I'm working through Alan Baker's book A Concise Introduction to the Theory of Numbers, and there's an assertion in there that confuses me. Here's the quote:

It is easily seen that no polynomial $f(n)$ with integer coefficients can be prime for all $n$ in $\mathbb{N}$, or even for all sufficiently large $n$, unless $f$ is constant. Indeed, by Taylor's theorem, $f(mf(n)+n)$ is divisible by $f(n)$ for all $m$ in $\mathbb{N}$

How is this an application of Taylor's theorem? It's entirely mysterious to me. Thanks in advance for any insight on this.

  • $\begingroup$ no idea. Try it for $f(n) = an+b$ $\endgroup$ – Will Jagy May 15 '15 at 16:42
  • $\begingroup$ I don't see how Taylor's theorem is used, but I do see the result. Let me give an example. The polynomial $n^2+n+41$ gives primes for a while, but can't give primes for every $n$, since $n=41$ will make every term divisible by 41. If there is no constant term to use, just factor until one appears in one of the factors. $\endgroup$ – Alfred Yerger May 15 '15 at 16:44
  • $\begingroup$ I see how it works, I'll post an answer below give me a minute $\endgroup$ – Gregory Grant May 15 '15 at 16:46

By Tayor's Theorem

$$f(x) = f(n) + f'(n)(x-n) + \frac{f''(n)}{2!}(x-n)^2 + \cdots + \frac{f^{(k)}(n)}{k!}(x-n)^k + h_k(x)(x-n)^k$$

Now take $k$ big enough so that $h_k(x)=0$ (possible since $f$ is a polynomial).


$$f(x) = f(n) + f'(n)(x-n) + \frac{f''(n)}{2!}(x-n)^2 + \cdots + \frac{f^{(k)}(n)}{k!}(x-n)^k$$

Now plug in $mf(n)+n$ for $x$. You get an $f(n)$ in every term.

  • $\begingroup$ That's wonderful. Thank you very much! $\endgroup$ – G Tony Jacobs May 15 '15 at 17:02
  • $\begingroup$ @GTonyJacobs You're welcome thanks for posting the question, I was unaware of this factoid till now. $\endgroup$ – Gregory Grant May 15 '15 at 17:21

${\rm mod}\ \color{#c00}{f(n)}\!:\,\ f(m\color{#c00}{f(n)}+n)\equiv f(\color{#c00}0+n)\equiv 0\,\ $ by the Polynomial Congruence Rule $\ \ $ QED

Remark $\ $ To explain the relationship to Taylor's Theorem, the above amounts to using only the first couple of terms of the Taylor series, and this amounts to using the Factor Theorem, namely

$$\begin{align} f(x)\, &=\, f(n)\, +\, (x\!-\!n)(f'(n) +\, \cdots),\ \ \ \text{Taylor series at }\ x=n\\[4pt] \Rightarrow\quad\ \, f(x)\, &=\, f(n)\, +\, (x\!-\!n)\, g(x)\ \text{ for some }\ g(x)\in \Bbb Z[x]\\[4pt] \Rightarrow\ \ m f(n)\, &=\, \underbrace{x\!-\!n\mid f(x)-f(n)}_{\rm Factor\ Theorem}\,\Rightarrow\, f(n)\mid f(x)\ \ {\rm for}\ \ x = mf(n)\!+\!n\end{align}\qquad$$

But, of course, it is a bit overkill to use Taylor's Theorem to derive the Factor Theorem. And the Factor Theorem is a special case of the Polynomial Congruence Rule just as above, i.e.

${\rm mod}\ x\!-\!n\!:\,\ \color{#c00}{x\equiv n}\ \Rightarrow\, f(\color{#c00}x)\equiv f(\color{#c00}n)\,\ $ by the Polynomial Congruence Rule.

See here for further discussion of the Factor Theorem from a congruence standpoint.

  • $\begingroup$ Nice idea, though the OP specifically asked how it follows from Taylor's Theorem. $\endgroup$ – Gregory Grant May 15 '15 at 17:27
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    $\begingroup$ @Gregory Taylor's theorem and the Congruence Rules are closely related, e.g. see here. They can be considered equivalent from a more general perspective. For a course in elementary number theory it is simpler and more natural to use congruences as above. $\endgroup$ – Bill Dubuque May 15 '15 at 18:00
  • $\begingroup$ Thanks, that's pretty deep. But there's a trivial route from Taylor to the result at hand, so maybe the direct proof is more parsimonious in this context. $\endgroup$ – Gregory Grant May 15 '15 at 18:04
  • $\begingroup$ @Gregory I suspect it has more to do with the fact that the textbook author is an analytic (vs. algebraic) number theorist. I have seen many cases like this where analysts present analytic-inspired proofs that are way more complicated than the natural algebraic-based proof. $\endgroup$ – Bill Dubuque May 15 '15 at 18:07
  • $\begingroup$ Good point, there's definitely a cultural divide there. I think more algebraically by nature, but I respect analytical number theory a lot, I love to read the proofs, but I never understand how people come up with them. $\endgroup$ – Gregory Grant May 15 '15 at 18:10

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