An exercise at the end of a chapter on primitive roots asked me to compute $1^k + 2^k + \ldots + (p-1)^k \pmod{p}$ for any positive integer $k$ and any prime $p$. Clearly, if $k$ is odd, the expression is congruent to $0$ modulo $p$. But how can I even approach this problem if $k$ is even? And how do primitive roots apply? Hints would be greatly appreciated.
I've been working through a bit of Ireland and Rosen's Number Theory for fun. Problem 4.11 of Ireland and Rosen asks     Prove that $1^k+2^k+\cdots+(p-1)^k\equiv 0\pmod{p}$ if $p-1\nmid k$, and $-1\pmod{p}$ if $p-1\mid k$.    My work is leading where I didn't expect to go. Here I assume $p$ id odd. First, if $p-1\mid k$, then we can wrike $k=(p-1)j$ for some $j$. But then $$ 1^k+\cdots+(p-1)^k\equiv 1^{(p-1)j}+\cdots+(p-1)^{(p-1)j}\equiv \underbrace{1+\cdots+1}_{p-1\text{ times}}=\frac{(p-1)p}{2}\equiv 0\pmod{p} $$ since $p-1$ is even.   Secondly, I suppose $p-1\nmid k$. WLOG, I can assume $0&lt;k&lt;p-1$, Since if $k=(p-1)j+r$ with $0&lt;r&lt;p-1$, then  $$ a^k\equiv a^{(p-1)j+r}\equiv a^r\pmod{p}. $$  Then if $g$ is a primitive root, $$ 1^k+\cdots+(p-1)^k=\sum_{i=1}^{p-1}(g^k)^i=\frac{g^k(1-g^{k(p-1)})}{1-g^k}. $$ Since $g$ is primitive, $p\nmid 1-g^k$, but $p\mid 1-g^{k(p-1)}$, so the above sum is again congruent to $0\pmod{p}$.  So in all cases, I get the sum is congruent to $0$ regardless. Have I messed up, or is there a typo? Thanks.
On really old Ubuntu 6.06 - How to upgrade “Airplane Mode” in OS X. Alternative command for `rfkill`? Mac OS X doesn't recognize rfkill from its terminal, and I am looking for an alternative command that has similar functionality (toggling "Airplane Mode").  NOTE: I don't wish to open an .app to toggle. I'd like to execute the command from command line or within script in Bash (or equivalents).    By "Airplane Mode", I meant to borrow the term from iOS support page . I would like to disable all incoming/outgoing network communications from my system. Can you be refused a visitor visa to US if you have bad Credit there? I did my 3 years of College in the US, 2010-2013 due to finance issues I was unable to pay my credit card bill of a certain bank, which still lie unpaid.  However I plan on visiting the US later this year; could I be denied visa for having an unpaid bill ? 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This problem is quite easily solved by using logarithms and derivative and forming the function $f(x,y) = x^y - y^x$, however there are assertions that this problem can be solved without using either of the two. I can not see how one would proceed to solve this problem in the absence of logarithmic manipulation and differentiation. Can anyone shed some light on this? Usage of "they" / "them" / "their" when the person's gender is not known nameref (hyperref) evaluating counter instead of section name What's the proper way to reference a section with a counter in the title? The TOC generates correctly, but the cross references aren't so lucky.  MWE:  \documentclass[letterpaper, 12pt, twoside]{exam}  \usepackage{hyperref}  \newcounter{KSRLessonNumberCounter} \setcounter{KSRLessonNumberCounter}{1} \newcommand{\KSRSection}[2][\value{KSRLessonNumberCounter}]{         \setcounter{KSRLessonNumberCounter}{#1}         \section*{Lesson \arabic{KSRLessonNumberCounter}: #2}         \addcontentsline{toc}{section}{Lesson \arabic{KSRLessonNumberCounter}: {#2}}         \stepcounter{KSRLessonNumberCounter} } \newcommand{\KSRSubsection}[1]{         \subsection*{#1}         \addcontentsline{toc}{subsection}{#1} } \newcommand{\KSRLessonNumber}{ }  \begin{document}  \KSRSection{Sample}  This is the first lesson. The second lesson is \nameref{sec:blah}.  \KSRSection{Blah} \label{sec:blah}  This is the second section  \KSRSection{Lorem Ipsum}  Dollars are Set in \nameref{sec:blah}.  \end{document} What happens if no one survives the Hunger Games? 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