1

The question is: "Calculate all polynomials P with integer coefficients s. t. for infinitely many integers $n$, $P(P(n)+n)$ is a prime."

After using google, I have found proof of contradiction for $P(n)$: Link

My questions are:

  1. Do you think that, is it sufficient to find a polynomial which does not fulfill the statement. Is there a famous polynomial for that, what I don't know? Sorry, I am not a mathematician.

  2. I posted a line to a contradiction. Can I also adapt that to solve my problem?

jester
  • 324
  • 4
    If the question is, find all polynomials with such and such a property, and you find one polynomial that doesn't have that property, then, no, that's not sufficient. It's like, if someone asked you to find all the primes, and you answered, "Four isn't prime", that wouldn't be a sufficient answer. – Gerry Myerson Oct 28 '20 at 22:51
  • Yes, I agree. Do you have a suggestion, how to start to solve this problem? – jester Oct 29 '20 at 11:46
  • 1
    We don't know many polynomials for which we can prove they are prime infinitely often. Basically, we know that if $\gcd(a,b)=1$ then there are infinitely many $n$ such that $an+b$ is prime; but if the degree of $f$ exceeds one, we can't prove it's prime infinitely often. So my suggestion would be to try to determine which linear $P(n)$ have the required property, and see whether it's the case that no $P$ of higher degree works. – Gerry Myerson Oct 29 '20 at 11:53
  • 1
    The link only shows that no nonconstant polynomial $P(n)$ can be prime for ALL $n$, but a polynomial $P(n)$ can still be prime for INFINITE MANY $n$. The other points were mentioned by Gerry. – Peter Oct 30 '20 at 08:43
  • So, have you looked at the linear case? – Gerry Myerson Oct 30 '20 at 12:13
  • I say, have you looked at the linear case??? – Gerry Myerson Nov 01 '20 at 00:35
  • https://en.wikipedia.org/wiki/Formula_for_primes#Prime_formulas_and_polynomial_functions - non-linear case: $n^2 - n + p$ where p must be a prime number... – jester Nov 01 '20 at 14:47
  • https://mathworld.wolfram.com/Prime-GeneratingPolynomial.html – jester Nov 01 '20 at 15:18
  • In 1752, Goldbach showed that no polynomial with integer coefficients can give a prime for all integer values (Nagell 1951, p. 65; Hardy and Wright 1979, pp. 18 and 22) – jester Nov 01 '20 at 17:35
  • Why are you so insistent on not seeing what happens if $P$ is linear? – Gerry Myerson Nov 02 '20 at 12:07
  • I don't find any linear case that works – jester Nov 02 '20 at 21:28
  • $P(n)=2*n+1$ and for the example $P(P(n)+n) = [4,7,10,13,16,19,22,25,28,31…] $ but as you said not all of them are prime. – jester Nov 02 '20 at 21:35
  • Your question doesn't ask for all of them to be prime, it asks for infinitely many of them to be prime. Infinitely many of the numbers $4,7,10,13,\dots$ are prime. – Gerry Myerson Nov 04 '20 at 00:45
  • So, for the fifth time, I suggest that you settle the linear case. – Gerry Myerson Nov 05 '20 at 12:20
  • Never mind, I did it for you. – Gerry Myerson Nov 08 '20 at 09:31
  • Any thoughts about my answer? – Gerry Myerson Nov 10 '20 at 11:43

1 Answers1

1

If $P(n)=an+b$, then $P(P(n)+1)=P(an+b+1)=a(an+b+1)+b=a^2n+ab+a+b$, so it will represent infinitely many primes if and only if $\gcd(a^2,ab+a+b)=1$, which is the same as $\gcd(a,b)=1$. This settles the case where $P$ is linear.

Gerry Myerson
  • 3,404