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Let $P(x)$ be a polynomial and $P(x)=a_n x^n+...+a_1 x +a_0$, with $a_0,a_1,...,a_n \in \{ 1,0,-1\}, a_n \neq 0$. Consider a sequence $S$ : $$P(0), P(1), ...,P(n)$$ Is it true that there are infinite prime numbers in the sequence $S$? If not, are there any conditions of $P(x)$ so that there are infinite prime numbers in the sequence $S$?

(Sorry, English is my second language)

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    The Bunyakowsky Conjecture would give us a good criterion for settling such questions, but not a single positive example (of degree $>1$) is known. Even a polynomial like $x^2+1$ seems too hard to handle with present methods. – lulu Nov 01 '18 at 13:54
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    To be sure, your statement is too broad. $P(x)=x^2-1$ clearly doesn't take many prime values, for example. You certainly want the polynomial to be irreducible. – lulu Nov 01 '18 at 13:56
  • Along the lines of polynomials having factors, it is important to consider that for example $P(x)=x^2-x+8$ has a factor $2$ for every integer $x$, despite having no linear polynomial factors. One way to get around “invisible constant divisor” problems like this in this context is to write the polynomial in “binomial coefficient” form, e.g., $P(x)=b_0+b_1\binom x1+b_2\binom x2\dots$. – abiessu Nov 01 '18 at 14:06
  • Just a curiosity: Euler found that $n^2+n+41$ gives distinct primes for the 40 consecutive integers ($n$=0 to 39). This page has a lot of interesting facts: http://mathworld.wolfram.com/Prime-GeneratingPolynomial.html – Saša Nov 01 '18 at 14:27

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