# Can I search for factors of  (11!)!+11!+1 \ (11!)!+11!+1\ efficiently?

Is the number $$(11!)!+11!+1(11!)!+11!+1$$ a prime number ?

I do not expect that a probable-prime-test is feasible, but if someone actually wants to let it run, this would of course be very nice. The main hope is to find a factor to show that the number is not prime. If we do not find a factor, it will be difficult to check the number for primality. I highly expect a probable-prime-test to reveal that the number is composite. “Composite” would be a surely correct result. Only if the result would be “probable prime”, there would remain slight doubts, but I would be confident with such a test anyway.

Motivation : $$(n!)!+n!+1(n!)!+n!+1$$ can only be prime if $$n!+1 \ n!+1\$$ is prime. This is because a non-trivial factor of $$n!+1 \ n!+1\$$ would also divide $$(n!)!+n!+1 \ (n!)!+n!+1\$$. The cases $$n=2,3 \ n=2,3\$$ are easy , but the case $$n=11 \ n=11\$$ is the first non-trivial case. We only know that there is no factor upto $$p=11!+1 \ p=11!+1\$$

What I want to know : Can we calculate $$(11!)!\mod \ p(11!)!\mod \ p$$ for $$\ p\ \ p\$$ having $$\ 8-12\ \ 8-12\$$ digits with a trick ? I ask because pari/gp takes relatively long to calculate this residue directly. So, I am looking for an acceleration of this trial division.

## Answer

I let $$p_1=1+11!p_1=1+11!$$ for convenience. By Wilson’s theorem if there’s a prime $$pp$$ that divides $$1+11!+(11!)! = p_1 + (p_1-1)!1+11!+(11!)! = p_1 + (p_1-1)!$$ then

$$(p-1)!\equiv -1\pmod p(p-1)!\equiv -1\pmod p$$

And also

$$(p_1-1)!\equiv -p_1(p_1-1)!\equiv -p_1$$

So

$$(p-1)(p-2)…p_1\cdot(p_1-1)!\equiv -1(p-1)(p-2)...p_1\cdot(p_1-1)!\equiv -1$$

$$(p-1)(p-2)…p_1\cdot p_1\equiv 1(p-1)(p-2)...p_1\cdot p_1\equiv 1$$

This way I was able to check all the primes from $$p_1p_1$$ to 74000000 in 12 hours. This gives a 3.4% chance of finding a factor according to big prime country’s heuristic. The algorithm has bad asymptotic complexity because to check a prime $$pp$$ you need to perform $$p-11!p-11!$$ modular multiplications so there’s not much hope of completing the calculation.

Note that I haven’t used that $$p_1p_1$$ is prime, so maybe that can still help somehow. Here’s the algorithm in c++:

// compile with g++ main.cpp -o main -lpthread -O3

#include <iostream>
#include <vector>
#include <string>

#include <boost/process.hpp>

#include <thread>

namespace bp = boost::process;

const constexpr unsigned int p1 = 1 * 2 * 3 * 4 * 5 * 6 * 7 * 8 * 9 * 10 * 11 + 1; // 11!+1
const constexpr unsigned int max = 100'000'000;                                    // maximum to trial divide
std::vector<unsigned int> primes;
unsigned int progress = 40;

void trial_division(unsigned int n) { // check the primes congruent to 2n+1 mod 16
for(auto p : primes) {
if(p % 16 != (2 * n + 1)) continue;
uint64_t prod = 1;
for(uint64_t i = p - 1; i >= p1; --i) {
prod = (prod * i) % p;
}
if((prod * p1) % p == 1) {
std::cout << p << "\n";
}
if(n == 0 && p > progress * 1'000'000) {
std::cout << progress * 1'000'000 << "\n";
++progress;
}
}
}

int main() {
bp::ipstream is;
bp::child primegen("./primes", std::to_string(p1), std::to_string(max), bp::std_out > is);
// this is https://cr.yp.to/primegen.html
// the size of these primes don't really justify using such a specialized tool, I'm just lazy

std::string line;
while (primegen.running() && std::getline(is, line) && !line.empty()) {
primes.push_back(std::stoi(line));
} // building the primes vector

// start 8 threads, one for each core for on my computer, each checking one residue class mod 16
// By Dirichlet's theorem on arithmetic progressions they should progress at the same speed
// the 16n+1 thread owns the progress counter
std::thread t0(trial_division, 0);
std::thread t1(trial_division, 1);
std::thread t2(trial_division, 2);
std::thread t3(trial_division, 3);
std::thread t4(trial_division, 4);
std::thread t5(trial_division, 5);
std::thread t6(trial_division, 6);
std::thread t7(trial_division, 7);

t0.join();
t1.join();
t2.join();
t3.join();
t4.join();
t5.join();
t6.join();
t7.join();
}


I only need to multiply integers of the order of $$11!11!$$ so standard 64 bit ints suffice.

EDIT: Divisor found! $$15904298891590429889$$

So first of all, the Wilson’s theorem trick slows down instead of speeding up after $$2p_12p_1$$. Secondly, the trial division function is nearly infinitely parallelizable, which means that it’s prone to being computed with a GPU. My friend wrote an implementation that can be found here. This can be run on CUDA compatible nvidia GPUs. Finding the factor took about 18 hours on a Nvidia GTX Titan X pascal.

Attribution
Source : Link , Question Author : Peter , Answer Author : Sophie