And also that: $\gcd(p^m-1,p^n-1)=p^{\gcd(m,n)}-1$. For the love of a God I don't see how to show these two identities.
For all $a, m, n \in \mathbb{Z}^+$,   $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$
What is a good non-technical introduction to theories of everything? What is the penalty for dual US citizens who lie on the ESTA form in order to enter under VWP using their second passport? Why are these estimates to the German tank problem different? Suppose that I observe $k=4$ tanks with serial numbers $2,6,7,14$.   What is the best estimate for the total number of tanks $n$?   I assume the observations are drawn from a discrete uniform distribution with the interval $[1,n]$. I know that for a $[0,1]$ interval the expected maximum draw $m$ for $k$ draws is  $1 - (1/(1+k))$. So I estimate $\frac {k}{k+1}$$(n-1)≈$ $m$, rearranged so $n≈$ $\frac {k+1 }{k}$$m+1$.  But the frequentist estimate from  is defined as:  $n ≈ m-1 + $$\frac {m}{k}$  I suspect there is some flaw in the way I have extrapolated from one interval to another, but I would welcome an explanation of why I have gone wrong! Is it possible to play D&D 5e solo? Is it at all possible to run a solo game where you act as the DM/GM for yourself? I know there was a 4th edition adventure published in an old issue of one of the D&amp;D magazines but recently I read somewhere (Reddit I think) that the games Outbreak: Undead and Eclipse phase were good games to play solo.  Can anyone with experience with running a game of D&amp;D 5e for themselves explain how to do so? Cursor is set to move by increments. how to reset to normal? I realize this isn't possible, but I can't see why not, especially if you change the model a little bit so that the balls simply travel through a tube of water on the way up, rather than exactly this model.  Please be clear and detailed.  I've heard explanations like "the balls wouldn't move" but that doesn't do it for me - I really do not see why the balls on the right would not be pulled/pushed up, and the rest of the chain wouldn't continue on. Choosing the best model from among different "best" models How do you choose a model from among different models chosen by different methods (e.g. backwards or forwards selection)?  Also what is a parsimonious model? Top of answer covered by sticky top bar after clicking direct link With the introduction of the network-wide sticky top bar , I noticed a bug that's happening on every site (except for Stack Overflow). When you click on an answer link (of the form https://site.stackexchange.com/a/####), the very top of the answer and the upvote button are covered by the top bar:    Stack Overflow shows the correct behavior: Is it possible to exclude questions I've already answered from the search criteria? Recently I asked if it were possible to find questions with just one answer and as a result we got an .  That's great, but ...     Is there a way to restrict the search to questions where that one answer isn't mine? That would help. :-)   It's certainly possible to search for , but it's not possible to exclude your own posts as far as I can tell:   -user:me searches for posts that . user:1-67 doesn't search for posts from user #1 to user #67, but searches for .  (And I'm not sure that will help anyway since I don't think it would be possible to simultaneously exclude 69-max(userID).) -Ericson searches for posts that , but since my last name is in the the metadata and not the searchable portion of posts: no joy.   And of course none of this will actually work to eliminate the questions that that I've answered already and want to ignore.  In general, If I'm involved in a Q&amp;A (or ) the notification procedures do a good job of keeping me up-to-date with developments.  But when I search, it's often to find questions I've not noticed before.  So being able to eliminate questions I have already answered would be a generally useful feature for me. How do I show that all continuous periodic functions are bounded and uniform continuous? In the tense of present perfect, Are 'recent past' and 'hot news' the same? Wow, the iOS app feed sure showed me an interesting question! I want to read more. Uh, how do I get there? In the iOS app, pulling up a question from the feed gives the standard question view, which provides no means of navigating elsewhere on that site, either to browse or search. This should be changed for feed questions.  For example, I tapped on a highlighted Seasoned Advice question today, and found that I wanted to make a comment pointing to a related question on the site. The best way I could think of finding that post was to leave the app and use my browser. Then I came back with the link to copy-paste in.  Similarly, a question in the feed might be so intriguing that I want to look around for more cool stuff on the site. That requires two taps back and then a search via the "All Sites" menu item.  I'd suggest that questions selected from the feed should get a special header bar (or maybe stuck in between the questions and the answers) that provides a "More posts!" link at least to the site's top view; ideally there would also be a search field that could pop out of it.  Another possibility that would work pretty well, although it's a small tap target on a phone, would be to link the site name that appears under the question title to the site's top view. What does it mean that GUIkit is being deprecated? Error: 'else' without 'if' In the Harry Potter series, are wizards capable of creating their own spells, or are they limited to the spells that they're taught/learn?  I seem to remember something about Snape creating sectumsempra, but I'm not sure if that really is the case. What is the relationship between symmetry and degeneracy in quantum mechanics? Let me remind you about the following classical  examples in quantum mechanics.     Example 1. Bound states in 1-dim potential V(x).   Let $V(x)$ be a symmetric potential i.e.   $$V(x) = V(-x)$$   Let us introduce the parity operator $\hat\Pi$ in the following way:   $$\hat\Pi f(x) = f(-x).$$   It is obvious that    $$[\hat H,\hat\Pi] = 0.$$   Therefore, for any eigenfunction of $\hat H$ we have:   $$\hat H|\psi_E(x)\rangle = E|\psi_E(x)\rangle = E\hat\Pi|\psi_E(x)\rangle,$$   i.e. state $\hat\Pi|\psi_E(x)\rangle$ is eigenfunction with the same eigenvalue. Is $E$ a degenerate level? No, because of linear dependence of $|\psi\rangle$ and $\hat\Pi|\psi\rangle.$   Consider the second example.     Example 2. Bound states in 3-dim a potential $V(r)$. Where $V(r)$ possesses central symmetry, i.e. depends only on distance to center.   In that potential we can choose eigenfunction of     angular momentum $\hat L^2$ for basis    $$|l,m\rangle,$$   where $l$ is total angular momentum and $m$ - its projection on chosen axis (usually $z$). Because of isotropy eigenfunction with different $m$ but the same $l$ correspond to one energy level and linearly independent. Therefore, $E_l$ is a degenerate level.      My question is if there is some connection between symmetries and degeneracy of energy levels. Two cases are possible at the first sight:   Existence of symmetry $\Rightarrow$ Existence of degeneracy Existence of degeneracy $\Rightarrow$ Existence of symmetry     It seems like the first case is not always fulfilled as shown in the first example. I think case 1 may be fulfilled if there is continuous symmetry. I think the second case is always true. Tablet or Phone - Android There is no "helpful flags" counter (with a link to ) on the mobile version of profile page. And I see no other way to reach flagging summary page and check your flagging history from mobile rather than switching to the full site view.  Desktop version   &nbsp; Mobile version Let $\{x_n\}$ and $\{y_n\}$ be sequences of real numbers which converge to $\ell$ and $m$ respectively. Show that  $$\lim_{n\to \infty} \frac{1}{n}\sum_{k=1}^{n}x_ky_{n+1-k}=\ell m$$  This is a problem I found in a book, it seemed to be routine, but now I can't solve it.  What I tried :     Let us take some $\epsilon&gt;0$. Now after a stage $N$, both $|x_n-\ell|&lt;\epsilon$ and $|y_n-m|&lt;\epsilon$. Now I wanted to break the sum $$\left|\frac{1}{n}\sum_{k=1}^{n}x_ky_{n+1-k}-\ell m\right|$$in two parts, one upto $N$, and other one for $N+1$ to $n$. Now we need this to be arbitrarily small. But here I am lost. I don't know how to proceed anymore. Because when I take $x_i's$ to make up a finite sum, then the $y_j's$ are there too with $j$ very large. I think this is the main idea, but unable to do anything further. Can someone help me? Thanks. Photoelectric effect intensity I understand the PE effect quite well but I'm failing to understand one thing. Intensity is the amount of energy per second incident to a given area.   So can you can increase the intensity by either increasing the electrical energy or making the area smaller? By increasing the electrical energy, what exactly is happening?  I assume that by increasing intensity you are increasing the number of photons incident to an area of the metal thus increasing the rate of photoelectric emission. But why doesn't increasing intensity via more electrical power lead to greater energy of the electrons, if by increasing intensity you are increasing the energy of the incident waves?  I understand the results but I'm trying to understand why the intensity works the way it is thanks