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Strange result when removing item from a list while iterating over it I've searched online, but I have not found any clear instructions how to permanently add .sty files to my TeXShop installation. I'm using a Mac with OSX (TeXShop is a mac only program).  What is the best way of doing this? The one comprehensive set of instructions I found .   I've got as far as extracting the files (running LaTeX on the .ins and . dtx files) but after that I don't know what to do. I also don't understand how to update the index of my distribution, and whether there are particular OSX specifics (the instructions are for Windows). Unique pair of positive integers $(p,n)$ satisfying $p^3-p=n^7-n^3$ where $p$ is prime Q. Find all pairs $(p,n)$ of positive integers where $p$ is prime and $p^3-p=n^7-n^3$.   Rewriting the given equation as $p(p+1)(p-1)=n^3(n^2+1)(n+1)(n-1)$, we see that $p$ must divide one of the factors $n,n+1,n-1,n^2+1$ on the $\text{r.h.s}$.   Now, the $\text{l.h.s}$ is an increasing function of $p$ for $p\ge1$. This implies that for any given $n\ge1$, there is exactly one real $p$ for which $\text{l.h.s}=\text{r.h.s}$. For $p=n^2$, we get $\text{l.h.s}=n^6-n^2&lt;n^7-n^3=\text{r.h.s}.$ This means that either $p&gt;n^2$ or $p&lt;n^2$ must hold.  Assuming $p&gt;n^2$, it follows that the prime $p$ cannot divide any of $n,n+1,n-1$. So $p$ must divide $n^2+1$ and hence $p=n^2+1\quad (\because p&gt;n^2)$.  Substituting the value of $p$ in the given equation we get, $n^2+2=n^3-n\implies n^3-n^2-n=2$. As the factor $n$ on the $\text{l.h.s}$ must divide $2$, the above equation has a unique integer solution $n=2$.  Finally, we get $(5,2)$ as the solution to the given equation.      But how do I conclude this is the only solution possible? Also, why does'nt $p&lt;n^2$ (the case which I ignored) hold? As a bonus question, I would like to ask for any alternative/elegant solution (possibly using congruence relations) to the problem. Stateful bash function Why is there a difference between urls submitted to a sitemap and urls in the google index? I've seen videos of some folks having several armor sets (4 is the most I've seen so far) standing about on their settlement. I'm wondering if having too many would cause some sets to disappear. They're not like guns/items that you can safely store in settlement crates, the suits kind of just... stand around. (I know there's the extra step of taking out the fusion cores, so that NPC's don't use them.)  Is there a limit to the number of Power Armor sets you can keep? Wine or virtual machine (VirtualBox) to run apps (eg. Photoshop) I am wondering if it will be better (stability, convenience and speed) to run apps like Photoshop with Wine or a virtual machine like VirtualBox? What does unwrapping a model do and why is it important? After discussion with colleagues regarding the use of the 'var' keyword in C# 3 I wondered what people's opinions were on the appropriate uses of type inference via var?  For example I rather lazily used var in questionable circumstances, e.g.:-  foreach(var item in someList) { // ... } // Type of 'item' not clear. var something = someObject.SomeProperty; // Type of 'something' not clear. var something = someMethod(); // Type of 'something' not clear.   More legitimate uses of var are as follows:-  var l = new List&lt;string&gt;(); // Obvious what l will be. var s = new SomeClass(); // Obvious what s will be.   Interestingly LINQ seems to be a bit of a grey area, e.g.:-  var results = from r in dataContext.SomeTable               select r; // Not *entirely clear* what results will be here.   It's clear what results will be in that it will be a type which implements IEnumerable, however it isn't entirely obvious in the same way a var declaring a new object is.  It's even worse when it comes to LINQ to objects, e.g.:-  var results = from item in someList               where item != 3               select item;   This is no better than the equivilent foreach(var item in someList) { // ... } equivilent.  There is a real concern about type safety here - for example if we were to place the results of that query into an overloaded method that accepted IEnumerable&lt;int&gt; and IEnumerable&lt;double&gt; the caller might inadvertently pass in the wrong type.  var does maintain strong typing but the question is really whether it's dangerous for the type to not be immediately apparent on definition, something which is magnified when overloads mean compiler errors might not be issued when you unintentionally pass the wrong type to a method. Orthogonal matrix and orthonormal columns Get most common value for each value of another column in SQL I have a table like this:   Column  | Type | Modifiers  ---------+------+-----------  country | text |   food_id | int  |   eaten   | date |    And for each country, I want to get the food that is eaten most often.  The best I can think of (I'm using postgres) is:  CREATE TEMP TABLE counts AS     SELECT country, food_id, count(*) as count FROM munch GROUP BY country, food_id;  CREATE TEMP TABLE max_counts AS     SELECT country, max(count) as max_count FROM counts GROUP BY country;  SELECT country, max(food_id) FROM counts     WHERE (country, count) IN (SELECT * from max_counts) GROUP BY country;   In that last statement, the GROUP BY and max() are needed to break ties, where two different foods have the same count.  This seems like a lot of work for something conceptually simple.  Is there a more straight forward way to do it? Combinatorial Proof Question: Vandermonde convolution identity I'm really iffy on combinatorial proofs in general and now that there is a sum, it's just confused me even more. Can someone try and walk me through this proof?  $$ \binom{m + n}{r} = \sum_{k=0}^r \binom{m}{k}\binom{n}{r - k} $$ Why are the longer edges being beveled more? Is Dirichlet function Riemann integrable? "Dirichlet function" is meant to be the characteristic function of rational numbers on $[a,b]\subset\mathbb{R}$.   On one hand, a function on $[a,b]$ is Riemann integrable if and only if it is bounded and continuous almost everywhere, which the Dirichlet function satisfies.  On the other hand, the upper integral of Dirichlet function is $b-a$, while the lower integral is $0$. They don't match, so that the function is not Riemann integrable.  I feel confused about which explanation I should choose... I am not able to play "Half-Life-2" on Steam + Ubuntu 14.04 . Error shown in message box on screen:      Could not find required OpenGL entry pint 'glGetError'! Either your video card is unsupported, or your OpenGL driver needs to be updated.   I have a Dell Inspiron Laptop.  My OS is Ubuntu 14.04 and my graphics card is AMD Radeon HD 8600M.  I have installed AMD properietry Driver.    The partial output of "glxinfo" is as follows:      name of display: :0     libGL: screen 0 does not appear to be DRI3 capable     libGL: pci id for fd 4: 8086:0a16, driver i965     libGL: OpenDriver: trying /usr/lib/i386-linux-gnu/dri/tls/i965_dri.so     libGL: OpenDriver: trying /usr/lib/i386-linux-gnu/dri/i965_dri.so     libGL: Can't open configuration file /home/arushi/.drirc: No such file or directory.     libGL error: failed to load driver: i965     libGL: OpenDriver: trying /usr/lib/i386-linux-gnu/dri/tls/swrast_dri.so     libGL: OpenDriver: trying /usr/lib/i386-linux-gnu/dri/swrast_dri.so     libGL: driver does not expose __driDriverGetExtensions_swrast(): /usr/lib/i386- linux-gnu/dri/swrast_dri.so: undefined symbol: __driDriverGetExtensions_swrast     libGL: Can't open configuration file /home/arushi/.drirc: No such file or directory.     libGL: Can't open configuration file /home/arushi/.drirc: No such file or directory.     libGL error: failed to load driver: swrast     display: :0  screen: 0     direct rendering: No (If you want to find out why, try setting LIBGL_DEBUG=verbose) Let $(X, d(x,y))$ be a complete metric space. Prove that if $A\subseteq X$ is a closed set, then $A$ is also complete.  My attempt: I tried to prove that every Cauchy sequence $(b_n)$ of points of $A$ converges to a point $b\in A$. However could not figure out the exit way. Maybe I am on the wrong track. Could you please help me?  edit: More from my attempt:  Suppose $A$ is a closed set and let $(x_n)$ be a sequence of points $A$ such that $\lim_n x_n\to b$.  Suppose now that $A$ has the property that $b\in A$, whenever $x_n$ converges to $b$. We know that every element of $x_n$ which is convergent in $X$ also converges to a point in $A$. Since $x_n$ is a Cauchy sequence in $A$, it must converge to a point $y\in A$. But the limit of a convergent sequence is unique.   Take $x\in A$ and select an appropriate $n$, which enables $x_n$ to converge to a point $x$ in $X$. Since the limit is unique, it must follow that $x=y$. Thus $x\in A$ and $A$ is closed.   If $A$ is a closed subset of $X$, then any Cauchy sequence of a point in $A$ is convergent in $X$ and hence converges to a point in $A$. Thus $A$ is complete. Where to start with the Star Wars books? When I launch certain programs from the command line like eclipse and document viewer in 11.10 it spews a load of information that seems inconsequential.   Also when they are run in the background they sometimes continue to produce output to the terminal which I am currently working on, which is irritating.  I would like them just launch and keep the background stuff in the background. My reasoning is that if you launch these programs through the GUI (eg double clicking on an icon) these messages are never shown to me, so I don't need them in the command line. Schengen Visa Application: what are some helpful tips for preparing? Is there an example of a ring $S$ with identity $1_S$ containing a non-trivial subring $R$ which itself has an identity $1_R$, but $1_R\neq 1_S$ (or equivalently $1_S\notin R$). I'd also like to know under what conditions the identities have to be equal. I know they must be equal if $S$ has no zero divisors, since for every $r\in R$ we have $(1_S-1_R)r=0$  In other words: Is the category of unital rings with identity-preserving morphisms a full subcategory of the category of rings (where morphism are only required to respect addition and multiplication)?