Would you please post your solution,my dear friend?I think it for more than 5 hours and no method.

I am wondering whether there is a typo in the problem statement: Perhaps q should also be prime? Or perhaps the feasible translations should be more restricted? Anyway, here is a solution to the current formulation: ========================================= Choose an arbitrary prime p of the form p=2004k+1 (with k>=2), and set q=p-2=2004k-1. Clearly: p is prime; p does not divide 2004; q is co-prime to p; q is co-prime to 2004. Consider the polynomial P(x)= (x-p+2)(x-p+1)(x-p+1002). Then P(q)=P(p-2)=0. And P(p)= 2*1*1002= 2004. There are infinitely many primes of the form 2004k+1, and the infinitely many corresponding polynomials are just shifts of each other. =========================================

The answer from gjw is correct as far as I know. From my point of view, remains only to prove that there are infinitely many primes of the form $ 2004k+1$ (if it is a well-known result, it is not well-known by me...). An alternative solution is taking $ P\left(x\right) =\left(\left(x-a\right)^{2}-2\right)\left(x-a-1\right)$, which is obviously equal to 0 for $ x = a+1$, and equal to $ \left(13^{2}-2\right)12 = 2004$ for $ x-a = 13$. Remains then "only" to prove that there are infinitely many prime numbers $ p = a+13$ such that $ q = p-12$ is prime with 167 (obviously $ p$ and $ q$ are coprime for infinitely many prime $ p$ since their common divider must divide 12, and obviously $ p-12$ is prime with $ 12 =\frac{2004}{167}$ for infinitely many primes $ p$). Assume that only a finite number of primes not congruent to 12 modulus 167 exists. Take the 166th power of each, and multiply their product by any prime that has remainder 12 modulus 167. Add one, and you find a number that is prime with all primes not congruent to 12 modulus 167, and that leaves remainder 13 when divided by 167. Or $ 13\equiv12^{a}\pmod{167}$ for some $ a$. If we show that no such $ a$ exists, we are done. We do so by finding that $ 2^{83}\equiv3^{83}\equiv1\pmod{167}$ but $ 13^{83}\equiv-1\pmod{167}$, or if $ 13\equiv12^{a}\pmod{167}$, then $ 13^{83}$ would leave simultaneously remainders $ 1$ and $ -1$ when divided by 167: $ 2^{9}= 512\equiv11\pmod{167}$, and $ 11^{3}= 1331\equiv-5\pmod167$ since $ 501 = 3\cdot167$, $ 1336 = 8\cdot167$. Then, $ 2^{81}\equiv11^{9}\equiv\left(-5\right)^{3}\equiv-125\pmod{167}$, or $ 2^{83}\equiv-500\equiv1\pmod{167}$. $ 3^{7}= 2187 = 13\cdot167+2^{4}$, or $ 3^{84}\equiv2^{48}\pmod{167}$. But $ 2^{12}= 4096 = 24\cdot167+88$, or $ 2^{48}\equiv2^{12}\cdot11^{4}\equiv2^{3}\cdot11^{5}\pmod{167}$. But $ 8\cdot121 = 968 = 1002-34$, or $ 3^{84}\equiv5\cdot34\equiv3\pmod{167}$, and $ 3^{83}\equiv1\pmod{167}$. Finally, since $ 13^{2}\equiv2\pmod{167}$, $ 13^{83}\equiv13\cdot2^{41}\equiv13\cdot32\left(2^{9}\right)^{4}\equiv13\cdot32\cdot11\cdot(-5)\pmod{167}$. Since $ -5\cdot32=-167+7$, and $ 7\cdot11\cdot13=1001=6\cdot167-1$, $ 13^{83}\equiv-1\pmod{167}$, and we are done.

daniel73 wrote: The answer from gjw is correct as far as I know. From my point of view, remains only to prove that there are infinitely many primes of the form $ 2004k + 1$ (if it is a well-known result, it is not well-known by me...). The statement "There are infinitely many primes of the form $ 2004k + 1$" is a special case of Dirichlet's theorem. See for instance: http://mathworld.wolfram.com/DirichletsTheorem.html If you know Dirichlet's theorem, you can also give a very short solution for IMO problem 1977/3: http://www.kalva.demon.co.uk/imo/imo77.html (Consider the product of four distinct primes of the form kn-1.)

For primes of type $ nk + 1$, see http://www.mathlinks.ro/viewtopic.php?t=126566 . For the IMO problem, see http://www.mathlinks.ro/viewtopic.php?t=61031 . For primes of type $ nk-1$ you can take a look at http://www.mathlinks.ro/viewtopic.php?t=100567 .