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I calculated for primes $p=3,5, \dots ,47$ and for $v=3,4, \dots ,30$ the number $n=\frac{1+v^p}{1+v}$, and I noticed that only in 4 cases I found non-squarefree numbers n: (p,v)=(3,19),(3,23),(7,26),(11,28).

I tried to explain it by looking at the Möbius function $\mu(n)$, iterated exponents, and Carmichael's Function $\lambda(n)$. But none of these approaches seems to indicate what I'm looking for. The only property I can 'guess' is that if $g>p$ is a prime divisor of $n$, then $g \equiv 1 \mod(p)$.

My two questions are: (i) any idea why non-squarefree numbers seem rare, and (ii) what properties we can prove for the prime divisors $g>p$ of $n$?

Rolandb
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2 Answers2

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A couple of remarks: Since $p$ is odd, you might as well look at $\frac{v^{p}-1}{v-1},$ and allow $v$ to be negative as well as positive. As you noticed, any prime $q \neq p$ which divides this quantity is congruent to $1$ (mod $p$) in partial answer to your second question. This is because $v + q \mathbb{Z}$ has order $p$ in the multiplicative group of $\mathbb{Z}/q \mathbb{Z}.$ Also, $\frac{v^{p}-1}{v-1}$ is only divisible by $p$ when $v \equiv 1$ (mod $p$), and in that case, it is divisible by $p,$ but not by $p^{2}.$ That begins to answer your first question. To continue answering your first question, I find it necessary ( at least in my own way of thinking) to use some algebraic number theory. I can't give (and don't at the moment see) a totally convincing answer, but here's some slight justification. In rings of algebraic integers,we need to work with prime ideals, not prime numbers. In the ring $R = \mathbb{Z}[\omega],$ where $\omega = e^{\frac{2 \pi i}{p}}, $ we have $\frac{v^{p}-1}{v-1} = \prod_{j=1}^{p-1}(v - \omega^{j}).$ The only prime ideal of $R$ containing the rational prime number $p$ is the principal ideal $(1- \omega)R$. We already know that $\frac{v^{p}-1}{v-1}$ is not divisible by $p^{2},$ so we can ignore this. If $I$ is any other prime ideal of $R,$ then there are no values of $k \neq j$ such that $v-\omega^{j}$ and $v - \omega^{k}$ both lie in $I,$ for otherwise we would have $\omega^{j} - \omega^{k} \in I,$ and then we would easily find that $I = (1-\omega)R,$ which we have deal with. Now $I \cap \mathbb{Z} = q \mathbb{Z}$ for some rational prime number $q,$ and if $q^{2}$ is to divide $\frac{v^{p}-1}{v-1},$ we need $v - \omega^{j} \in I^{2}$ for some $j,$ and we might expect this to be rare.

I just thought of a reasonably elementary explanation. In order that $v + \mathbb{q^{2}\mathbb{Z}}$ should have multiplicative order $p$ in $\mathbb{Z}/q^{2}\mathbb{Z}$ ( this last ring is not a field, so we need to take care). But this multiplicative group is cyclic of order $q^{2}-q.$ However, only $q-1$ elements of this multiplicative group have order not divisible by $q$. So, loosely speaking, if we choose $v$ at random, the probability that its image has multiplicative order not divisible by $q$ is about $\frac{1}{q}.$ Since we know that $q \equiv 1 $ (mod $2p$), this means that the "probability" that $\frac{v^{p}-1}{v-1}$ is divisible by $q^{2},$ given that it is divisible by $q$, should be around $\frac{1}{q},$ so certainly at worst around $\frac{1}{2p+1}.$

Geoff Robinson
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Here are some asymptotic results in addition to Geoff Robinson's answer:

If $g$ is a prime with $g\mid \frac{1+v^p}{1+v}$ then $v^p \equiv -1 \bmod g$. There are three cases:

  • $p=g$: in this case, $\frac{1+v^p}{1+v}$ is not divisible by $p^2$

  • $g\neq p\not \mid g-1$. Then $p\neq 2$ and by consideration of the multiplicative group $(\Bbb Z/ g^n \Bbb Z)^\times \cong (\Bbb Z/g^{n-1}\Bbb Z)\times \Bbb Z/ (g-1)\Bbb Z$ we can conclude from $v^p\equiv -1 \bmod g^n$ that $v\equiv -1 \bmod g^n$. Hence, $v_g(v^p+1) = v_g(v+1)$ and thus $g \not \mid \frac{1+v^p}{1+v}$.

  • $p\mid g-1$

Consequently, the only case in which possibly $g^2 \mid \frac{1+v^p}{1+v}$ is $p \mid g-1$. In this case, the solutions have the following form: for each $n\geq 2$ there are exactly $p$ solutions $\bmod g^n$ to the equation $$v^p\equiv -1 \bmod g^n.$$ They can be lifted from the $p$ solutions to $v^p \equiv -1 \bmod g$ using Hensel's Lemma. Consequently, $$g^2 \mid \frac{1+v^p}{1+v} \Leftrightarrow g^2 \mid 1+v^p \text{ and } g \not \mid 1+v.$$

Therefore, for a fixed prime $p$, a pretty good approximation for the number of $f(v,p)$ with $v\leq N$ for which there is a prime $g$ with $g^2 \mid f(v,p)$ is given by $$N(p-1)\sum_{p\mid (g-1) \leq N, g \text{ prime}}\frac{1}{g^2}.$$

benh
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