Prove that $\gcd(a^2, b^2) = \gcd(a, b)^2$
Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$
"Open in Terminal" in Ubuntu 18.04 This option seems to have gone away after upgrading from Ubuntu 16.04. I tried installing nautilus-open-terminal but it is unable to locate the package for 18.04. Also trying to install nautilus-actions gives:  Package nautilus-actions is not available, but is referred to by another package. This may mean that the package is missing, has been obsoleted, or is only available from another source     How to fix this? How to mount an image file without root permission? Is your MAC address revealed when you hit up a website? When you hit up a site, such as cnn.com, does the TCP/IP packet contain the MAC address of your network card?  Could cnn.com theoretically record my MAC address? I installed the latest Ubuntu today, 15.04. Spotify has stopped working. How can it be made to work again?  The Unity bar flashes green 4 or 5 times and then stops flashing. There are no error messages. It was installed successfully and played well on 14.10 till the upgrade a few hours ago. It just does not appear on the screen. Suppose $X$ is a real-valued random variable and let $P_X$ denote the distribution of $X$. Then $$ E(|X-c|) = \int_\mathbb{R} |x-c| dP_X(x). $$  of $X$ are defined as any number $m \in \mathbb{R}$ such that $P(X \leq m) \geq \frac{1}{2}$ and $P(X \geq m) \geq \frac{1}{2}$.  Why do the medians solve $$ \min_{c \in \mathbb{R}} E(|X-c|) \, ? $$ The following is the residuals vs predicted scatter plot for a regression model with two IVs.   Initially, I thought it was evidence of heteroskedasticity. But, I reasoned that although there is a visible pattern in the plot, the variance across different levels of predicted values is same.  To clear my doubts, I saved the standardized residual and predicted values and ran a bi-variate correlation test. The correlation is zero, as expected. I am, nevertheless, intrigued by this observation. Any idea why I might have obtained this pattern?  PS:- My dependent variable is a sum of two likert-type items (Response scale: 1-5). So, it's theoretical range is 2-10 and it has no absolute zero value. What is the Convergence Radius and Domain of $\sum\limits_{n=0}^\infty \frac{1}{2^{n}} z^{2^n}$? So I need to find the Convergence Radius and Convergence Domain of the complex power series: $\sum\limits_{n=0}^\infty \frac{1}{2^{n}} z^{2^n}$  I've tried looking at the individual parts of the series and rewriting it somehow but couldn't find a way.  It can probably be solved, after a step or two, by the formula $R=\frac{1}{\displaystyle\limsup_{n\rightarrow \infty}\sqrt[n]{|a_n|}}$ .  Thanks. booting windows 10 spends too long Pie chart labels I want to create a pie chart with labels (one blank an one with percentage for the teacher, the mwe is the teachers version) so that my students have to fill the percentage in by themselves.  So far I created a pie chart using the code given in the answer here    and then I used labels from the code given in the answer here    Now I have this    The node in the violett part is the one I created manually and the other ones are implemented in the code, but I don't know what to change so that the automatic generated ones look like the one I created manually  My MWE      \documentclass[11pt]{article}   \usepackage{tikz} \usetikzlibrary{arrows}  \usepackage{xcolor}  \usepackage[utf8]{inputenc}  \definecolor{rosso}{RGB}{220,57,18} \definecolor{giallo}{RGB}{255,153,0} \definecolor{blu}{RGB}{102,140,217} \definecolor{verde}{RGB}{16,150,24} \definecolor{viola}{RGB}{153,0,153}  \makeatletter  \tikzstyle{chart}=[     legend label/.style={font={\scriptsize},anchor=west,align=left},     legend box/.style={rectangle, draw, minimum size=5pt},     axis/.style={black,semithick,-&gt;},     axis label/.style={anchor=east,font={\tiny}}, ]  \tikzstyle{pie chart}=[     chart,     slice/.style={line cap=round, line join=round, very thick,draw=white},     pie title/.style={font={\bf}},     slice type/.style 2 args={         ##1/.style={fill=##2},         values of ##1/.style={}     } ]  \pgfdeclarelayer{background} \pgfdeclarelayer{foreground} \pgfsetlayers{background,main,foreground}   \newcommand{\pie}[3][]{     \begin{scope}[#1]     \pgfmathsetmacro{\curA}{90}     \pgfmathsetmacro{\r}{1}     \def\c{(0,0)}     \node[pie title] at (90:1.3) {#2};     \foreach \v/\s in{#3}{         \pgfmathsetmacro{\deltaA}{\v/100*360}         \pgfmathsetmacro{\nextA}{\curA + \deltaA}         \pgfmathsetmacro{\midA}{(\curA+\nextA)/2}          \path[slice,\s] \c             -- +(\curA:\r)             arc (\curA:\nextA:\r)             -- cycle;         \pgfmathsetmacro{\d}{max((\deltaA * -(.5/50) + 1) ,1)}           \begin{pgfonlayer}{foreground}         \draw[*-,thin] \c -- (\midA:\r) node[above,near end, pos=\d,pie values,values of \s]{$\v\%$};         \end{pgfonlayer}               \global\let\curA\nextA     }     \end{scope} }  \newcommand{\legend}[2][]{     \begin{scope}[#1]     \path         \foreach \n/\s in {#2}             {                   ++(0,-10pt) node[\s,legend box] {} +(5pt,0) node[legend label] {\n}             }     ;     \end{scope} }   \begin{document}   \begin{tikzpicture} [     pie chart,     slice type={comet}{blu},     slice type={legno}{rosso},     slice type={coltello}{giallo},     slice type={sedia}{viola},     slice type={caffe}{verde},     pie values/.style={font={\small}},     scale=2 ]  \pie{Sommerjobs in Österreich}{52/comet,12/legno,20/sedia,16/coltello}   \legend[shift={(-1cm,-1cm)}]{{Eisverkäufer}/comet, {Schwimmbad}/legno, {Zettelverteiler}/coltello} \legend[shift={(1cm,-1cm)}]{{Kellner}/sedia}     \draw [*-,thin] (0.6,-0.2)  -- (0.7,-0.2 + 0.5ex) -- (2,0.5ex) node[above, near end]{20\%};    \end{tikzpicture}   \end{document} Why does HMAC use two different keys? Suppose $H$ is a hash function; why is $$H(k\mathbin\|H(k\mathbin\|m))$$ not secure?  See this . In there, indeed two keys are used and the mac algorithm is $$H(k_1\mathbin\|H(k_2\mathbin\|m)).$$ Why don't we use $$H(k\mathbin\|H(k\mathbin\|m)),$$ which has only one key? How to change the shadow color in Blender Cycles? I want to change the color of the shadow casting by a cube to green. While it's very easy to do it in Blender Internal (you select the lamp, go to the Lamp data header and change the shadow's color in the Shadow panel) in Cycles there're no such options. How to achieve it in Cycles? If we remove the square in the upper right corner of a $8\times 8$ chessboard.  The question is: Is it possible to cover the entire remaining area, with $1\times 3$ chocolate bars? (they can be laid on the chessboard vertically and horizontally aswell)  For the answer: We know that there will be $63$ squares left on the chessboard, of which $32$ are white and $31$ are black, since the one in the upper right corner was black. $63$ is divisible by $3$, so it could be possible to cover the area, but that doesn't prove that an arrangement exists to do it. I'm not sure how to continue and am seeking help. Thanks. Was Voldemort a Better Wizard Than Dumbledore? Pursuant to , I have a somewhat related question. The majority of Harry Potter readers automatically assume that Dumbledore must be a better wizard than Voldemort, because Dumbledore is &quot;good&quot; and Voldemort is &quot;evil.&quot; However, would Tom Riddle's prematurely developed skills demonstrate that Voldemort was inherently a better wizard than Dumbledore? In chapter 17, The Heir of Slytherin, Harry challenges Tom Riddle when he claims to be the greatest sorcerer in the world, retorting that Albus Dumbledore is the greatest sorcerer in the world. Harry reminds Tom Riddle that Voldemort is frightened of Dumbledore. Conversely, in Half-Blood Prince, Dumbledore himself states that Voldemort knows perhaps more magic than any other wizard alive. Yet, as Harry is J.K. Rowling's most direct voice, do we conclude that Dumbledore was, in fact, the superior wizard? Who was the better¹ wizard: Voldemort or Dumbledore? ¹Better meaning on a technically skilled level. Why does ice make such peculiar sounds? I've come across a couple of videos where some interesting sounds are produced using ice.  (Click on the images to see the video.)      Here, they drop a block of ice into a deep crevice and as the block falls, you can hear some strange sounds. These sounds remind me of some sound effects in old cartoons and sci-fi movies. The pitch of the sounds start high and end up low. Is this some kind of doppler effect?      Here, there's a man performing 'nordic skating', making the same sounds as he skates on top of a thin ice sheet. Maybe they are caused by the cracking ice. But why does that sound like this?    There are some other videos too, where they skip stones on frozen lakes.    Here the ice makes some high-frequency sounds which correspond to the chirping of birds.  What are these sounds and why do they sound the way they sound? Find full path of the Python interpreter? Equivalent of *Nix 'which' command in PowerShell? How do I ask PowerShell where something is?    For instance, "which notepad" and it returns the directory where the notepad.exe is run from according to the current paths. In &quot;Introduction to Mathematical logic&quot; by Elliot Mendelson , page $59$ . I couldn't understand the basic preliminary notations for the rigorous definition of Satisfiability. For example , at a line they said, &quot;Moreover, instead of talking about the $n$-tuples of objects that satisfy a wf that has $n$ free variables, it is much more convenient from a technical standpoint to deal uniformly with denumerable sequences. What we have in mind is that a denumerable sequence $s = (s_1, s_2 , s_3 , ... )$ is to be thought of as satisfying a wf $\beta$ that has ${x_j}_1 , {x_j}_2 ,  ... , {x_j}_n$ as free variables (where $j_1 &lt; j_2 &lt; ... &lt; j_n$ ) if the $n$-tuple $\langle {s_j}_1 , {s_j}_2 ,  ... , {s_j}_n \rangle$ satisfies $\mathscr B$ in the usual sense.&quot; I didn't quite understand why they used ${s_j}_1$ and ${x_j}_1$ instead of $s_1$ and $x_1$ . Maybe there is a critical detail I am missing out? Also , Another passage for the rigorous definition: &quot;Let $M$ be an interpretation of a language $\mathscr L$ and let $D$ be the domain of $M$. Let $\sum$ be the set of all denumerable sequences of elements of $D$. For a wf $\mathscr B$ of $\mathscr L$ , we shall define what it means for a sequences $s= (s_1, s_2, ... )$ in $\mathscr L$ to satisfy $\mathscr B$ in $M$. As a preliminary step, for a given s in $\sum$ we shall define a functions $s^*$ that assigns to each term $t$ of $\mathscr L$ an element $s^{*}(t)$ in $D$.&quot; They did say that for each $s = (s_1, s_2, ... )$ , there is a defined function $s^*$ , but even after further reading the text afterwards , I still did not understand how exactly $s^*$ is defined from $s$ . Can someone tell me where the gap in my understanding here is? Was Thanos really committed or just a hypocrite? The title is extremely vague, because this question contains HEAVY spoilers for Avengers: Infinity War, specifically the very last scenes of the movie. I can't reasonably put the entire question in spoiler tags, so consider this your warning.  At the end of the movie, Thanos pretty much wins. He gets all the Infinity Stones, Gamora and Vision have been brutally murdered, and the rest of the cast completely beaten. Nobody could stop him. So he executes his plan.  As far as I know (unless I missed something), his plan is to eradicate exactly half of all mortals in the universe, and from his own words, in a fair lottery (random chance), without taking into account a person's wealth, social status, or anything else. Pure chance decides whether you are sacrified for the greater good or live through this greater good (according to him).  During the movie, Thanos and one of his underling (Ebony Maw) say themselves that being sacrified is an honor and should be appreciated. As you would expect, most people do not appreciate being sacrified.  At the end of the movie, Thanos uses all 6 stones to do a finger snap, deleting half the universe.  The question is: Did he exclude himself from the targets? Was there a 50% chance that Thanos himself would disappear right there?  As far as I know, this is not explicitly answered in the movie itself. I am looking for similar situations in the comics (or in the movie if I happened to miss something) where you could know, or have an educated guess, as to Thanos being the kind of person who follows through to the end, even if it means dying, or the kind of person who gets scared of death when it comes down to it, and comes off as an hypocrite.  Some people also mentioned his "deal" with Strange would make him spare Tony from the snap too. I'm also wondering about that, if there is a link between the two, but it's not the focus of the question.  The reason I'm wondering is that Thanos seems to me like a very extreme utilitarist. Even though he might be wrong about the value of his plan (Do the remaining half really become happy?), in-universe it seemed successful at least once (Gamora's home planet), so his reasoning is not bad, even if the premise might be horribly flawed. Excluding himself from the snap would make it into a straight villain, while considering himself equal to the rest would make him into a wannabe hero with very very evil methods. This makes a huge difference in the interpretation of the movie and his character. (a.k.a. "god rays" and many other names) can be created in Blender Internal by checking the Halo checkbox in the lamp properties box.  How can I create a similar effect using Cycles? How do simulacra behave?