It seems like there is no polynomial with finite variables known, which could generate all prime numbers, by integer assignments. Is there a proof that such polynomial can not exist and does anyone have one in his/her stack?
How to demonstrate that there is no all-prime generating polynomial with rational cofficents?
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number-theory
polynomials
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0Related: https://math.stackexchange.com/questions/59846/proof-of-no-prime-representing-polynomial-in-2-variables – 2016-10-24
1 Answers
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In fact there does not even exist a non-constant polynomial $f$ (I assume you want integer coefficients) which only takes prime values with integer inputs. It suffices to prove this for polynomials in one variable. By the hypothesis that $f$ is non-constant, it takes arbitrarily large values, so without loss of generality $|f(0)| > 1$; in particular, $f(0)$ is divisible by some prime $p$. Then $f(kp)$ is always divisible by $p$, hence cannot be prime for sufficiently large $k$.
However, remarkably there do exist polynomials in more than one variable all of whose positive values are prime.
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1@Gerry: oh, I see; rational is specified in the _title._ That's confusing. In any case, rational coefficients is not a problem: just ensure in addition that $k$ is divisible by the least common denominator of the coefficients. In your example, $f(4)$ is divisible by $2$. – 2012-09-11