Project Euler

Perfect Right-angled Triangles

A right-angled triangle with sides a, b, and hypotenuse c is called perfect if: 1. (a, b, c) is a primitive Pythagorean triple. 2. The hypotenuse c is a perfect square. How many perfect right-angle...

Source sync May 21, 2026

Problem #0218

Level Level 08

Solved By 3,427

Languages C++, Python

Answer 0

Length 291 words

geometrynumber_theorymodular_arithmetic

Consider the right angled triangle with sides $a=7$, $b=24$ and $c=25$. The area of this triangle is $84$, which is divisible by the perfect numbers $6$ and $28$.

Moreover it is a primitive right angled triangle as $\gcd (a,b)=1$ and $\gcd (b,c)=1$.

Also $c$ is a perfect square.

We will call a right angled triangle perfect if

it is a primitive right angled triangle
its hypotenuse is a perfect square.

We will call a right angled triangle super-perfect if

it is a perfect right angled triangle and
its area is a multiple of the perfect numbers $6$ and $28$.

How many perfect right-angled triangles with $c \le 10^{16}$ exist that are not super-perfect?

Problem 218: Perfect Right-angled Triangles

Mathematical Development

Theorem 1 (Parametrization of primitive Pythagorean triples). Every primitive Pythagorean triple $(a,b,c)$ with $b$ even is given by

a = m^2 - n^2, \qquad b = 2mn, \qquad c = m^2 + n^2,

where $m > n > 0$ , $\gcd(m,n)=1$ , and $m \not\equiv n \pmod 2$ .

Proof. This is the classical Euclid parametrization. $\square$

Lemma 1 (Area formula). The area is

A = mn(m-n)(m+n).

Proof. Since $A = ab/2$ ,

A = \frac{(m^2-n^2)(2mn)}{2} = mn(m-n)(m+n). \qquad \square

Theorem 2 (Every perfect triangle has area divisible by 84). For every perfect right-angled triangle,

84 \mid A.

In particular, its area is divisible by both 6 and 28.

Proof. Since the hypotenuse is a perfect square, say

c = m^2 + n^2 = k^2,

the triple $(m,n,k)$ is itself a primitive Pythagorean triple. Hence one may write

m = u^2 - v^2, \qquad n = 2uv

or the symmetric variant.

Factor 3: if neither $m$ nor $n$ were divisible by 3, then $m^2 + n^2 \equiv 1 + 1 \equiv 2 \pmod 3,$ impossible for a square. So $3 \mid mn$ .
Factor 4: the even parameter is $2uv$ , and one of $u,v$ is even, so in fact $4$ divides that factor.
Factor 7: substituting the second parametrization and checking residues modulo 7 shows that one of $m,\ n,\ m-n,\ m+n$ is always divisible by 7.

Therefore $3 \cdot 4 \cdot 7 = 84$ divides

mn(m-n)(m+n) = A. \qquad \square

Editorial

There is no search left once the second parametrization is used. A perfect triangle already starts as a primitive Pythagorean triple, and the extra condition that the hypotenuse is a square forces the parameters $(m,n)$ to form another primitive Pythagorean triple. That second layer is what injects the extra divisibility.

From the resulting area formula

A = mn(m-n)(m+n),

the factors 3 and 4 are immediate, and a short residue check supplies the factor 7. So every perfect triangle has area divisible by 84, which means no triangle can fail both the divisibility-by-6 and divisibility-by-28 requirements.

Pseudocode

Use the primitive Pythagorean parametrization for (a, b, c).
Use the condition c = square to parametrize (m, n, k) as another primitive triple.
Deduce from the area formula that every candidate area has factors 3, 4, and 7.
Since every perfect triangle has area divisible by 84, the number of exceptions is 0.
Return 0.

Complexity Analysis

Time: $O(1)$ .
Space: $O(1)$ .

Answer

\boxed{0}

C++ project_euler/problem_218/solution.cpp

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

long long countNonSuperPerfect(long long /*limit*/) {
    // The proof shows every perfect right-angled triangle has area divisible by 84,
    // so none can fail both the divisibility-by-6 and divisibility-by-28 tests.
    return 0;
}

int main() {
    cout << countNonSuperPerfect(10000000000000000LL) << '\n';
    return 0;
}

Python project_euler/problem_218/solution.py

"""
Problem 218: Perfect Right-angled Triangles

Every perfect right-angled triangle has area divisible by 84.
Therefore no perfect triangle can fail both the divisibility-by-6
and divisibility-by-28 tests.
"""


def count_non_super_perfect(limit):
    # The proof is independent of the hypotenuse bound.
    return 0


def solve():
    print(count_non_super_perfect(10**16))


if __name__ == "__main__":
    solve()