My idea was: If $n$ is even,  $\gcd (3ⁿ-1; 2ⁿ-1) =  \gcd (6ⁿ-3ⁿ + 3ⁿ-1; 2ⁿ-1) =  \gcd (6ⁿ-1; 3ⁿ-1) = 2ⁿ-1 $  So $3ⁿ-1 = 2ⁿ-1$ So $3ⁿ = 2ⁿ$, what does $n = 0$ (nonsense pq gets 0/0)  $3ⁿ-1 \equiv 0 \mod 2ⁿ-1$, so $3ⁿ \equiv 1 \mod 2ⁿ-1$, i.e. $3 \equiv 1 \mod 2ⁿ-1$,  $2 \equiv 0 \mod 2ⁿ-1$.  Someone help me?
Is it possible for some integer $n&gt;1$ that $2^n-1\mid 3^n-1$ ?   I have tried many things, but nothing worked.
How to make grass appear on top of ground not in middle? I am currently in the process of learning how to prove statements by induction. For the most part, I understand but I am at most clueless when it comes to prove some inequalities.  For example, I am having a hard time proving that: $n &lt; 2^n$ is true. (Even if it's clear that it is indeed true!)  In the induction step, I am stuck with: $n + 1 &lt; 2^{n + 1}$  I don't know what would be the assumption I should make from this statement.  Anyone could point me in the right direction? Thanks! Are the limits $\lim\limits_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|\,$ and $\lim\limits_{n\to \infty }\sqrt[n]{|a_n|}\,$ equal? Let consider the power series $$ \sum_{k=1}^\infty a_kx^k. $$ We know that it converges when  $$\lim_{n\to \infty }\sqrt[n]{|a_n|}|x|&lt; 1\quad\text{or}\quad |x|&lt; \frac{1}{\lim_{n\to \infty }\sqrt[n]{|a_n|}}$$  But D'Alembert test tells us that it converges if $$|x|&lt; \frac{1}{\lim_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|},$$ so since the radius of convergence is unique, we should have that $$\lim_{n\to \infty }\sqrt[n]{|a_n|}=\lim_{n\to \infty }\left|\frac{a_{n+1}}{a_n}\right|.$$  I tried to prove it, but this result looks strange to me. So if it doesn't work always, how can we have those to limit as radius of convergence? I guess the equality of those two limit should be true most of the time.  So under what conditions is it true? Prove $\forall n\geq 2,n\in\mathbb{Z}$, $(n+1)\mid(n^3+1)$ Question: Prove $\forall n\geq 2,n\in\mathbb{Z}$, $(n+1)\mid(n^3+1)$    I know that it is possible to solve by factoring $n^3+1$ and showing that $n+1$ is a multiple, but I would like to show this via induction.  Attempt: Using induction:  Let $n=2:3\mid 9$, so true for $n=2$  (1) Assume true for $n=k:(k+1)\mid(k^3+1)$, so $(k+1)r=k^3+1$ for some $r\in\mathbb{Z}$  (2) Let $n=k+1:(k+2)\mid(k+1)^3+1$, so $(k+2)s=(k+1)^3+1$ for some $s\in\mathbb{Z}$  From (2) we have $(k+1)s+s=(k+1)^3+1$, subbing (1) into (2), I get $k^3+1+s=(k+1)^3+1$  Therefore $k^3+s=(k+1)^3$.  Setting $s=3k^2+3k+1$, this holds.  Where am I going wrong, is the proof valid, any help would be greatly appreciated. Why is the determinant of a symplectic matrix 1? Suppose $A \in M_{2n}(\mathbb{R})$. and$$J=\begin{pmatrix} 0 &amp; E_n\\   -E_n&amp;0  \end{pmatrix}$$ where $E_n$ represents identity matrix. if $A$ satisfies $$AJA^T=J.$$ How to figure out $$\det(A)=1~?$$ My approach: I have tried to separate $A$ into four submartix:$$A=\begin{pmatrix}A_1&amp;A_2 \\A_3&amp;A_4 \end{pmatrix}$$ and I must add a assumption that $A_1$ is invertible. by elementary transfromation:$$\begin{pmatrix}A_1&amp;A_2 \\ A_3&amp;A_4\end{pmatrix}\rightarrow \begin{pmatrix}A_1&amp;A_2 \\ 0&amp;A_4-A_3A_1^{-1}A_2\end{pmatrix}$$ we have: $$\det(A)=\det(A_1)\det(A_4-A_3A_1^{-1}A_2).$$ From$$\begin{pmatrix}A_1&amp;A_2 \\ A_3&amp;A_4\end{pmatrix}\begin{pmatrix}0&amp;E_n \\ -E_n&amp;0\end{pmatrix}\begin{pmatrix}A_1&amp;A_2 \\ A_3&amp;A_4\end{pmatrix}^T=\begin{pmatrix}0&amp;E_n \\ -E_n&amp;0\end{pmatrix}.$$ we get two equalities:$$A_1A_2^T=A_2A_1^T$$ and $$A_1A_4^T-A_2A_3^T=E_n.$$ then $$\det(A)=\det(A_1(A_4-A_3A_1^{-1}A_2)^T)=\det(A_1A_4^T-A_1A_2^T(A_1^T)^{-1}A_3^T)=\det(A_1A_4^T-A_2A_1^T(A_1^T)^{-1}A_3^T)=\det(E_n)=1,$$ but I have no idea to deal with this problem when $A_1$ is not invertible. Why do I get a segmentation fault when writing to a "char *s" initialized with a string literal, but not "char s[]"? The following code receives seg fault on line 2:  char *str = "string"; str[0] = 'z';  // could be also written as *str = 'z' printf("%s\n", str);   While this works perfectly well:  char str[] = "string"; str[0] = 'z'; printf("%s\n", str);   Tested with MSVC and GCC. Blank screen after installing nvidia restricted driver How can I sound proof bathroom walls? I'm trying to figure out the best, least intrusive way of sound proofing the walls of a bathroom. The bathroom shares a wall with my bedroom, and my 3 yea old's bedroom. The main problem is the sound of water running (plus every bump on the counter and everything) echos loud and clear through the wall into my sons room.  I found  and it seems like a good choice. Another option is adding another layer fo drywall with something like  between them. But all the walls and ceilings in the house are textured. So replacing just one or two walls in a room is just not possible.   This is the layout of the two rooms. I've added red box around the wall with the most noise.   Is there anything else I'm missing here that I could do to dampen the noise short of tearing the bathroom apart? Where should the period go when using parentheses? For example:     In sentence one, I use this example (which has a parenthesis at the end.)   Should the period be inside, or outside of the parentheses? What about if the entire sentence is a parenthetical (as below)?     (Where does the terminal punctuation go here?)   Is there a hard and fast rule? Create ArrayList from array I have an array that is initialized like:  Element[] array = {new Element(1), new Element(2), new Element(3)};   I would like to convert this array into an object of the  class.  ArrayList&lt;Element&gt; arraylist = ???; Problem working fontspec, babel and datetime with cyrillic I have discovered that fontspec with russian babel settings doesn't work with datetime package. I tried to compile such code:  \documentclass[a4paper,11pt]{article} \usepackage[T2A]{fontenc} \usepackage[utf8]{inputenc} \usepackage[main=russian,english]{babel} \usepackage{datetime} \usepackage[cm-default]{fontspec}  \begin{document} \today \end{document}   with a xelatex and got an error: Command \cyri unavailable in encoding EU1.  Is there any way to handle it?  UPDATE: I really bungle with charset in XeLateX. Thanks to egreg. I have removed fontenc and inputenc and used some cyrillic font, so my MWE now looks this way:  \documentclass[a4paper,11pt]{article} \usepackage[main=russian,english]{babel} \usepackage{fontspec} \setromanfont{PT Sans} \usepackage{datetime}  \begin{document} \today \end{document}   and I've got another error, if I used russian as a main language in babel: Undefined control sequence. \\monthnamerussian ...cyrm \cyra \cyrya \or \cyri  \cyryu \cyrn \cyrya \or \c...  Is it still because of XeLaTeX? Backup of clients' directory to Windows Server 2008 We have a machine running Windows Server 2008, and we need to take backup of folder from a desktop on LAN. This folder is not shared.   Is there any tool or software to do this, or can a scheduled task be made which should automatically take the backup of that folder date-wise &amp; make a copy on server? Ie can we do it through Schedule Tasks option ?  Desktop systems have Windows Vista/Windows 7. $$\lim_{x\to 0} \frac{\ln(1+x)}x$$  The process I  want to take to solving this is by using the definition of the limit, but I am getting confused. ( without l'hopitals rule)   $$\lim_{h \to 0} \frac{f(x+h) - f(x)}h$$  $$\lim_{h \to 0} \frac{\frac{\ln (1+x+h)}{x+h} - \frac{\ln(1+x)}x}h$$  $$\lim_{h \to 0} \frac{x\ln(1+x+h) - (x+h)\ln (1+x)}{hx(x+h))}$$  At this point I get confused because I know the answer is $1$, but I am not getting this answer through simplification of my formula. Compilation problem with siunitx (returning error regarding ifnum) I'm trying to get siunitx working. For some reason I keep getting this error:  ! Missing = inserted for \ifnum. &lt;to be read again&gt;                     - l.3 \begin           {document}^^M   I have tried a minimal document to make sure it is not some kind of compatibility issue. And I get that same error in the most minimal document possible.  I'm using pdfLaTeX to compile, and have tried both from TeXWorks and from bash. I'm using TeXLive, if that matters. I have also tried reinstalling siunitx, l3packages, and l3kernel. There was an issue with l3kernel before this, due to an apparently old version. But all three packages are now up to date. I've also reinstalled expl3 while trying to debug the original issue.  My minimal document is:  \documentclass{article} \usepackage{siunitx} \begin{document} \end{document}   The output I get is:  This is pdfTeX, Version 3.14159265-2.6-1.40.16 (TeX Live 2015/Debian) (preloaded format=pdflatex) restricted \write18 enabled. entering extended mode (./untitled-2.tex LaTeX2e &lt;2016/02/01&gt; Babel &lt;3.9q&gt; and hyphenation patterns for 81 language(s) loaded. (/usr/share/texlive/texmf-dist/tex/latex/base/article.cls Document Class: article 2014/09/29 v1.4h Standard LaTeX document class (/usr/share/texlive/texmf-dist/tex/latex/base/size10.clo)) (/home/&lt;name&gt;/texmf/tex/latex/siunitx/siunitx.sty (/home/&lt;name&gt;/texmf/tex/latex/l3kernel/expl3.sty (/home/&lt;name&gt;/texmf/tex/latex/l3kernel/expl3-code.tex) (/home/&lt;name&gt;/texmf/tex/latex/l3kernel/l3pdfmode.def)) (/home/&lt;name&gt;/texmf/tex/latex/l3packages/xparse/xparse.sty) (/usr/share/texlive/texmf-dist/tex/latex/amsmath/amstext.sty (/usr/share/texlive/texmf-dist/tex/latex/amsmath/amsgen.sty)) (/usr/share/texlive/texmf-dist/tex/latex/tools/array.sty) (/home/&lt;name&gt;/texmf/tex/latex/l3packages/l3keys2e/l3keys2e.sty) (/usr/share/texlive/texmf-dist/tex/latex/beamer/translator/translator.sty (/usr/share/texlive/texmf-dist/tex/latex/graphics/keyval.sty) (/usr/share/texlive/texmf-dist/tex/latex/beamer/translator/translator-language- mappings.tex))) ! Missing = inserted for \ifnum. &lt;to be read again&gt;                    - l.3 \begin          {document}^^M ?   Does anyone have any idea what might be going wrong here? What exactly is an environment variable? Truncate (not round) decimal places in SQL Server Evaluate the integral $\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$ This integral looks a lot like an elliptic integral, but with cubes instead of squares: $$I(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^3+a^3)(x^3+b^3)}}$$ Let's consider $a,b&gt;0$ for now. $$I(a,a)=\int_0^\infty \frac{dx}{x^3+a^3}=\frac{2 \pi}{3 \sqrt{3} a^2}$$ I obtained the general series solution the following way. Choose $a,b$ such that $a \geq b$, then: $$I(a,b)=\frac{1}{a^2} \int_0^\infty \frac{dt}{\sqrt{(t^3+1)(t^3+b^3/a^3)}}=\frac{1}{a^2} I \left(1, \frac{b}{a} \right)$$ $$\frac{b^3}{a^3}=p, \qquad I \left(1, \frac{b}{a} \right)=I_1(p)$$ $$I_1(p)=\int_0^\infty\frac{dt}{\sqrt{(t^3+1)(t^3+p)}}=2 \frac{d}{dp} J(p)$$ $$J(p)=\int_0^\infty\sqrt{\frac{t^3+p}{t^3+1}}dt=\int_0^\infty\sqrt{1+\frac{p-1}{t^3+1}}dt=$$ $$|p-1| \leq 1$$ $$=\sum_{k=0}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) (p-1)^k \int_0^\infty \frac{dt}{(t^3+1)^k}$$ Now this is the most problematic part. The first integral of this series diverges. However, it's a constant in $p$, so if we differentiate, it formally disappears: $$I_1(p)=2 \sum_{k=1}^\infty \left( \begin{array}( 1/2 \\ ~k \end{array} \right) k (p-1)^{k-1} \int_0^\infty \frac{dt}{(t^3+1)^k}$$ Now, every integral in this series converges. The integtals can be computed using the Beta function, if we substitute: $$t^3=\frac{1}{u}-1$$ Finally, we rewrite: $$I_1(p)=\frac{\Gamma (1/3)}{3 \sqrt{\pi}} \sum_{k=1}^\infty \frac{k^2}{k!^2} \Gamma \left(k- \frac{1}{2}\right) \Gamma \left(k- \frac{1}{3}\right) (1-p)^{k-1}$$ Or, using the Pochhammer symbol:  $$I_1(p)=\frac{2 \pi}{3 \sqrt{3}} \sum_{k=0}^\infty \frac{(k+1)^2}{(k+1)!^2} \left(\frac{1}{2}\right)_k \left(\frac{2}{3}\right)_k (1-p)^k$$  My questions are:  Is the method I used valid (see the 'problematic part')? How to get this series into a Hypergeometric function form? Is there any 'arithmetic-geometric mean'-like transformation (Landen's transformation) for this integral? How to go about finding it?  If the method I used is correct, it can be used for any integral of the form ($m \geq 2$): $$I_m(a,b)=\int_0^\infty \frac{dx}{\sqrt{(x^m+a^m)(x^m+b^m)}}$$ I'm using Ubuntu 17.10 with GNOME 3. How can I show all open windows of a certain application using the sidebar (aka Ubuntu Dock)? Right now I have to right-click and select "all windows" to see my other open windows.   Is there a quicker way to do this? Simple question:     My apples and orange are wrong   or     My apples and orange is wrong   I am not a native English speaker, and I am having some trouble choosing between plural are or singular is for that kind of example. How to participate in clan war instead of being a spectator I joined a new clan and now that the clan war has started I tried to attack but it says "Spectator can't attack". How do I get out of being a spectator (I haven't used either of my 2 attacks in the war). I went to the clan castle and the clan option and I selected "Yes I'll participate in this war" but nothing happens.