for $r = O(\sqrt{N})$ iterations. For each $r$, the inner loop over $f$ runs while $f^3 \le 8rN$, i.e., $f \le (8rN)^{1/3}$. The total number of $(r, f)$ pairs examined is bounded by $\sum_{r=1}^{O(\sqrt{N})} (rN)^{1/3} = O(N^{1/3} \cdot N^{1/2} \cdot N^{1/6}) = O(N)$, giving an overall time complexity of $O(N)$ with small constants.

Space complexity: $O(1)$ auxiliary storage.

Answer

/*
 * Project Euler Problem 251: Cardano Triplets
 *
 * A Cardano Triplet (a, b, c) satisfies:
 *   cbrt(a + b*sqrt(c)) + cbrt(a - b*sqrt(c)) = 1
 *
 * Fundamental equation: (a+1)^2 * (8a-1) = 27 * b^2 * c
 * Parametrization: a = 3m-1 yields m^2 * (8m-3) = b^2 * c
 *
 * We write b = e*f, m = e*r with gcd(f,r) = 1, f^2 | (8m-3).
 * For each (r, f), valid e values form an arithmetic progression mod f^2.
 *
 * Answer: 18946051
 */

#include <bits/stdc++.h>
using namespace std;

int main() {
    const long long N = 110000000LL;
    long long count = 0;

    for (long long r = 1; r * (r + 3) <= N; r++) {
        // f = 1 case
        {
            long long coeff = 3 * r + 1 + 8LL * r * r * r;
            long long offset = 3LL * r * r + 1;
            if (coeff <= N + offset) {
                long long emax = (N + offset) / coeff;
                if (emax >= 1) count += emax;
            }
        }

        // f >= 3, odd, gcd(f, r) = 1
        for (long long f = 3;; f += 2) {
            if (f * f * f > 8LL * r * N) break;
            if (__gcd(f, r) != 1) continue;
            long long f2 = f * f;

            long long val = (8LL * r) % f2;
            if (__gcd(val, f2) != 1LL) continue;

            // Compute modular inverse of val mod f2 via extended GCD
            long long inv_val;
            {
                long long a = val, b0 = f2, x0 = 1, x1 = 0;
                while (a > 0) {
                    long long q = b0 / a;
                    b0 -= q * a; swap(a, b0);
                    x1 -= q * x0; swap(x0, x1);
                }
                inv_val = ((x1 % f2) + f2) % f2;
            }
            long long e0 = (3LL * inv_val) % f2;
            if (e0 == 0) e0 = f2;

            long long m = r * e0;
            long long h = 8LL * m - 3;
            long long s = (3 * m - 1) + e0 * f + r * r * (h / f2);
            if (s <= N) {
                long long delta = f2 * (3 * r + f) + 8LL * r * r * r;
                long long kmax = (N - s) / delta;
                count += kmax + 1;
            }
        }
    }

    cout << count << endl;
    return 0;
}

"""
Project Euler Problem 251: Cardano Triplets

A Cardano Triplet (a, b, c) satisfies:
    cbrt(a + b*sqrt(c)) + cbrt(a - b*sqrt(c)) = 1

This reduces to (a+1)^2 * (8a-1) = 27 * b^2 * c.
Setting a = 3m-1 yields m^2 * (8m-3) = b^2 * c.

Parametrize: b = e*f, m = e*r with gcd(f,r) = 1 and f^2 | (8m-3).
Then c = r^2*(8m-3)/f^2, and a+b+c = (3m-1) + e*f + r^2*(8m-3)/f^2 <= N.

For each (r, f) pair, valid e values form an arithmetic progression mod f^2.

Answer: 18946051
"""

from math import gcd


def solve():
    N = 110_000_000
    count = 0

    r = 1
    while r * (r + 3) <= N:
        # f = 1 case (closed form)
        coeff = 3 * r + 1 + 8 * r * r * r
        offset = 3 * r * r + 1
        if coeff <= N + offset:
            emax = (N + offset) // coeff
            if emax >= 1:
                count += emax

        # f >= 3, odd, gcd(f, r) = 1
        f = 3
        while True:
            if f * f * f > 8 * r * N:
                break
            if gcd(f, r) != 1:
                f += 2
                continue
            f2 = f * f
            val = (8 * r) % f2
            if gcd(val, f2) != 1:
                f += 2
                continue

            inv_val = pow(val, -1, f2)
            e0 = (3 * inv_val) % f2
            if e0 == 0:
                e0 = f2

            m = r * e0
            h = 8 * m - 3
            s = (3 * m - 1) + e0 * f + r * r * (h // f2)
            if s <= N:
                delta = f2 * (3 * r + f) + 8 * r * r * r
                kmax = (N - s) // delta
                count += kmax + 1

            f += 2
        r += 1

    print(count)


if __name__ == "__main__":
    solve()