6

This question raised from discussions around my previous question. This may seem trivial or easy, but I am so confused and can't see the answer. So I will be so grateful if you would help me please.

For the natural number $n$ and prime number $p$, it is likely possible that there exists a polynomial $f(x)\in\mathbb{Z}[x]$ of the form $f(x)=\prod_{i=1}^{m}(x^{\lambda_i}-1)$ such that $deg(f)>n$ and $f(p)\mid p^n-1$. For example, take $p=3$, $n=2$ and $f(x)=(x-1)^3$ we have $(3^1-1)(3^1-1)(3^1-1)\mid 3^2-1$. However, with assumptions $p\geq 3$ and $n\geq 4$, I could not find any such polynomial $f$ with $\deg(f)>n$ and $f(p)\mid p^n-1$. So it made me to claim that the following statement might be true:

Let $p\geq 3$ be a prime number and $n\geq 4$. Then for every $f(x)\in\mathbb{Z}[x]$ of the form $f(x)=\prod_{i=1}^{m}(x^{\lambda_i}-1)$ such that $\deg(f)>n$, we have $f(p)\not\mid p^n-1$.

Is the above assertion true? If yes, would you please hint me how to prove it (or refer me to a reference)?
Thank you in advance.

Community
  • 1
Math_Student
  • 123
  • 1
    Isn't $f(p)$ simply too large a number to be a factor of $p^n-1$ in most cases? Bernoulli's inequality implies that $f(p)\ge(p-1)^{n+1}>p^n-1$ when $p>n+1$. And that is not a tight estimate. Anyway, $p$ needs to be small, so if you rule out small cases, then it probably will work, but also may not be as interesting as the case $\deg f=n$. – Jyrki Lahtonen Aug 17 '16 at 21:43
  • @JyrkiLahtonen From Bernouli's inequality we have $((p-2)+1)^{n+1}>1+(n+1)(p-2)$. I didn't get that how the assumption $p>n+1$ will imply that $(p-1)^{n+1}>p^n-1$. Would you please explain it a bit more? – Math_Student Aug 17 '16 at 22:36
  • 1
    Using Bernoulli I got $$\frac{(p-1)^{n+1}}{p^n}=(p-1)(1-\frac1p)^n\ge(p-1)(1-\frac np).$$ If $p>n+1$ then $1-\dfrac np\ge\dfrac2p$, and the right hand side of the displayed inequality will be $>1$ implying $(p-1)^{n+1}>p^n$. The moral: Bernoulli gives more useful information when the base is close to $1$. Anyway, for large enough values of the input the value of a given degree $n+1$ polynomial will be larger than a given degree $n$ polynomial, and it is only a matter of determining exactly what "large enough" means with the given pair of polynomials. – Jyrki Lahtonen Aug 18 '16 at 05:49
  • @JyrkiLahtonen Thank you for explanation. Just another point: What can be said when $degf$=n? Can we find a tight estimate for $p$ which ensures us that $f(p)∤p^n−1$ when $degf=n$? For example I noticed that if $p>2n$, then $f(p)\geq(p-1)^n>p^n/2>p^n-1/2$ (In fact I used a similar argument as you mentioned) . This implies that $f(p)$ can not be a divisor of $p^n-1$. But Is there any other estimate for $p$ which is more tight than $p>2n$ when $degf=n$? – Math_Student Aug 18 '16 at 09:34

0 Answers0