Show that if $\gcd(a,b)=1$ and $p$ is an odd prime, then ${\gcd(a+b,}\frac{a^p +b^p}{a+b}$$) = 1$ or $p$     Sorry about the duplicate  In another answer, however, the sum $\sum\limits_{k=0}^{n-1} (-1)^{k}a^{n-1-k}b^{k}$ was expressed as $\left(\sum\limits_{k=0}^{n-2} (-1)^{k}(k+1)a^{n-2-k}b^{k}\right)(a+b) + (-1)^{n-1}nb^{n-1}$ How was it done?
Let $p$ be prime number ($p\gt2$) and $a,b\in\mathbb Z$ ,$a+b\neq0$ ,$\gcd(a,b)=1$ how to  prove that $$\gcd\left(a+b,\frac{a^p+b^p}{a+b}\right)=1~~\text{or}~~ p$$  Thanks in advance .
If I have a Type C visa from the Norwegian embassy valid for February 9-28, and my Type D national visa from the German embassy valid from March 1 - August 31, 2017, do I need to leave/fly out of the Schengen area on the 28th and fly back on the 1st day of March? Closing as "off topic" has just gotten painful I know that useability and friendliness are given priority in the features implemented on StackExchange, but after using the new close-as-off-topic flow for the first time I would have to say I'm somewhat underwhelmed.  Why do we not have a simple It's off topic type reason? And why do I have to write half a book chapter as my custom close reason, especially when I can go back and edit the automatically posted comment so that it's way shorter?   This question complaint is prompted by  here on Meta. It's off topic, plain and simple. We now have a bunch of people wanting to close it, all potentially writing their own custom close reasons, it's just messy suboptimal. Regardless of how nice and friendly you want to be to the user, this particular question shouldn't need an explanation.    Does the off topic flow have to stay this way? Can we allow higher rep users to close as off topic without having to specify a (really long) custom reason, maybe have some canned reasons like we used to?  Edit: just to be clear, there are still three canned close reasons, but none of them were applicable to this example. The second one is closest, but does not appear to seek input... is quite different to It's a question, but doesn't belong here in any way, shape, or form. If $(1), (log_yx), (log_zy), (-15log_xz)$ are in Arithmetic Progression... If $(1), (log_yx), (log_zy), (-15log_xz)$ are in AP then...   (A)$z^3=x$   (B)$x=y^{-1}$   (C)$z^{-3}=y$   (D)$x=y^{-1}=z^3$   (Multiple answers may be correct)   I tried many approaches, one of them being: $1+log_zy=2log_yx$ $log_zz+log_zy=log_yx^2$ $log_z(yz)=log_yx^2$ But the fact that the bases are different is causing the main problem as I don't know how to proceed further. It would be great if anybody could give me a hint for solving such questions. Let $A =$ {$(x,y,z) \in \mathbb{R}^3 |$  $x,y,z \geq 0$ and $x+y+z \leq 1$}.   Compute the integral $$\int_{A} x^m y^n z^p (1-x-y-z)^q dx dy dz$$  using new variables $u = x+y+z$, $v=\frac{y+z}{x+y+z}$, $w=\frac{z}{y+z}$.  Also, although it was not stated in the problem, I'm assuming that $m,n,p,q \in \mathbb{Z}$.   Since $f(x,y,z)=(x+y+z, \frac{y+z}{x+y+z}, \frac{z}{y+z}) = (u,v,w)$, $f^{-1}(u,v,w) = (u(1-v), uv(1-w), uvw)$  Thus J $= \begin{vmatrix} 1-v &amp; -u &amp; 0\\  v(1-w) &amp; u(1-w) &amp; -uv\\  vw &amp; uw &amp; uv\\ \end{vmatrix} = u^2v$  Skipping the part where I find the new limits of integration, and skipping some algebraic simplification we have:  $$\int_{A} x^m y^n z^p (1-x-y-z)^q dx dy dz$$   $$= \int_{0}^1 \int_{0}^1 \int_{0}^1 u^{m+n+p+2}(1-u)^q v^{n+p+1}(1-v)^mw^p(1-w)^n du dv dw$$  However, I have no  idea where to go from here. Any help would be much appreciated. For x < 5 what is the greatest value of x It can't be $5$. And it can't be $4.\overline{9}$ because that equals $5$. It looks like there is no solution... but surely there must be? How can I find hard links on Windows? I've created some hard links on my Windows 7 file system using mklink. It was some time ago and I can't remember for sure where, or which files. When I use Explorer, all files look the same. When I use the command line and type "dir", they all look the same.  How do I find hard links? Or how do I determine whether a specific file IS a hard link? ceiling fan from a switch with only black and red wires? The current switch controls an outlet in the corner of the room for a lamp. I pulled it out and noticed three sets of individual wires: two set of wires coming up into the box from the bottom, and one set with a red wire coming down into box.  All the white and grounding wires are capped.  Also all the black wires are capped; one set I’m assuming is the hot since I see it contains a red wire. On the back of the switch, the top two ports have black wires and the bottom right has the one red wire.  There are two black pigtail wires coming off the black capped wires. I tried to connect a separate switch to control a ceiling fan with remote. I ran new wire making the old switch box into a two switch. I ran 14/2 gauge wire to ceiling for fan / light remote combo. Fan didn’t work after multiple re-wiring  combinations. The breaker did pop and half click when it popped. Then I thought to re-wire the ceiling fan wires as if I was going to make it into outlet only and use a tester to indicate an open ground, or a wiring error but the two light on tester are both orange indicating correct wiring. I heard that sometimes the ceiling fan remote transformer goes bad. Should I check batteries and/or unit placement? I’m really confused at this point why the light is not working at least with remote after follow wiring installation instructions that were included with the fan. I think it’s my wiring with the switch. The way I left the wiring is that it has constant power. As if we’re a outlet the current switch does not turn it off. Proving that $\sum\limits_{n = 0}^{2013} a_n z^n \neq 0$ if $a_0 > a_1 > \dots > a_{2013} > 0$ and $|z| \leq 1$ I'm going to teach a preparation course for the complex analysis qualifying exam from my university (which basically consists of me doing some problems from past exams) and I'm trying to solve some questions from previous exams. One of the questions is the following:           Prove that $\displaystyle{\sum_{n = 0}^{2013} a_n z^n \neq 0}$ if the coefficients  satisfy $a_0 &gt; a_1 &gt; \dots &gt; a_{2013} &gt; 0$ and $|z| \leq 1$.      I tried to approach the problem by using the reverse triangle inequality to try to isolate the leading term, which I thought should be the constant coefficient, as follows:  \begin{array} . \left|  \sum_{n = 0}^{2013} a_n z^n \right| &amp;= \left| a_0 +  \sum_{n = 1}^{2013} a_n z^n \right| \\ &amp;\geq |a_0| - \left|  \sum_{n = 1}^{2013} a_n z^n \right|\\ &amp;\geq |a_0| -  \sum_{n = 1}^{2013} |a_n z^n|\\ &amp;\geq a_0 -  \sum_{n = 1}^{2013} a_n |z^n|\\ &amp;\geq a_0 -  \sum_{n = 1}^{2013} a_0 |z^n|\\ &amp;= a_0\left( 1  -  \sum_{n = 1}^{2013} |z^n| \right) \end{array}  Then for $|z| &lt; 1$, we have that this last expression is equal to  $$ a_0\left( 1  -  |z| \frac{|z|^{2013} - 1}{|z| - 1} \right) $$  and my hope was to prove that this was always positive, nevertheless, after failing to prove it, I plotted the expression inside the parenthesis as a function of $|z|$, and found out that it changes sign in the interval $(0, 1)$.  Question  Can my approach somehow be made to work? And if not, how can I prove that the sum $\displaystyle{\sum_{n = 0}^{2013} a_n z^n \neq 0}$ under the given conditions?  Thank you very much for any help. I have a polygon shape file that has many overlapping features. Each feature represents an area used by polar bears. So some of the polygons overlap and some don't, and there are separate clusters of overlapping areas.   I would like to get a measure of how much overlap there is in a particular area. But I have no clue how to do this. I'm pretty new to GIS and am still learning the ropes, it just seems I can't find an answer. Dividing polynomial $f(x)$ by $x-3$ and $x+6$ leaves respective remainders $7$ and $22$. What's the remainder upon dividing by $(x-3)(x+6)$? I am trying to create a schema page in LaTeX. It shall look like this.   How can I create schema code which look like this image? There should be also a click to go function. When the user clicks on the user_schema link he should be redirected to that page.  What classes I can use for this? Any suggestions would be appreciated.    \documentclass{article} \usepackage[utf8]{inputenc}  \begin{document} \maketitle  \section{User schema} \mySchema{Users} {     _id: ObjectId - Id do registro     name: String - Nome do usuário     date: Date - Data de Cadastro     ren_date: Date - Data de renovação do Cadastro     email: String - Lorem ipsum dolorsitamet, consecteturadipisicingelit, sed do eiusmodtemporincididunt ut labore et dolore magna aliqua. Ut enim ad minimveniam, exercitationullamcolaborisnisi ut aliquipexeacommodoconsequat. Duisautereprehenderit in voluptatevelit esse cillumdolore eu fugiatnullapariatur. Occaecatcupidatat non proident, sunt in culpa quiofficiadeseruntmollit.     access_log: Array [          {              date: String - Data da ocorrência.             user: ObjectId - Usuário que executou.             action: String - Ação Executada.         }         {              date: String - Data da ocorrência.             user: ObjectId - Usuário que executou.             action: String - Ação Executada.         }     ]     \link{group}{group_id}: ObjectId - Id do grupo. } \end{document}   Here is the image how should it appear: Why isn't there a potential difference across a disconnected diode? I know this question sounds silly, as if there was a potential difference a current would be created when the terminals are connected together and this would mean energy has come from somewhere.  The reason I ask this though is that from my understanding of the depletion region and built in potential of a diode it seems like if you connected a voltmeter across the whole diode it would show the value of the built in potential.  This is explained in the image below:    At first, electrons flow from the n type to the p type because there are a higher concentration in the n type, and holes do vise versa. This is called the diffusion current. The first electrons and holes to cross the pn boundary are the ones which are closest to it; these carriers recombine when they meet each other and are then no longer a carrier. This means there is a depletion region of no carriers near the pn boundary. because electrons have left the n type material, and holes have left the p type material, there is a surplus of positive and negative charge on the n and p side of the pn boundary respectively. This causes an electric field that opposes the diffusion current, and so no more electrons or holes cross the boundary and combine. In short, only the electrons and holes near the boundary combine, because after they have done that an electric field is formed that prevents any more carriers from crossing. The current due to this electric field is called drift current, and when in equilibrium this will equal the diffusion current. Because there is an electric field at the boundary (pointing from the positive charge to the negative charge) there is an associated voltage. This is called the built in potential.  If you sample the electric field at each point along the diode from left to right, you would start with 0 in the p region because there are an equal number of protons and electrons. As you approach the depletion region you would see a small electric field pointing back towards the p region, caused by acceptor impurities which now have an extra electron (due to recombination) and therefore now have a net negative charge. This electric field would increase in strength as you get closer to the boundary, and then die away as you get further away.  This electric field means there is a voltage, as shown in graph (d). The p side is at an arbitrary potential, and the n side is at a potential higher than this because there is an electric field between them. This means there is a potential difference across the depletion region; this is known as the built-in potential.  But why when I connect a volt-meter across the whole diode will I not see this built in potential? Analysis Question Involving Real Numbers Change the Ubuntu program that runs when a command can't be found Can you grammatically end a sentence with "with"? Do you want to come with? Can I come with?  I seem to hear this construction more often in recent years, but it still grates on my ear. I know it's often said that one shouldn't , but I class that rule along with  (i.e. — pedantic tosh). Is it always informal/casual speech to end a sentence with &quot;with&quot;? Was it ever thus? How to avoid using sudo when working in /var/www? I read many places that the rule of thumb for swap space is to double the amount of physical RAM. However, 32 GB does seem a LOT. Do I need that much? Do I need it at all with this high amount of physical RAM? calculating mean of a random variable raised to a power Redirecting from http to https in Apache This is a  about redirecting from http to https in Apache Related:     I have an Appache web server which serves both http://example.com/ and https://example.com/. I want to redirect all http requests to https. Currently I'm using this .htaccess rule to redirect http requests to https. RewriteEngine On RewriteBase / RewriteCond %{HTTPS} off RewriteRule (.*) https://%{HTTP_HOST}%{REQUEST_URI}   It's working as expected for example.com but the same rule is not working for sub links, when I access existing links, like example.com/about it will still load in http, no redirection happens for existing links. How can I make Apache redirect all http requests to https? set of almost complex structures on $\mathbb R^4$ as two disjoint spheres The set of almost complex structures on $\mathbb R^{2n}$ is given by $$ M_n = \frac{GL(2n,\mathbb R)}{GL(n,\mathbb C)} = \mathcal C_+ \sqcup \mathcal C_-,$$ taking into account that $\det = \pm 1$ gives two disjoint sets.  How do we show that $M_2=S^2 \sqcup S^2$? Moreover, I'm a bit confused by dimensions: how does $\dim_{\mathbb R}M_2=8$ relates with the latter decomposition?  This is exercise 1.2.1 from the book Complex geometry by Huybrechts.