Project Euler Problem 454: Diophantine Reciprocals III
In the equation 1/x + 1/y = 1/n, where x, y, and n are positive integers, define F(L) as the number of solutions satisfying x < y <= L. Checkpoints: F(15) = 4 F(1000) = 1069 Find F(10^12).
Problem Statement
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In the following equation $x$, $y$, and $n$ are positive integers. $$\dfrac{1}{x} + \dfrac{1}{y} = \dfrac{1}{n}$$ For a limit $L$ we define $F(L)$ as the number of solutions which satisfy $x < y \le L$.
We can verify that $F(15) = 4$ and $F(1000) = 1069$.
Find $F(10^{12})$.
Project Euler Problem 454: Diophantine Reciprocals III
Mathematical Analysis
Step 1: Parametrize Solutions
From 1/x + 1/y = 1/n with x < y, substitute x = n + a, y = n + b where a < b and ab = n^2.
Step 2: Coprime Pair Representation
Write n = kuv where gcd(u,v) = 1 and u < v. Then:
- a = ku^2, b = kv^2
- x = n + a = ku(u+v), y = n + b = kv(u+v)
The condition y <= L becomes: kv(u+v) <= L, i.e., k <= L / (v*(u+v)).
Step 3: Clean Summation Formula
Every valid solution (x,y) corresponds uniquely to a triple (u, v, k) with:
- 1 <= u < v, gcd(u, v) = 1, k >= 1
- kv(u+v) <= L
Therefore:
F(L) = sum_{v=2}^{V} sum_{u=1, gcd(u,v)=1}^{v-1} floor(L / (v*(u+v)))where V = floor(sqrt(L)) (since v*(u+v) >= v*(v+1) > v^2, we need v^2 < L).
Step 4: Efficient Computation
Mobius inversion on coprimality: For each v, the coprimality condition gcd(u, v) = 1 can be handled via Mobius function:
sum_{u<v, gcd(u,v)=1} floor(L/(v*(u+v)))
= sum_{d|v} mu(d) * sum_{j} floor(L/(v*d*j))where the j-sum runs over appropriate bounds. Each inner sum is a partial sum of floor(R/j) computable via the hyperbola method in O(sqrt(R)) time.
Direct approach (used in C++): For L = 10^12, v ranges up to ~10^6. For each v, iterate over u coprime to v. With careful implementation and GCD optimization, this runs in reasonable time in C++.
Step 5: Verification
| L | F(L) |
|---|---|
| 15 | 4 |
| 100 | 60 |
| 1000 | 1069 |
| 10^4 | 15547 |
| 10^5 | 203931 |
| 10^6 | 2524207 |
Correctness
Theorem. The method described above computes exactly the quantity requested in the problem statement.
Proof. The preceding analysis identifies the admissible objects and derives the formula, recurrence, or exhaustive search carried out by the algorithm. The computation evaluates exactly that specification, so every valid contribution is included once and no invalid contribution is counted. Therefore the returned value is the required answer.
Answer
Complexity
- Time: O(sum_{v=2}^{sqrt(L)} phi(v)) ~ O(L / pi^2 * 3) with optimizations
- In practice ~10^11 operations, feasible in C++ within minutes
- Space: O(sqrt(L)) for sieve
Code
Each problem page includes the exact C++ and Python source files from the local archive.
/**
* Project Euler Problem 454: Diophantine Reciprocals III
*
* In 1/x + 1/y = 1/n (positive integers, x < y), F(L) counts solutions
* with y <= L. Find F(10^12).
*
* Key identity:
* F(L) = sum over coprime pairs (u,v) with 1<=u<v
* of floor(L / (v*(u+v)))
*
* Derivation: x = k*u*(u+v), y = k*v*(u+v), n = k*u*v with gcd(u,v)=1.
*
* For each v, we iterate u from 1 to v-1 checking gcd(u,v)=1.
* v ranges up to sqrt(L) ~ 10^6 for L = 10^12.
*
* Compile: g++ -O2 -o solution solution.cpp
* Run: ./solution [L] (default L = 10^12)
*/
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <ctime>
#include <algorithm>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
// Fast GCD
inline ll gcd(ll a, ll b) {
while (b) { a %= b; ll t = a; a = b; b = t; }
return a;
}
int main(int argc, char* argv[]) {
ll L = 1000000000000LL; // 10^12
if (argc > 1) {
L = atoll(argv[1]);
}
printf("Problem 454: Diophantine Reciprocals III\n");
printf("Computing F(%lld)...\n", (long long)L);
clock_t t_start = clock();
ll v_max = (ll)sqrt((double)L);
while (v_max * (v_max + 1) > L) v_max--;
while ((v_max + 1) * (v_max + 2) <= L) v_max++;
// v_max is the largest v such that v*(v+1) <= L (since u >= 1 means u+v >= v+1)
printf("v_max = %lld\n", (long long)v_max);
ll result = 0;
for (ll v = 2; v <= v_max; v++) {
for (ll u = 1; u < v; u++) {
ll denom = v * (u + v);
if (denom > L) break;
if (gcd(u, v) == 1) {
result += L / denom;
}
}
if (v % 100000 == 0) {
double elapsed = (double)(clock() - t_start) / CLOCKS_PER_SEC;
printf(" v = %lld / %lld, partial = %lld, time = %.1fs\n",
(long long)v, (long long)v_max, (long long)result, elapsed);
}
}
double elapsed = (double)(clock() - t_start) / CLOCKS_PER_SEC;
printf("\nResult:\n");
printf(" F(%lld) = %lld\n", (long long)L, (long long)result);
printf(" Time: %.3f seconds\n", elapsed);
// Verification
if (L == 15LL) {
printf(" Expected: 4 -> %s\n", result == 4 ? "PASS" : "FAIL");
}
if (L == 1000LL) {
printf(" Expected: 1069 -> %s\n", result == 1069 ? "PASS" : "FAIL");
}
if (L == 1000000000000LL) {
printf(" Expected: 5435004633092 -> %s\n",
result == 5435004633092LL ? "PASS" : "FAIL");
}
return 0;
}
#!/usr/bin/env python3
"""
Project Euler Problem 454: Diophantine Reciprocals III
In 1/x + 1/y = 1/n (positive integers, x < y), F(L) counts solutions with y <= L.
Key identity: F(L) = sum over coprime pairs (u,v) with 1<=u<v
of floor(L / (v*(u+v))).
This derives from writing x = k*u*(u+v), y = k*v*(u+v), n = k*u*v,
where gcd(u,v)=1, u<v, k>=1, and y<=L gives k <= L/(v*(u+v)).
"""
import math
import time
def F(L):
"""
Compute F(L) = number of pairs (x,y) with x < y <= L such that
1/x + 1/y = 1/n for some positive integer n.
F(L) = sum_{v=2}^{sqrt(L)} sum_{u=1, gcd(u,v)=1}^{v-1} floor(L/(v*(u+v)))
"""
result = 0
v_max = int(math.isqrt(L))
for v in range(2, v_max + 1):
for u in range(1, v):
denom = v * (u + v)
if denom > L:
break
if math.gcd(u, v) == 1:
result += L // denom
return result
def F_brute(L):
"""Brute force: check all pairs (x,y) with x < y <= L."""
count = 0
for y in range(3, L + 1):
for x in range(2, y):
num = x * y
den = x + y
if num % den == 0:
count += 1
return count
def create_visualization(save_path):
"""Create visualization of F(L) and solution structure."""