In a paper about Zsigmondy's theorem, there is the following statement as a remark, without proof, nor reference:

Let $n, a $ be integers $(\gt1)$ and $q$ a prime divisor of $n=q^{i}.r$, such that $q$ does not divide $r$, let $b=a^{q^{i-1}}$, let $\Phi_n$ be the nth cyclotomic polynomial and $\phi$ the Euler totient function. Then $$\Phi_n(a)\ge\left(\frac{b^q+1}{b+1}\right)^{\phi(r)}$$

In the same paper, a weaker result: $$\Phi_n(a)\gt\left(\frac{b^q-1}{b+1}\right)^{\phi(r)}$$ is easily proved, thanks to a quite simple geometric argument, and used for a nice and quick proof that $$\Phi_n(a)\gt a^{\frac{\phi(n)}{2}}$$

I have tried, but failed, to prove the first inequality: this one seems to me much more difficult. Any help?

Edit: this where I am: for $q$ a prime, $b\gt1$, and $r$ not divisible by $q$ one has to demonstrate that $$\frac{\Phi_r(b^q)}{\Phi_r(b)}\ge\left(\frac{b^q+1}{b+1}\right)^{\phi(r)}$$ because the left side is $\Phi_n(a)$.

Actually, when $r=2$, since $\phi(2)=1$, this inequality is true because it is obviously an equality. Then I think one should try to show that, when $r\gt2$ $$\frac{\Phi_r(x)}{(x+1)^{\phi(r)}}$$ is a growing function of $x$ (from $x\ge1$) ? but i am stuck here.

One could also try to prove that it is decreasing from $1$, as $x$ varies from $0$ to $1$ and use the fact that this function is unchanged when $x$ is changed for $1/x$ (because $x^{\phi(r)}.\Phi_r(1/x)=\Phi_r(x)$). It must tend to $1$ as $x$ grows to infinity (as the ratio of two polynomials with same degree). Should also be minimal at $x=1$.


1 Answer 1


I'll venture a proof of the last claim in your edit. First, we recall the definition of the $r$th cyclotomic polynomial is the product of $\phi(r)$ terms of the form $x-\xi$ where $\xi$ is a primitive $r$th root of unity. Therefore we may write $$\frac{\Phi_r(x)}{(1+x)^{\phi(r}}=\prod_{\xi\in\text{ prim.}} \left|\frac{x-\xi}{x+1}\right|.$$ Here we have assumed $x>0$ and used the fact that primitive roots occur in conjugate pairs to eliminate the phases. (Note that if $-1$ is a primitive root then it cancels in this product and so we ignore it in the set of primitive roots; the $r=2$ case has this as the only factor.).

Now consider some particular $\xi$, and note that $|x-\xi|$ and $|x+1|$ represent the distances to $\xi$ and $-1$. Since $x$ is on the positive real axis and $\xi$ is on the unit circle, $x$ is always closer to $\xi$ than $-1$ and so the ratio is always less than 1. But as $x$ increases, their relative separations are becoming smaller and smaller and so the ratio increases towards 1 from below. Since this occurs for every such factor, we conclude that $\dfrac{\Phi_r(x)}{(1+x)^{\phi(r)}}$ is indeed an increasing function of $x$.

EDIT: To give the argument in the last paragraph more rigorously, consider a point $e^{i\phi}\neq -1$ on the unit circle. Then $$1-\left|\frac{x-e^{i\phi}}{x+1}\right|^2 =1-\frac{x^2-2x\cos \phi+1}{(x+1)^2} =\frac{2x}{(x+1)^2}(1+\cos \phi)\sim \frac{2}{x^2}(1+\cos\phi)$$ which is positive and decreasing with increasing $x$ as claimed above.

  • $\begingroup$ @ReneGy: let me know if I've misinterpreted your question or something isn't clear $\endgroup$ Jul 28, 2014 at 22:04
  • $\begingroup$ I like your proof because it is geometric, but it seems to me that at some point you may need to assume $x\ge1$, don't you? $\endgroup$
    – René Gy
    Jul 28, 2014 at 22:15
  • $\begingroup$ @RenéGy) Oh, certainly: you have to be past $1$ for the geometry to work. But that's exactly what the ratio does---it has a global minimum at $x=1$ and is concave up for positive $x$. So any proof that it's increasing had better assume $x\geq 1$. (Whether that's enough to prove what you need is a different question...) $\endgroup$ Jul 28, 2014 at 22:40

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