<< problem 130 - Composites with prime repunit property | Large repunit factors - problem 132 >> |

# Problem 131: Prime cube partnership

(see projecteuler.net/problem=131)

There are some prime values, p, for which there exists a positive integer, n, such that the expression n^3 + n^2 p is a perfect cube.

For example, when p = 19, 8^3 + 8^2 * 19 = 123.

What is perhaps most surprising is that for each prime with this property the value of n is unique, and there are only four such primes below one-hundred.

How many primes below one million have this remarkable property?

# My Algorithm

The equation n^3 + n^2 p = k^3 can be rewritten:

n^3 (1 + dfrac{p}{n}) = k^3

Which becomes:

n * sqrt[3]{1 + dfrac{p}{n}} = k

And:

n * sqrt[3]{dfrac{n+p}{n}} = k

The only way that n and k are integers is when the cube root is rational:

sqrt[3]{dfrac{n+p}{n}} = sqrt[3]{dfrac{a^3}{b^3}} = dfrac{a}{b}

Then the equation would be:

n * dfrac{a}{b} = k

The new variables a and b are:

a^3 = n+p

b^3 = n

Solving for p:

p = a^3 - n = a^3 - b^3

The binomial expansion a^3 - b^3 = (a - b)(a^2 + ab + b^2) tells us:

p = (a - b)(a^2 + ab + b^2)

All values a, b and p must be integers. Moreover, p must be a prime. Remember: a prime number can only be factorized into two numbers: 1 and itself.

Obviously a - b < a^2 + ab + b^2 (for positive values of a and b). Then it follows for the factors 1 and p:

1 = a - b

p = a^2 + ab + b^2

The first equation is interesting: a and b are consecutive numbers:

a = b + 1

Which can be inserted in the second equation:

p = a^2 + a(a+1) + (a+1)^2

p = a^2 + a^2 + a + a^2 + 2a + 1

p = 3a^2 + 3a + 1

Whenever p = 3a^2 + 3a + 1 is prime, then a valid value was found.

I didn't see it at first, but (a + 1)^3 - a^3 = 3a^2 + 3a + 1, too → p is the difference of two consecutive cubes.

This time I use my Wheel-based primality test (because it's the fastest and I haven't used it in a long time).

It's part of my toolbox.

## Modifications by HackerRank

Hackerrank has a huge amount of input values: I have to split my program into two parts. First, I find all solutions up to a `limit`

,

then I scan those solutions and print the result.

This time all solutions are generated in ascending order. `std::lower_bound`

finds the largest matching prime below the input very fast (binary search) and returns its position.

For example: querying 100 returns position 3 because 61 is the largest prime below 100. `std::distance`

returns 4 because index counting starts at zero.

My primality test is too slow to handle all primes below 25 * 10^12. About 10^12 is the most I can process in two seconds.

## Note

I increased `limit`

to 10^8 for the live test. You only need `limit = 1000000`

for the original problem.

# My code

… was written in C++11 and can be compiled with G++, Clang++, Visual C++. You can download it, too.

#include <iostream>
#include <vector>
#include <algorithm>
// wheel-based prime test from my toolbox

bool isPrime(unsigned int x)
{
// prime test for 2, 3 and 5
if (x % 2 == 0 || x % 3 == 0 || x % 5 == 0)
return x == 2 || x == 3 || x == 5;
// wheel with size 30 (=2*3*5):
// test against 30m+1, 30m+7, 30m+11, 30m+13, 30m+17, 30m+19, 30m+23, 30m+29
// their deltas/increments are:
const unsigned char Delta[] = { 6, 4, 2, 4, 2, 4, 6, 2 };
// start with 7, which is 30*0+7
unsigned int i = 7;
// 7 belongs to the second test group
unsigned int pos = 1;
// check numbers up to sqrt(x)
while (i*i <= x)
{
// not prime ?
if (x % i == 0)
return false;
// skip forward to next test divisor
i += Delta[pos];
// next delta/increment
pos = (pos + 1) & 7;
}
// passed all tests, must be a prime number
return x > 1;
}
int main()
{
// find all matching primes up to this limit
unsigned int limit = 100000000;
// store all matching primes
std::vector<unsigned int> matches;
// start with a=1
unsigned int a = 1;
while (true)
{
// (a+1)^3 - a^3 = 3a^2 + 3a + 1
auto p = 3*a*a + 3*a + 1;
// too big ?
if (p >= limit)
break;
// found one more prime ?
if (isPrime(p))
matches.push_back(p);
// keep going ...
a++;
}
// process STDIN
unsigned int tests = 1;
std::cin >> tests;
while (tests--)
{
std::cin >> limit;
// find highest position located before the limit (matches are sorted, should use binary search)
auto lower = std::lower_bound(matches.begin(), matches.end(), limit);
// count number of primes
auto result = std::distance(matches.begin(), lower);
std::cout << result << std::endl;
}
return 0;
}

This solution contains 12 empty lines, 22 comments and 3 preprocessor commands.

# Interactive test

You can submit your own input to my program and it will be instantly processed at my server:

This is equivalent to`echo "1 100" | ./131`

Output:

*Note:* the original problem's input `1000000`

__cannot__ be entered

because just copying results is a soft skill reserved for idiots.

*(this interactive test is still under development, computations will be aborted after one second)*

# Benchmark

The correct solution to the original Project Euler problem was found in 0.01 seconds on a Intel® Core™ i7-2600K CPU @ 3.40GHz.

(compiled for x86_64 / Linux, GCC flags: `-O3 -march=native -fno-exceptions -fno-rtti -std=c++11 -DORIGINAL`

)

See here for a comparison of all solutions.

**Note:** interactive tests run on a weaker (=slower) computer. Some interactive tests are compiled without `-DORIGINAL`

.

# Changelog

June 27, 2017 submitted solution

June 27, 2017 added comments

# Hackerrank

see https://www.hackerrank.com/contests/projecteuler/challenges/euler131

My code solves **4** out of **7** test cases (score: **50%**)

I failed **0** test cases due to wrong answers and **3** because of timeouts

# Difficulty

Project Euler ranks this problem at **40%** (out of 100%).

Hackerrank describes this problem as **hard**.

*Note:*

Hackerrank has strict execution time limits (typically 2 seconds for C++ code) and often a much wider input range than the original problem.

In my opinion, Hackerrank's modified problems are usually a lot harder to solve. As a rule thumb: brute-force is rarely an option.

# Links

projecteuler.net/thread=131 - **the** best forum on the subject (*note:* you have to submit the correct solution first)

Code in various languages:

Python: www.mathblog.dk/project-euler-131-primes-perfect-cube/ (written by Kristian Edlund)

Go: github.com/frrad/project-euler/blob/master/golang/Problem131.go (written by Frederick Robinson)

# Heatmap

green problems solve the original Project Euler problem and have a perfect score of 100% at Hackerrank, too.

yellow problems score less than 100% at Hackerrank (but still solve the original problem).

gray problems are already solved but I haven't published my solution yet.

blue problems are solved and there wasn't a Hackerrank version of it at the time I solved it or I didn't care about it because it differed too much.

red problems are solved but exceed the time limit of one minute or the memory limit of 256 MByte.

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I scored 13,183 points (out of 15300 possible points, top rank was 17 out of ≈60000 in August 2017) at Hackerrank's Project Euler+.

Look at my progress and performance pages to get more details.

My username at Project Euler is

**stephanbrumme**while it's stbrumme at Hackerrank.

# Copyright

I hope you enjoy my code and learn something - or give me feedback how I can improve my solutions.

All of my solutions can be used for any purpose and I am in no way liable for any damages caused.

You can even remove my name and claim it's yours. But then you shall burn in hell.

The problems and most of the problems' images were created by Project Euler.

Thanks for all their endless effort.

<< problem 130 - Composites with prime repunit property | Large repunit factors - problem 132 >> |