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I recently learned about The polynomial $p(n)=n^2-n+41$, and how for $1\le n\le40, n\in\mathbb Z$, $p(n)$ is prime. I understand that primes are very difficult to find, so from that I can conclude that we haven't found a polynomial $q(m)$ where $m\in\mathbb N$ with no upper bound is prime, but is it proven that no such polynomial exists?

Jacob Claassen
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  • what if the constant is $0$? – Jacob Claassen Jul 08 '17 at 02:26
  • See also here and here. – Bill Dubuque Jul 08 '17 at 02:27
  • As I said under my answer, when it got talked about it doesn't work for constant polynomials. –  Jul 08 '17 at 02:27
  • Actually - I'm going to retract my comment. I assumed the coefficients were integers and I missed a few special cases. So none of what I've said applies. – Shuri2060 Jul 08 '17 at 02:28
  • The title asks about polynomials which "return" only prime numbers. The body seems to be asking a different question, about "a polynomial $q(m)$ where $m\in\mathbb N$ with no upper bound is prime", i.e., whether there are polynomials which "return" (to use your jargon) infinitely many prime values; there are probably lots of those, but the only proven examples are polynomials of degree one. – bof Jul 08 '17 at 02:41
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    I think the questioned asked is entirely different than the title and the cited previous answer. The title asks for a polynomial that only returns primes. But the question asked asks for a polynomial with no upper limit of prime values. (Which if such existed would NOT help the discovery of primes in any way). – fleablood Jul 08 '17 at 02:41

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No. This is because of things like, $f(c)-f(0)= ac^n+\ldots+c$ has all terms divisible by c. In general, $f(f(d)+d)-f(d)$ will divide by f(d). Even more generally $f(k\cdot f(d)+d)$ should also divide by d.

  • That doesn't show $f(n)$ isn't prime? – Shuri2060 Jul 08 '17 at 02:20
  • @Shuri2060 okay you got me. It does for any polynomial not repeating values though. –  Jul 08 '17 at 02:21
  • Assuming $f(c)\ne c$. this is possible in constant polynomials though. –  Jul 08 '17 at 02:25
  • Let's simplify the answer. We have n^2 -n +41 =(n-7)(n+6)+42+41=(n-7)(n+6) +83. 83 is prime and n-7 may be divisible by is ,so is x+6 with especial values of x. So this polynomial does not always give primes. – sirous Jul 08 '17 at 10:55
  • We, are talking about a general topic, not the specific polynomial. –  Jul 08 '17 at 17:27