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

Question

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

What I want to know : Can we calculate $$(11!)!\mod \ p$$ for $\ p\ $ having $\ 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 1

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

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

And also

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

So

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

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

This way I was able to check all the primes from $p_1$ 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 $p$ you need to perform $p-11!$ modular multiplications so there's not much hope of completing the calculation.

Note that I haven't used that $p_1$ 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 $11!$ so standard 64 bit ints suffice.

EDIT: Divisor found! $1590429889$

So first of all, the Wilson's theorem trick slows down instead of speeding up after $2p_1$. 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.

Answer 2

By Mertens' theorem, we have

$$\prod_{p < n} \left(1 - \frac{1}{n}\right) \sim \frac{e^{-\gamma}}{\log(n)},$$

In particular, if you do "trial division" of a large number $N \gg b^2$ for $a < p < b$ with $a$ and $b$ very large, you expect to fail to find a factor approximately

$$\prod_{a < p < b} \left(1 - \frac{1}{p} \right) \sim \frac{\log(a)}{\log(b)}$$

of the time. In your case, you have $a \sim 11!$. So, for example, to have a 50% chance of finding a factor, you would want to take $\log(b) \sim 2 \log(a)$, or $b \sim a^2$. For $b = 11!$, this would involve trial division to primes well over $10^{15}$, and in particular (estimating using the prime number theorem) more than 10 trillion primes. That seems a little unlikely to ever be possible.

Note that $11!$ is about $39$ million. If you wanted merely to check the next 10 million primes after $11!$ (involving taking $b$ around $230$ million or so), your chance of finding a factor would be less than 10%.

In particular, even if you speed up your computation of $(11!)! \pmod p$ for $p \sim 10^{10}$ to one second (on pari currently it seems to take around 13 seconds), it would then take 80 days to have a 10% chance of finding an answer.