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This problem is taken from "Mathematics for Computer Science" (Lehman, Leighton, Meyer, 2018).

Problem

For $n = 40$, the value of the polynomial $p(n) := n^2 + n + 41$ is not prime, as noted in Section 1.1. But we could have predicted based on general principles that no non-constant polynomial can generate only prime numbers.

In particular, let $q(n)$ be a polynomial with integer coefficients, and let $c:=q(0)$ be the constant term of $q$.

(a) Verify that $q(cm)$ is a multiple of $c$ for all $m \in \mathbb{Z}$.

(b) Show that if $q$ is nonconstant and $c > 1$, then as $n$ ranges over the nonnegative integers $\mathbb{N}$ there are infinitely many $q(n) \in \mathbb{Z}$ that are not primes. Hint: You may assume the familiar fact that the magnitude of any nonconstant polynomial $q(n)$ grows unboundedly as $n$ grows.

(c) Conclude that for every nonconstant polynomial $q$ there must be an $n \in \mathbb{N}$ such that $q(n)$ is not prime. Hint: Only one easy case remains.

Solution attempt

(a) The polynomial can be expressed as $q(n) = c + a_1n + a_2n^2 + \cdots + a_kn^k$. So, $q(cm) = c + a_1cm + a_2c^2m^2 + ... + a_kc^km^k$. Since all terms of $q(cm)$ are divisible by $c$, $q(cm)$ is a multiple of $c$ for all $m \in \mathbb{Z}$.

(b) As $n$ ranges over the nonnegative integers, it will range over infinitely many values of the form $n=cm$ ($m \in \mathbb{Z}$). As shown in (a), for each $n=cm$, $q(cm)$ is a multiple of $c$. Therefore, assuming that the magnitude of $q(n)$ grows unboundedly as $n$ grows, this means that $q(n)$ will take infinitely many non-prime values.

(c) Item (b) covered the cases where $c > 1$. For nonconstant $q$, two cases remain: $c < -1$ and $-1 \leq c \leq 1$.

  • For $c < -1$, a similar argument to (b) applies: as $n$ ranges over the negative integers, it will range over infinitely many values of the form $n=cm$ (where $m$ is a negative integer). For each of these values, $q(n)$ is a multiple of $c$. Therefore, assuming that the magnitude of $q(n)$ grows unboundedly as $n$ grows, this means that $q(n)$ will take infinitely many non-prime values.

  • For $c -1 \leq c \leq 1$, $q(0) = c$ is an example of root that is not prime.

Therefore, for every nonconstant polynomial $q$, there must be an $n \in \mathbb{N}$ such that $q(n)$ is not prime.

Is this proof correct?

Thank you in advance.

Community
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    You handle the cases $c=\pm 1$ by arguing that $\pm 1 =p(0)$ is not prime. That's ok, but in discussing this result it is more common to prove that $p(n)$ takes on infinitely many composite values. After all, a polynomial that took on only prime values for sufficiently large $n$ would be almost as useful as one which only took on prime values. You've almost done that, but you still have to show this in the case $c=\pm 1$. – lulu Feb 19 '20 at 12:58
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    this duplicate and others you can easily find by searching might give you ideas if you get stuck on that last detail. – lulu Feb 19 '20 at 13:00
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    A different approach. If $q(x)$ produces only prime numbers then $q(x+1)$, which can be written $q(x+1)=xp(x)+q(1)$ for some $p(x)$, produces only prime numbers. Compute $q(nq(1)+1)$ and use the fact that $q(1)$ is a prime to conclude. – jijijojo Feb 19 '20 at 16:43
  • @lulu isn't it mentioned in part C that n belongs to only set of natural numbers? How can we put n=0 there? – Alex Aug 20 '21 at 06:17
  • @Alex Many people, including (apparently) the OP, say that $0\in \mathbb N$. But, really, the important point here is that the goal is to show that $p(n)$ is composite for infinitely many natural numbers...not to just find a single composite $p(n)$. – lulu Aug 20 '21 at 10:44

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I think your proof is fine. As the commenters say, there are other proofs for it. But it seems clear that the the scheme that the book set up intended for you to find the "easy case" that you found.