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In Conrad's paper on Euclidean proofs of Dirichlet's theorem he states (on page 2),

note that the divisibility condition $p|h(n)$, for some $n$, is equivalent to $h \pmod p$ having a root.

This is not clear to me, maybe I am missing something?

Here $h(x)\in\mathbb{Z}[x]$, so I assume $h \pmod p$ is attained by reducing the coefficients modulo $p$?

He also doesn't state what kind of root $h \pmod p$ should have; of course it will have complex roots, so does $p|h(n)$ imply that $h \pmod p$ will have real roots/roots in $\mathbb{Z}$/roots in $\mathbb{Z_p}$?

Mee Seong Im
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Isabel
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  • what is $h(n)$? – Mastrem Jun 23 '17 at 12:20
  • @Mastrem in this statement, $h(x)$ is defined only as a polynomial with integer coefficients, and $n$ is an arbitrary integer, so $h(n)$ is a integer. As $p$ is a prime divisor of $h(n)$ you could write $h(n)=pd$ for some integer $d$. – Isabel Jun 23 '17 at 12:28
  • Have you studied any ring theory, e.g. polynomial rings and quotient rings? – Bill Dubuque Jun 23 '17 at 13:47

2 Answers2

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By definition $\ p\mid h(n)\iff \color{#c00}{h(n)\equiv 0}\pmod p$

The latter says that, $\bmod p\!:$ $\ \color{#c00}{n\ \text{is a root of }\ h(x)}.\,$ By Congruence Sum and Product Rules this remains true for any $\,n'\equiv n,\,$ and any $\,h'\equiv h\pmod{p},\,$ where the latter means they have congruent coefficients $\,h_i'\equiv h_i$ for all $i.\, $ In particular it's true when $\,h'_i := h_i\bmod p,\,$ i.e. for $\,h' := h\bmod p,\,$ the polynomial obtained by reducing $\!\bmod p$ all coefficients of $h.$

Bill Dubuque
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A friend and I think we may have come up with a solution to this question, if we assume that Conrad meant that $h(n)$ as a root in $\mathbb{Z_p}$;

$p|h(n)$ $\Rightarrow$ $h\mod p$ has root in $\mathbb{Z_p}$ :

For $h\in\mathbb{Z}[x]$, write $h'$ for $h$ reduced $\mod p$ (so $h' \in \mathbb{Z_p}$).

Then $h(n)=pd$, some $d\in\mathbb{Z}$, so $h'(n)=0$, i.e. $h'(x)=0$ for some $x \in \mathbb{Z_p}$, which is equivalent to saying $h'(x)$ has a root in $\mathbb{Z_p}$.

$h\mod p$ has root $\mathbb{Z_p}$ $\Rightarrow$ $p|h(n)$:

As $h'(x)$ has a root in $\mathbb{Z_p}$ we can write $h'(x)=(x-a)(a_s x^s +...+a_1 x+a_0)$, then for $n\in \mathbb{Z_p}$, $h'(a)=0$.

Now, $h(x)=(x-a)(a_s x^s +...+a_1 x+a_0)+(b_k x^k +...+b_1 x+b_0)$, where $p|b_i$ $\forall$ $0\leq i \leq k$.

So $h(n)=0+(b_k x^k +...+b_1 x+b_0)$ which is divisible by $p$.

Does this seem correct?

Isabel
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  • We have an integer polynomial $h(x) = \sum_{m=0}^d a_m x^m \in \mathbb{Z}[x]$. The statement $\exists n \in \mathbb{Z}, p | h(n) $ is equivalent to $\exists n \in \mathbb{Z}, h(n) \equiv 0 \bmod p$. But $h(n+p) \equiv h(n) \bmod p$ so this is equivalent to $\exists n \in \mathbb{Z}/p\mathbb{Z}, h(n) \equiv 0 \bmod p$ ie. $\overline{h}(x)= \sum_{m=0}^d a_m x^m \in (\mathbb{Z}/p\mathbb{Z})[x]$ has a root – reuns Jun 23 '17 at 13:23