# Application of Taylor's Theorem in Number Theory

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.

• no idea. Try it for $f(n) = an+b$ – Will Jagy May 15 '15 at 16:42
• 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. – Alfred Yerger May 15 '15 at 16:44
• I see how it works, I'll post an answer below give me a minute – 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).

So

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

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

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

• Nice idea, though the OP specifically asked how it follows from Taylor's Theorem. – Gregory Grant May 15 '15 at 17:27
• @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. – Bill Dubuque May 15 '15 at 18:00
• 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. – Gregory Grant May 15 '15 at 18:04
• @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. – Bill Dubuque May 15 '15 at 18:07
• 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. – Gregory Grant May 15 '15 at 18:10