Let $m$ and $n$ be natural numbers, where $n>m\geq 1$ and $p$ be an odd prime. Could we determine all natural numbers $j$ such that $p^n\mid ((p^{n-m}+1)^j-1)$?

  • $\begingroup$ Are you asking if this is true for all natural numbers $n,m,j$ and every prime $p\not=2$? $\endgroup$ – Bobson Dugnutt Sep 16 '16 at 20:42
  • $\begingroup$ @Lovsovs I think what he wants is to find all $j$ in function of $m$ and $n$ so that its frue for every $p>2$ $\endgroup$ – HeatTheIce Sep 16 '16 at 20:44
  • $\begingroup$ Dear Lovsovs, $m$, $n$ and $p$ are fixed and I want to find such $j$'s with this property. $\endgroup$ – sebastian Sep 16 '16 at 20:45
  • $\begingroup$ I am sorry! There is a mistake in my question. I edited it. $\endgroup$ – sebastian Sep 16 '16 at 20:53
  • $\begingroup$ Would the LTE Lemma help somehow? $\endgroup$ – HeatTheIce Sep 16 '16 at 21:06

I hope this is enough.

Using the LTE Lemma lets define $\alpha:=v_p((p^{n-m}+1)^j-1)\Leftrightarrow p^\alpha\|(p^{n-m}+1)^j-1$. So $p^\alpha$ exactly divides $(p^{n-m}+1)^j-1$. Using now the Lemma we get $\alpha=v_p((p^{n-m}+1)-1)+v_p(j)=v_p(p^{n-m})+v_p(j)=n-m+v_p(j)$. It is enough that $\alpha\ge n$, bcs then we know that $p^n$ will divide $(p^{n-m}+1)^j-1$. So $v_p(j)\ge m$, which means $j$ is independent of $n$. Now we just have that $j=p^m\cdot t, t\in\mathbb{N}$ (this follows from the fact that $j=p^b\cdot k, k,b\in\mathbb{N}$ and $k\ge m$ and $p\nmid k$ $\Rightarrow$$b=m+h,t=p^h\cdot k$).


Use the binomial theorem on $(p^{n-m}+1)^j$. You'll see that the 1's will cancel and then you can look at the exponent to determine which $j$'s satisfy division.

  • 2
    $\begingroup$ @ Bo Rel, of course I did it but I could not determine such $j$'s $\endgroup$ – sebastian Sep 16 '16 at 20:57

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