Other words, are there infinitely many primes $p$ such that $p-1$ is square-free. This seems likely to be true, but I can't seem to give an easy argument why. This set of primes and all primes 1 more an integer with a square divisor make up the set of primes, so at least one of these is infinite. I looked at this question for enlightenment but to no avail :  as 2 could be included in this product.
How to prove that $p-1$ is squarefree infinitely often, where $p$ is prime?  I had thought of using Dirichlet's theorem on arithmetic progressions.  The number of squareful values of $p-1$ less than or equal to $n$ is bounded by $\sum_{q}{\pi_{q^2,1}(n)}$, each term $\pi_{q^2,1}(n) \approx \frac{\pi(n)}{q \cdot (q-1)}$, and $\sum_{q}{\frac{1}{q \cdot (q-1)}} \lt 1$.  But using for example , the possibility is open that the error in each approximation for $\pi_{q^2,1}(n)$ may continue to increase with $q$, and since the sum runs up to $\sqrt{n}$, this argument is invalid.  Is there a more refined version of Dirichlet's theorem that can be used to circumvent this issue?  Or an entirely different way to prove this statement?
Proving that these two sets are denumerable. (a) $S_k=\{A\subset\mathbb{N}: |A|=k\}$ for $k\in\mathbb{N}$  (b) $S = \bigcup_{k=1}^\infty S_k$   Work: For (a), I am not too sure about what approach I should use. I think finding a bijective function between the $S_k$ and $\mathbb{N}$ is hard to do . I also thought about using the Schroeder-Bernstein theorem and find injective functions $f:\mathbb{N}\rightarrow S_k$and $g:S_k\rightarrow\mathbb{N}$ in order to prove that $S_k$ and $\mathbb{N}$ are numerically equivalent. However, I have a hard time finding and proving these injective functions. I haven't really looked at (b) as I have yet to complete (a). Ubuntu 20.04 run ok on Lenovo ThinkPad E15 laptop? Using first person or third person in papers? php SimpleXML check if a child exists How can I fix/repair a corrupted PDF file? Zip folder without including path to folder and require only one argument I'm trying to make a zip function that'll do all the steps I typically do when zipping a file or folder. Say my current directory is ~, and I want to zip a folder named Folder1, whose file path is ~/Folder2/Folder1.  Issuing  cd ~/Folder2 &amp;&amp; zip -r Folder1 Folder1 &amp;&amp; cd -   does exactly what I want. Namely, it zips so that the output Folder1.zip does not contain folders that lead up to the folder I'm zipping (namely, /home/user/Folder2), it zips recursively, and it takes in the first argument of zip with the exact name as the file or folder I'm zipping (Folder1.zip).  So I want to make a function in my .zshrc something akin to  coolerzip() { cd /path/to/file &amp;&amp; zip -r file file &amp;&amp; cd - }   where it need only take in one argument, which in this case, it will be ~/Folder2/Folder1. Then Folder1 will be the name for both files in coolerzip, and I will cd right into ~/Folder2 so Folder1 lies in my current directory before zipping. Is there any way I can achieve this so that coolerzip only requires one argument rather than three? Since I am a regular Stack Overflow follower, I came across this and was shocked a bit regarding the design and user experience perspective.  Attaching picture 1 - it states that I can click any site and view a synopsis about it. Clear!!!    Attaching Picture 2 - you can see that when I clicked Webmasters, the design wants me to scroll horizontally to view all information and basically it hides information.    It would be nice, if it did something like below : Let us denote by $\chi(G)$ the chromatic number, which is the smallest number of colours needed to colour the graph $G$ with $n$ vertices. Let $\bar{G}$ be the complement of $G$. Show that  (a) $\chi(G)+\chi(\bar{G}) \leq n+1$  (b) $\chi(G)\chi(\bar{G}) \geq n$  I was able to prove (a) using induction. Any hints on proving (b)? How to properly handle collision in a component-based game? Trying to wrap my head around the ways to properly handle collision in a game designed around components.  I see many examples have some sort of PhysicsComponent that gets added to the entity's list of components but the actual implementation is confusing me.  For this to work, the PhysicsComponent would need access to the world around it. This doesn't make intuitive sense to me. Shouldn't a component be unaware of not only its container (the entity), but its container's container (the world)?  To me, it sounds like the level or scene should maintain a list of these entities and every game update, loop through the entities to determine which collide.  My question is firstly, whether or not this is good design, and secondly, how to determine which entities can collide. I suppose solid entities could implement an empty IRigidBody interface so the level could determine which entities in the list support collision. But is this breaking the component design?  Instead, should they contain an empty RigidBody component? This may actually be better because it may not always be empty and this approach is more future-proof. The only problem with this is the complexity. The scene would have to loop through not only every entity, but also every entity's components to determine if it had this RigidBody component.  Thirdly, when they do collide both entities should be informed somehow and I am unsure on how to accomplish this either.  Let's say that both entities contained a HealthComponent and when they collided both their healths would be decreased by some arbitrary value, 5. I suppose it would be the scene's responsibility to handle this when it detects a collision between two entities?  But then is the scene responsible for too much? I could see this possibly getting out of hand and becoming unwieldy when the scene is responsible for many things entities should not (?) have access to.  Edit: Question updated with more details. A scene I made is not rendering, at all. This grey screen stays the same and nothing changes.  Here are my settings:  [ How to calculate Network Service Areas in QGIS? I want to make a bootable USB thumb drive to install Windows on my computer. I don't want to use my old DVD to create an image and then upgrade to SP1, so I decided to download an untouched .iso file that already has SP1 in it. My question is this, is there a difference between these two files files?     en_windows_7_home_premium_with_sp1_x64_dvd_620851.iso   vs     en_windows_7_home_premium_with_sp1_x64_dvd_u_676549.iso   What does dvd_u mean, is it somehow different from plain dvd? Also will these work with my retail Home Premium key or do these work only for thoe with TechNet/MSDN accounts? What happens if a person can't die of a heart attack? This is all in iTerm2 on OS X.  I used to have  export TERM='xterm-256color'   in my .bashrc. This meant Vim in tmux did use 256 colors.  And once I added  set -g xterm-keys on   then keyboard shortcuts with modifiers worked fine in Vim. Namely: shift+left/right that I mapped to switch Vim tabs, ctrl+up/down that I mapped to move ("bubble") lines, and shift+left/right which worked out of the box to jump by word in the command-line mode (e.g. when typing something like :e foo bar baz).  However, this setup had the problem that the Vim background color only shows behind text, as mentioned .  So I removed  export TERM='xterm-256color'   from my .bashrc and instead put this in my .tmux.conf:  set -g default-terminal "screen-256color"   That fixed the Vim background color, but broke the keyboard shortcuts - they do unexpected things (move the cursor, delete text) instead.  By using ctrl+v to insert the verbatim output from the key combinations (as described ), I was able to work around it:  map [1;5A &lt;C-Up&gt; map [1;5B &lt;C-Down&gt; map [1;2D &lt;S-Left&gt; map [1;2C &lt;S-Right&gt; cmap [1;2D &lt;S-Left&gt; cmap [1;2C &lt;S-Right&gt;   This makes the shortcuts work, but it doesn't feel like the right solution. Could anyone tell me what's happening here and how to fix it? $a$ and $b$ are natural numbers and $p$ is a prime number. Prove that if $a^{p}-b^{p}$ is divisible by $p$, then it is also divisible by $p^{2}$.  My attempt: Based on Fermat's theorem $(a^{p}-a)$ and $(b^{p}-b)$ are divisible by p, therefore their difference is divisible by $p$. I.e. $(a^{p}-a)-(b^{p}-b)=(a^{p}-b^{p})-(a-b)$ is divisible by $p$. Since the left hand side of this equation is divisible by $p$, the right hand side also should be divisible by $p$. Based on the problem's assumption $(a^{p}-b^{p})$ is divisible by $p$ therefore we conclude $(a-b)$ must be divisible by $p$. Now We can factorize $(a^{p}-b^{p})$ $$(a^{p}-b^{p})=(a-b)(a^{p-1}+a^{p-2}b+....+b^{p-1})$$ we have to prove $(a^{p}-b^{p})$ is multiple of $p^2$ which means we have to show the right hand side should be multiple of $p^{2}$. But $(a-b)$ is multiple of p therefore we have to show  $(a^{p-1}+a^{p-2}b+\cdots+b^{p-1})$ is multiple of $p$ and I am stuck here. Any help would be appreciated. Suppose that $X_1,X_2,\ldots,X_n$ is a random sample from a distribution with probability function:  $$p_X(x)=\begin{cases} 1/2 &amp;x=-1,1 \\ 0 &amp; \text{otherwise} \end{cases}$$  Now if we define the average   $\bar{X}= \frac{1}{n} \sum X_i$, I need to show that that for an odd $n$ the probability function for $\bar{X}$ is:  $$p_{\bar{X}} (x)= \frac{\binom{n}{\frac{n}{2} (x+1)}}{2^n}$$ for $x=\pm 1/n,\pm 3/n,\ldots,\pm 1$, $0$ otherwise.    Since $P[\sum X_i=k] =P [\bar{X}=k/n ]$, I thought it would be easier to first derive the probability function for the sum. Since they are still $2^n$ n-tuples and each one is equilikely, I need to count the ones whose sum is $k$ for:  $k=\pm 1,\pm 3,\ldots \pm n-2,\pm n$ (since even numbers are not a feasible combination)  The problem here though is that I do not immediately recognize how I could count all these tuples. Could you please help me with this? If it is easier to proceed another way, I am of course all ears. Prove the summation:  $$\sum_{i=1}^n i! \cdot i = (n+1)! - 1$$ using induction.   base case:  $n=1$: \begin{align*} \sum_{i=1}^1 i! \cdot i &amp;= (1+1)! - 1 \\ 1 &amp;= 2 - 1 \\ 1 &amp;= 1 \end{align*}   This is a question from my test review packet, currently have the base case completed and I am a bit lost on where to go from there. Any help/hints are appreciated. When I visit  I see this:     The "Congratulations! You've earned (badge)" section has dropped down completely beneath the tabs under it. The "You've earned a new privilege!" section has a gap at the bottom:     Shog9 says in comments how to reproduce this:     Gotta have earned a badge, not dismissed the notification, and be looking at your own profile. Working in Cycles.  Say I have a "disco ball" with a glossy shader. I'm pointing a spot light to it, and I'm trying to have the reflections of that light illuminate the diffuse material on the walls with the patterns from the ball.    I get this:    but I'd like to get something like:    For the illustration I placed a point light inside the ball an removed faces. But what I'd like to do is move the light around the ball so that the light reflected from it changes like it would do in real life.   Is that possible at this point?  EDIT:  Just for the record here's the image rendered in LuxRender (yet another program to learn...) What does "All things X" mean? I've seen the expression in some company names; for example , .  Is the expression "All things X" a parody of some other popular phrase, or does it have its own meaning? Or is it just a misuse of grammar?