Find all positive integers $n$, such that $n2^n+1$ is divisible by $3$ I was trying to solve with normal Diophantine equation but couldn't get and don't know much about modular arithmetic . Please help me out in this
Question :   Determine all natural numbers n such that 3   divides $n\cdot2^{n}+1$  Actually I don't have any ideas to approach but  my efforts :   I see $n=1,2,7,8,13,14$ so I think :   $n=6k+1$ and $n=6k+2$ $k\in \Bbb{N} $  If I'm not wrong but I don't know how ? I prove it ?
To Find Large Powers in C++ If $x_1 = 3$, $x_{n+1} = \frac{1}{4-x_n}$ for $n \geq 1$, prove the sequence is bounded below by $0$, above by $4$. If $x_1 = 3$, $x_{n+1} = \dfrac{1}{4-x_n}$ for $n \geq 1$, prove the sequence is decreasing and bounded below by $0$, above by $4$.  Want to show: $0 &lt; x_{n+1} &lt; x_n &lt; 4$  I decided to do the bounded part first.    Base Case, $n=1$  $0 &lt; 3 &lt; 4$ [works]  IH  Suppose $0 &lt; x_k &lt; 4$, for some $k\in\mathbb{N}$  IS  Show $0 &lt; x_{k+1} &lt; 4$  We know $x_{k+1} = \dfrac{1}{4-x_k}$  We also know that $0 &lt; x_k &lt; 4$, by IH.  So we have:  $$\dfrac{1}{4} &lt; x_{k+1} &lt; \infty$$  But this doesn't show the boundness. Where have I went wrong? Why does Lang describe a field as a union and compositum of its subfields in this manner? I am reading Serge Lang's Algebra, revised 3rd edition, and on page 226, the author makes the following definitions and observations:   If $E,F \subset L$, the compositum of $E$ and $F$ in $L$ is defined to be the smallest subfield of $L$ containing both $E$ and $F$. It is denoted $EF$. If $k \subset E$ and $\alpha_1,\dots,\alpha_n \in E$, then $k(\alpha_1,\dots,\alpha_n)$ is defined to be the smallest subfield of $E$ containing $k$ and $\alpha_1,\dots,\alpha_n$. Observe that $$ E = \bigcup k(\alpha_1,\dots,\alpha_n), $$ where the union is taken over all finite families $\{ \alpha_1,\dots,\alpha_n \}$ of elements of $E$. The compositum of an arbitrary family of subfields of a field $L$ is defined as the smallest subfield of $L$ containing all fields in the family. $E$ is finitely generated over $k$ if there is a finite family $\{ \alpha_1,\dots,\alpha_n \}$ of elements of $E$ such that $E = k(\alpha_1,\dots,\alpha_n)$. Observe that $E$ is the compositum of all its finitely generated subfields over $k$.   My question is regarding the two observations (point nos. 3 and 6). It appears to me that it is enough to take union (and compositum) over all subfields generated by a single element. For example for point no. 3, I can write $$ E = \bigcup_{\alpha \in E} k(\alpha) $$ because $E$ is clearly contained in the RHS, and each $k(\alpha)$ is contained in $E$ and hence so is the union, implying that $E$ contains the RHS. Similarly for point no. 6.  So, why does Lang emphasise to take the union and compositum over all finitely generated subfields? Is there some perspective that he wishes to emphasise that I am missing? Any help in understanding this will be appreciated. Show that if $f(z)=\sum\limits_{n\geq0}a_nz^n$ is analytic in $\{z\in\mathbb{C}:|z|&lt;1\}\cup\{1\}$ and $\forall n\geq0:a_n\geq0$ then the radius of convergence of the power series is strictly larger than $1$.  My approach:   Let $D=\{z\in\mathbb{C}:|z|\leq1\}$.  We have that $f(1)$ converges, so $\sum\limits_{n\geq 0} a_n&lt;\infty$. Moreover since the $a_n$'s are all positive we have that $|a_n|=a_n$, therefore $\sum\limits_{n\geq 0} |a_n|=\sum\limits_{n\geq 0} a_n&lt;\infty$.  Let $f_n(z)=a_nz^n$, then $\forall z\in D, n\geq 0$ we have:  $$|f_n(z)|=|a_nz^n|=|a_n|\cdot|z^n|=a_n\cdot |z|^n\leq a_n\cdot 1^n\leq a_n$$  Then, applying Weierstrass M-test we get that the power series is convergent in $D$.  Now, I need to "find" an open ball $B$ such that $D\subset B$ and the power series is convergent in it.  Edit  $f$ is analytic in $1$, so there is an open set $U$ that contains $1$ such that   $$f(z)=\sum\limits_{n\geq 0} b_n(z-1)^n$$  Does it mean that the series is convergent in $U$?   I can see that we have that the two series coincide in $D\cap U$. How to prevent page breaking for lstinputlistings? Why does the CRT formula yield a solution of a congruence system? Must a husband and wife make separate online applications for a UK Standard Visitor visa, and so attend separate appointments for biometrics? If $\gcd(a,b)=1$, then $\gcd(a+b,a^2 -ab+b^2)=1$ or $3$. Hint:  $a^2 -ab +b^2 = (a+b)^2 -3ab.$  I know we can say that there exists an $x,y$ such that $ax + by = 1$.  So in this case, $(a+b)x + ((a+b)^2 -3ab)y =1.$  I thought setting $x = (a+b)$ and $y = -1$ would help but that gives me $3ab =1.$  Any suggestions? If $f(x)$ is continuous on $[a,b]$ and $M=\max \; |f(x)|$, is $M=\lim \limits_{n\to\infty} \left(\int_a^b|f(x)|^n\,\mathrm dx\right)^{1/n}$? I've been using the == operator in my program to compare all my strings so far. However, I ran into a bug, changed one of them into .equals() instead, and it fixed the bug.  Is == bad? When should it and should it not be used? What's the difference? Additional horizontal space when the last line of an environment is full I have created an environment env which just adds a \medskip and some text (say "Blablabla") after the text inside \begin{env} ... \end{env}. The problem is that if the last line is full, then I get an additional horizontal space between the text and "Blablabla".  Here is an MWE:  \documentclass{article} \newenvironment{env}{ ENVIRONMENT:\par }{ \medskip \par\noindent Blablabla }  \begin{document}  \section*{First case: the last line is not full}  \begin{env} Lorem ipsum dolor sit amet, consectetur adipiscing elit. Nulla malesuada porta tortor non vestibulum. Vivamus vel tincidunt purus. Curabitur quam arcu, suscipit sed diam sed, ornare efficitur felis. Duis at arcu iaculis, viverra ligula sit amet, hendrerit eros. Proin finibus venenatis sollicitudin. Nunc ullamcorper sem ut lacinia posuere. Integer pulvinar nulla lorem, ut pulvinar ipsum egestas at. Duis quam ipsum, posuere ac pulvinar vitae, auctor non leo. \end{env}  \section*{Seconde case: the last line is full}  \begin{env} Lorem ipsum dolor sit amet, consectetur adipiscing elit. Quisque suscipit vitae neque eu gravida. Quisque et enim quam. Integer elementum diam a ex laoreet, ac egestas lacus volutpat. Nunc ullamcorper non urna tincidunt ultrices. Etiam eu ultricies diam. Proin mollis, nunc sit amet volutpat varius, ipsum tellus bibendum nulla, in eleifend lacus felis pellentesque nisl. Phasellus laoreet, libero quis auctor congue, nulla nulla aliquam metus, in hendrerit arcu justo quis tortor. Nunc nec nisi euismod, consectetur tellus nec, suscipit libero.  \end{env}  \end{document} Proficiency with specific weapons vs 'simple and martial'? In the PHB on pg. 146 it states that there are two weapon categories in which the players can be proficient: simple and martial.  As per  answer to another question, being proficient in either category makes the players character proficient in using all weapons of the respective category.   However in the race descriptions the players are given proficiency for specific weapons. Dwarfs, for example, have the trait Dwarven Combat Training giving proficiency with battleaxes, handaxes, throwing hammers and warhammers.   How do the two types of proficiencies interact with each other? Where can I ask Android related questions in Stack Exchange? I would like to know where I can ask Android apps and other mobile related questions. Is there a Stack Exchange site for such questions? How to check if file exists on FTP before FtpWebRequest What is Renly's basis for removing Joffrey? Related question:   I get that he considers himself a better king than Stannis, but why isn't he ?  Did Renly know about Joffrey being illegitimate? I always thought he didn't and was distrustful that Cersei would accept Ned as Regent.  I think in both the books and the series, Renly finds out Joffrey is illegitimate from Stannis' letters, but rewatching that clips makes me think Renly was with Ned, Varys and LF in the plan to arrest Joffrey and Cersei.  [looks up something]  Oh also . So Renly did know? How to mount a RAM disk on startup? I'm trying to mount a RAM disk on startup with this:  /Users/Aram/Development/Tools/ramdisk.sh  diskutil erasevolume HFS+ "RamDisk" `hdiutil attach -nomount ram://800000`   com.aram.ramdisk.plist  &lt;?xml version="1.0" encoding="UTF-8"?&gt; &lt;!DOCTYPE plist PUBLIC "-//Apple//DTD PLIST 1.0//EN"          "http://www.apple.com/DTDs/PropertyList-1.0.dtd"&gt; &lt;plist version="1.0"&gt;     &lt;dict&gt;         &lt;key&gt;Label&lt;/key&gt;         &lt;string&gt;com.aram.ramdisk&lt;/string&gt;         &lt;key&gt;ProgramArguments&lt;/key&gt;         &lt;array&gt;             &lt;string&gt;/bin/sh&lt;/string&gt;             &lt;string&gt;/Users/Aram/Development/Tools/ramdisk.sh&lt;/string&gt;         &lt;/array&gt;         &lt;key&gt;RunAtLoad&lt;/key&gt;         &lt;true/&gt;     &lt;/dict&gt; &lt;/plist&gt;   And then running:   sudo chown root:wheel /Library/LaunchDaemons/com.aram.ramdisk.plist sudo launchctl load -w /Library/LaunchDaemons/com.aram.ramdisk.plist   But I get these in the console:  1/08/12 1:29:25.982 PM fseventsd[64]: could not open &lt;&lt;/Volumes/RamDisk/.fseventsd/fseventsd-uuid&gt;&gt; (No such file or directory) 1/08/12 1:29:25.982 PM fseventsd[64]: log dir: /Volumes/RamDisk/.fseventsd getting new uuid: 102D7293-F1F1-4640-AA50-D547C365339F  1/08/12 1:29:24.561 PM sudo[1193]:     Aram : TTY=ttys000 ; PWD=/Library/LaunchDaemons ; USER=root ; COMMAND=/bin/launchctl load -w /Library/LaunchDaemons/com.aram.ramdisk.plist  1/08/12 1:40:17.052 PM com.apple.launchd[1]: (com.aram.ramdisk) Throttling respawn: Will start in 8 seconds   It works if I set KeepAlive to true, but then it creates a RAM disk every 10 seconds.  Is there a way to delay the script for 10 seconds before running? I'm working through Rotman (1994). It's the 2nd chapter where my curiosity lead me to a question that is not from the book. Here's the problem that inspired the question:      2.8. Imbed $S_n$ as a subgroup of $A_{n+2}$, but show, for $n\ge2$, that $S_n$ cannot be imbedded in $A_{n+1}.$    Since the alternate subgroup has the property of even parity, I thought that a little investigation of parity for the second and third symmetric groups was in order. So I constructed a table for each of them, grouped by even and odd number of disjoint products.   I'll forego writing the actual table and simply provide the answers. In symmetric groups with an even idegree, two even permutations come together under composition to produce an even permutation. Two odds also make an even. Two different parities produce odd elements.   When the symmetric group has an odd degree, e.g. $S_3$, two odds make an odd, two evens make an odd, and two different parities make an even element. Exactly the opposite results from the symmetric groups with even index.   Two questions: is this enough to prove the $A_{n+1}$ case? And can this be generalized modulo 2 e.g. $A_{2n+1}$? How to draw tangent vectors and component vectors on a curve I am searching for a way to have a figure drawn using LaTeX. I am aware of some packages TikZ and PSTricks that could accomplish this, and have been tinkering around with them recently but do not have enough skill set at the moment to accomplish this drawing. Is there a way someone can accomplish the drawing, preferably with TikZ but could be any package your most comfortable with in a reduced amount of coding (non-advanced) to where I could understand what is happening from the code to the picture. If this is not possible, then I will have to study the package more and work with it until I understand the method (code) to create the graphic. The drawing is below.  Note: This is a rough sketch. The optimal output would have smooth quadratic curves and all quantities are vectors. Installed Ubuntu 13.04 on a Dell laptop I had laying around.  Install goes well but when I'm done, I only get a command line, no GUI at all.  Tried several versions, tried 32 vs 64 bit all give me the same result.  Any ideas on what to do/try? Galois group of irreducible cubic equation The polynomial $f(x) = x^3 -3x + 1$  is irreducible, and I'm trying to find the splitting field of the polynomial.  We've been given the hint by our lecturer to allow an arbitrary $\alpha$ to be a root and then to check that $\alpha$ is a root, then $2-\alpha^2$ is also root. I've done this with no problem.  My issue is since this is a polynomial of degree $3$, there must be another root that is somehow related to $\alpha$ so that I can create the splitting field and in turn the Galois group of my polynomial but I have no idea what this last root could be.