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In my previous questions it is shown that $f(x)=x^2+ax+a$ , where $a\in\mathbf{Z^+}$\ $\left \{ 4 \right \}$ is irreducible and that gcd$(f(1),f(2),f(3).....)=1$

So, according to Bunyakovsky conjecture this polynomial produces infinitely many primes for all $a\in\mathbf{Z^+}$\ $\left \{ 4 \right \}$

My question : Can we find such coefficient $a$ that $x^2+ax+a$ is composite for all natural x ?

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    Hm, I should learn my notation properly, sorry about that. Well unless my code's wrong I can't find any $a < 500000$ and almost all $a$ with a long chain of composites are even.2011-11-12
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    @Sp3000,what programming language you use...? May you post your code as answer ?2011-11-12
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    Bit of a quick hack in Python, but here: http://pastebin.com/0SVACHsa2011-11-12
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    @Andre,right...do yo know any quadratic counterexample to the Bunyakovsky conjecture ?2011-11-12
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    I know no counterexample. As far as I know, no one does, or at least no one did a day or two ago. The news would spread quickly!2011-11-12
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    So if I understand this correctly you are asking for a counterexample to the Bunyakowsky conjecture which has not beed found until now?2011-11-12
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    @Listing,if one finds such coefficient a, it may be considered as an counterexample to the Bunyakovsky conjecture2011-11-12
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    I don't see the point of the question. It is so close to the notorious unsolved Buniakowski problem that I don't see what is gained by asking it. You're not seriously expecting someone to post a counterexample to Buniakowski, so what are you hoping to get out of posting the question?2011-11-12
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    @Gerry,I have got exactly what I expected...source code that examines counterexamples..2011-11-12

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There is a good reason for the Bunyakovsky conjecture to be open, it is very hard to find a counterexample and even for polynomials that seem to be one, there are cases where you have to wait for very high $x$ until you encounter a prime. See for exampe the one given at wolfram: $x^{12}+488669$ it produces the first prime for $x=616980$, the first prime is therefore bigger than $10^{69}$.

Using GMP you can try if there might be a possible counterexample in your sequence:

#include 
#include 
using namespace std;

int Rabin_Miller(mpz_class n) { /* given an integer n >= 3 returns 0 composite, 1 probably prime */
    return mpz_probab_prime_p(n.get_mpz_t(), 10);
}

#define MAX_A 1000000
#define MAX_X 1000

int main() {
    mpz_class a;
    for(a = 4; a < MAX_A; ++a)
    {
        for(unsigned int x = 1; x <= MAX_X; ++x)
        {
            if(Rabin_Miller(x*x+a*x+a) > 0)
                x = MAX_X+1;    
            if(x == MAX_X)
                cout << "For a = " << a << " the first " << MAX_X << " values are composite!" << endl;
        }
    }
}

I used it to see (takes over 10 minutes to compute) that if there is a counterexample then $a>10.000.000$ (the above source is for 1 million, also note gmp uses the rabin miller test which only tells you if a number is probably prime but for prime numbers that are small like the ones in this program, it works correct [I think no example where it fails is known]).