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Let $f(x)$ denote a polynomial of degree at least 1 with integer coefficients and positive leading coefficient.

(i) Show that if $f(x_0)=m>0$, then $f(x)\equiv 0 \pmod{m}$ whenever $x\equiv x_0\pmod{m}$.

(ii) Show that there are infinitely many $x\in\mathbb{N}$ such that $f(x)$ is not prime.

Proof. (i) Since $f(x_0)=m$ we certainly have $f(x_0)\equiv 0\pmod m$, so it will be sufficient to show that $f(x)\equiv f(x_0)\pmod m$, if $x\equiv x_0\pmod m$. So, suppose $x\equiv x_0\pmod m$, then we know that, \begin{align} x^k&\equiv x_0^k\;\;\;\;\;\;\;\;\pmod m\\ 0&\equiv x_0^k-x^k\pmod m, \;\;\;\;\;\;\;(1) \end{align} where $k$ is a natural number. Now let, \begin{align*} f(x_0)&=a_nx_0^n+\cdot\cdot\cdot+a_1x_0+a_0=m\\ f(x)&=a_nx^n+\cdot\cdot\cdot+a_1x+a_0. \end{align*} Notice that, \begin{align*} f(x_0)-f(x)&=a_nx_0^n+\cdot\cdot\cdot+a_1x_0+a_0-(a_nx^n+\cdot\cdot\cdot+a_1x+a_0)\\ &=a_n(x_0^n-x^n)+\cdot\cdot\cdot+a_1(x_0-x)+a_0-a_0. \end{align*} From equation (1) we know that, \begin{align*} a_n(x_0^n-x^n)+\cdot\cdot\cdot+a_1(x_0-x)+a_0-a_0&\equiv a_n(0)+\cdot\cdot\cdot a_1(0)+0\pmod m\\ &\equiv 0 \;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\pmod m. \end{align*} Hence $f(x)\equiv f(x_0)\pmod m$ whenever $x\equiv x_0\pmod m$, as asserted.

(ii) There are infinitely natural numbers $x$ such that $x\equiv x_0\pmod m$, and from part (i) we know that if $x\equiv x_0 \pmod m$ then $f(x)\equiv 0\pmod m$, or in other words $m|f(x)$ whenever $x$ is congruent to $x_0$ modulo $m$. Since $m|f(x)$ it cannot be prime and hence there are infinitely many natural numbers $x$ such that $f(x)$ is not prime. Q.E.D.

Bryce Smith
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  • Your argument seems to implicitly presume that $\,f(x_0 + mk),\ k\in\Bbb Z\,$ takes infinitely many values. You need to prove this and make that inference more precise. – Bill Dubuque Feb 02 '20 at 20:04
  • Above is for (ii). And (i) is just a special case of the [Polynomial Congruence Rule](https://math.stackexchange.com/a/879262/242), i.e. $\ x\equiv x_0\,\Rightarrow\, f(x)\equiv f(x_0)\ [\equiv 0]\ \ \ $ – Bill Dubuque Feb 02 '20 at 20:08
  • @BillDubuque I wasn't sure if I should prove that since it seems trivial. The arithmetic progression $x_0+mk$ is infinite and each term is an integer so $f(x_0+mk)$ is defined over the entire progression. – Bryce Smith Feb 02 '20 at 21:46
  • "It seems trivial" is not a rigorous proof. Such claims are the point of failure of many arguments. – Bill Dubuque Feb 02 '20 at 21:54
  • @BillDubuque Do you have any suggests as to how to prove that rigorously then because I’m kind of at a loss. – Bryce Smith Feb 02 '20 at 23:35
  • As [here](https://math.stackexchange.com/a/817110/242) you could use that a nonzero polynomial over field has finitely many roots, or you could use that the polynomial's values are unbounded, etc. – Bill Dubuque Feb 02 '20 at 23:50
  • @BillDubuque Ah that makes sense, thank you so much – Bryce Smith Feb 02 '20 at 23:53

2 Answers2

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Your proof of part (ii) needs improving because there is nothing in what you have written that would prevent $m$ being, say, $1$. You could use the following.

$f$ has degree at least $1$ and a positive leading coefficient and therefore $f(x)$ tends to $+\infty$ as $x$ tends to $+\infty$. Choose an $x_0$ such that $f(x_0)=m>0$.

For positive integers $k$, all $f(x_0+km)$ are divisible by $m$ and they tend to $+\infty$ as $k$ tends to $+\infty$. Therefore infinitely many of them are a non-trivial multiple of $m$ and are therefore not prime.

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It's been a long time since I have doen anything with modular-aritmetic, but it looks good to me. My only concern is that in the beginning you prefice that $f(x)$ has a positive leading coëfficient, which is not used agiain. It seems like you have proved a more general case then first set out to be.

9sven6
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  • Would it then be good to say that $a_n$ is positive, I don't really think that would change the proof at all. – Bryce Smith Feb 02 '20 at 20:02
  • @DavidOrndorf The question is more, why is it in the assignment if you don't use it in your proof? Does your proof work with an example function with $a_n = -1$? These are the things to try. – 9sven6 Feb 02 '20 at 20:07