Show that infinitely many positive integer pairs $(m,n)$ exist such that $\frac{m+1}{n}+\frac{n+1}{m} \in \mathbb{N}$.  I couldn't solve it but I did make an observation which might or might not be helpful. WLOG assume that $m &lt; n$ (since the term is symmetric wrt $m,n$ along with the fact that ignoring the $m=n$ case wouldn't create any trouble). Write $n=mq+r$ for some $r\in \{[0,m-1]\cap \mathbb{N}\}$. Now, the question boils down to showing $\frac{m+1}{n}+\frac{r+1}{m} \in \mathbb{N}$. Note that each of the summand is $\leq 1$. Since the equality case wouldn't be helpful in generating infinitely many pairs of $(m,n)$, we can safely say that the sum is equal to $1$.  Now I don't know how to proceed from here.
The question is:     Show that there are an infinite number of pairs $(m,n): m, n \in \mathbb{Z}^{+}$, such that: $$\frac{m+1}{n}+\frac{n+1}{m} \in \mathbb{Z}^{+}$$   I started off approaching this problem by examining the fact that whenever the expression was a positive integer, the following must be true:  $$\frac{m+1}{n}+\frac{n+1}{m} - \left\lfloor\frac{m+1}{n}+\frac{n+1}{m}\right\rfloor = 0$$  However, I was unable to do much more with this expression, so I abandoned it and started the problem again from a different angle.  Next I re-arranged the expression to state that:  $$\frac{m^2+m+n^2+n}{mn} \in \mathbb{Z}^{+} \implies \frac{m(m+1) + n(n+1)}{mn} \in \mathbb{Z}^{+}$$  And therefore:  $$mn \mid (m(m+1) + n(n+1))$$  However, I'm unsure how to demonstrate there are infinitely many occurances of $(m, n)$ for which this is true. So I'd appreciate any help.   Thanks in advance
The laptop is dual booting with windows 7 and ubuntu 11.10.  For 11.10 to boot, I had to add 'nolapic' in GRUB boot options.  I downloaded 12.04LTS LiveCD yesterday and made a CD and a bootable USB stick.  With both USB and LiveCD, occasionally I can boot into 12.04.  But many times, the boot process stuck at some point.  For USB boot, sometime it works.  But other times, it seems to stuck right after seeing the message:  NMI watchdog enabled. takes one hw-pmu counted [0.408590] #2   The number in [] changes.  Some times it stuck at #2, and sometimes at #3.  For LiveCD boot, sometime it works.  But when it stuck, it stuck at different places.  At times, the boot process go directly to a blank screen with a blink '_' at up left corner.  Sometime it went a bit further, after showing options to try/install ubuntu then stuck with a blank screen with blink '_'.  Anything I can do to get around of this? Is there a website with historical temperatures and weather data? I am interested in hiking this fall in France and camping along the way. In order to figure what gear I need to pack to camp (particularly for the sleeping bag), I would like to know what temperatures to expect.  I am interested in historical temperatures to figure the usual temperature and I might look at them for other seasons. If that matters, I am thinking about hiking in the Pays Basque (South West of France), the mountainous part.  Where (ideally online and for free) can I find such data? Are there resources that cover Europe, or even the whole world? How to run find -exec? I'd like to find the files in the current directory that contain the text "chrome".  $ find . -exec grep chrome find: missing argument to `-exec'   What am I doing wrong? In 1D why are vectors, such as, e.g., velocity $v$ and acceleration $a$, not written with a vector arrow? In my textbook instantaneous velocity and acceleration are wrote only as $v$ and $a$ in the chapter of One dimensional motion. Instead, in two dimensional motion chapter they are written with the "arrow" symbol upon them, $\vec{v}$ and $\vec{a}$.  Why if they are both vectors they don't have an arrow in that particular chapter? I just can't get it. Proving that $\lim_{k\to \infty}\frac{f(x_k)-f(y_k)}{x_k-y_k}=f'(c)$ where $\lim_{k\to \infty}x_k=\lim_{k\to \infty}y_k=c$ with $x_k<c<y_k$ Question.  Suppose that a function $f$ is defined on an interval $I$, $c$ is a point of $I$, and {$x_k$} and {$y_k$} are any two sequences in $I$ such that $\lim_{k\to \infty}x_k=\lim_{k\to \infty}y_k=c$ with $x_k&lt;c&lt;y_k$ for all $k$. Prove that, if $f'(c)$ exists, then $\lim_{k\to \infty}\frac{f(x_k)-f(y_k)}{x_k-y_k}=f'(c).$   My Attempt.  Let $\epsilon_k=\frac{1}{k}$ be given.  For $\epsilon_1=1,$ there exists $\delta_1&gt;0$ such that $x_1,y_1\in I, x_1,y_1\in N'(c;\delta_1)$ implies $\vert \frac{f(x_1)-f(y_1)}{x_1-y_1} -f'(c) \vert&lt;\epsilon_1.$  For $\epsilon_2=\frac{1}{2},$ there exists $0&lt;\delta_2&lt;$min$\{\vert x_1-c \vert,\vert y_1-c \vert,\epsilon_2\}$ such that $x_2,y_2\in I, x_2,y_2\in N'(c;\delta_2)$ implies $\vert \frac{f(x_2)-f(y_2)}{x_2-y_2} -f'(c) \vert&lt;\epsilon_2.$  Similarly, for $\epsilon_{k}=\frac{1}{k},$ there exists $0&lt;\delta_k&lt;$min$\{\vert x_{k-1}-c \vert,\vert y_{k-1}-c \vert,\epsilon_k\}$ such that $x_k,y_k\in I, x_k,y_k\in N'(c;\delta_k)$ implies $\vert \frac{f(x_k)-f(y_k)}{x_k-y_k} -f'(c) \vert&lt;\epsilon_k.$  Then by taking the limit to infinity, we obtain $\lim_{k\to \infty}\frac{f(x_k)-f(y_k)}{x_k-y_k}=f'(c).$   It really took me a lot of time to finish the proof and I need this to be checked by an expert:) Is the proof okay? Any response would be appreciated! Why can't photons have a mass? Could you explain this to me in a short and mathematical way? What we've gelost — why doesn't English use the prefix "ge-"? The Germanic languages that I'm familiar with all use a prefix similar to ge- on past participles:  German: Ich habe mir den Fuß gebrochen. Dutch: Ik heb mijn voet gebroken.  But English doesn't do this at all:  English: I've broken my foot.  Where did this prefix come from? Did English ever have the ge- past participle prefix? If so, why was it lost? Photon energy - momentum in matter $E = h\nu$ and $P = h\nu/c$ in vacuum. If a photon enters water, its frequency $\nu$ doesn't change. What are its energy and momentum: $h\nu$ and $h\nu/c$ ? Since part of its energy and momentum have been transferred to water, it should be less.  If water's refractive index is $n$, are the energy and momentum equal to $h\nu/n$ and $h\nu/c/n$ ? How do I submit .jar file to the ubuntu software center I am writing an app using JAVA Swing for all platforms. After that I want to add this app to Ubuntu Software Center for free Is it possible?   My app is just one *.jar file, its helping to play card games.  I dont know Python language and saying the truth - dont want to learn it now.  What should I do to be able to submit my app? Let's say we have this code for a LWC:  &lt;!-- helloExpressions.html --&gt; &lt;template&gt;     &lt;div class="slds-m-around_medium"&gt;         &lt;lightning-input name='firstName' label="First Name" onchange={handleChange}&gt;&lt;/lightning-input&gt;         &lt;lightning-input name='lastName' label="Last Name" onchange={handleChange}&gt;&lt;/lightning-input&gt;         &lt;p class="slds-m-top_medium"&gt;Uppercased Full Name: {uppercasedFullName}&lt;/p&gt;     &lt;/div&gt; &lt;/template&gt;   and this is the JS file:  // helloExpressions.js import { LightningElement, track } from 'lwc';  export default class HelloExpressions extends LightningElement {     @track firstName = '';     @track lastName = '';      handleChange(event) {         const field = event.target.name;         if (field === 'firstName') {             this.firstName = event.target.value;         } else if (field === 'lastName') {             this.lastName = event.target.value;         }     }      get uppercasedFullName() {         return `${this.firstName} ${this.lastName}`.toUpperCase();     } }   firstName and lastName have been decorated with @track. Now as per the documentation, decorating a property with @track makes it private and reactive, while to mark a property public and reactive, use @api.  So if I just change the decorator to @api for both these variables, it should work. I am referring these from the same component only. But it's not working.  UPDATE: 27th Feb -------------------- Somehow it's working today. I just changed the decorator of the variables to api.  import { LightningElement, api } from 'lwc';  export default class HelloExpressions extends LightningElement {     @api firstName = '';     @api lastName = '';      handleChange(event) {         const field = event.target.name;         if (field === 'firstName') {             this.firstName = event.target.value;         } else if (field === 'lastName') {             this.lastName = event.target.value;         }     }      get uppercasedFullName() {         return `${this.firstName} ${this.lastName}`.toUpperCase();     } }   So it's working. "I'm not sure what the right way is" I believe the sentence in the title is grammatically correct.   Recently I've seen too many people writing it this way:     I'm not sure what is the right way.   Is it grammatically correct as well? Maybe in a different sense? Or simply misuse by non-native speakers? AES Mix Column Transformation Section 3.4.3 "The MixColumns Step"(page 39) of , it states     The columns of the state are considered as polynomials over   $\operatorname{GF}(2^8)$ and multiplied modulo $x^4 + 1$ with a fixed polynomial $c(x)$.      The polynomial $c(x)$ is given by   $c(x) = \mathtt{03} \cdot x^3 + \mathtt{01} \cdot x^2 + \mathtt{01} \cdot x + \mathtt{02}$      Let $b(x) \equiv c(x) \cdot a(x) \pmod{x^4 + 1}$. Then         How this polynomial multiplication modulo $x^4+1$ is converted to Matrix Multiplication? What is the C# equivalent of friend? Possible Duplicate:          I'd like the private member variables of a class to be accessible to a Tester class without exposing them to other classes.   In C++ I'd just declare the Tester class as a friend, how do I do this in C#? Can someone give me an example? Bash scripts ran from from gnome/nautilus don't have environment variables In my ~/.bashrc and ~/.profile I have some variables set to some filepaths:  export MY_VAR1="/path/to/somewhere"   In my ~/bin/ (which is added to my PATH) I have a few scripts that rely on those variables being set.  If I run those commands from the terminal, they work as expected.  However, if I try to double click on a script in Nautilus and run it, it runs as if those variables were not set.  Same thing happens if I add one of the scripts to Gnome's "Startup Applications" using the GUI utility or if I add them to the Gnome menu using the "Main Menu" GUI utility.  I would've thought that since I included these variables in my ~/.profile, which is run at login that gnome and nautilus and processes spawned by them would have access to those variables.  Both Gnome and Nautilus are run under my login user. All my scripts in ~/bin have the bash shebang at the top.  I am new to bash scripting, sorry if I am missing something obvious. I am using Debian Wheezy and Gnome3. This looks like a question asked , but it isn't.  $$T(n)=\begin{cases} T (\sqrt{n}) + 1 \quad &amp; \text{ if } n&gt;1 \\  1 &amp; \text{ if }n=1\end{cases}$$  My professor gave this to me in class yesterday. This is where I'm stuck:   $$T(n) =  T(n^{1/2})+1 =  T(n^{1/4})+2 = T(n^{1/8})+3 = \dots =   T(n^{1/ 2 ^k})+k$$  Now this will continue till  $$n^{1/ 2 ^k}=1$$  How do I proceed ahead? If I take log on both sides, then RHS becomes zero. How do I solve the relation?  I don't want to use the Master theorem, I want to know where and why am I stuck. Prove that, $\binom{3n}{r}=\binom{3n-1}{r} + \binom{3n-1}{r-1}$ Reading through my textbook I came across the following problem and I am looking for some help solving it.  Prove that,  $$\binom{3n}{r}=\binom{3n-1}{r} + \binom{3n-1}{r-1}$$  In a previous problem I solved the following but I wasn't sure how to use it to prove the question above. Prove,  $$\binom{n}{r}=\binom{n}{n-r}$$  Solution:  left Side,  $$\binom{n}{r}=\frac{n!}{(n-r)!r!}$$  right side,  $$\binom{n}{n-r}=\frac{n!}{(n-(n-r))!(n-r)!} = \frac{n!}{(n-r)!r!}$$  Therefore completing the proof,  but how do I solve my first question? thanks! Is $0$ a natural number? Is there a consensus in the mathematical community, or some accepted authority, to determine whether zero should be classified as a natural number?  It seems as though formerly $0$ was considered in the set of natural numbers, but now it seems more common to see definitions saying that the natural numbers are precisely the positive integers. I am facing problem to enable wifi on my Lenovo G50-30 Below are the command out put please help me out   lspci | grep -i wireless:  02:00.0 Network controller: Realtek Semiconductor Co., Ltd. RTL8723BE PCIe Wireless Network Adapter     nm-tool:  NetworkManager Tool  State: disconnected  - Device: eth0    Type:              Wired   Driver:            r8169   State:             unavailable   Default:           no   HW Address:        28:D2:44:81:36:0B    Capabilities:     Carrier Detect:  yes    Wired Properties     Carrier:         off   - Device: wlan0   Type:              802.11 WiFi   Driver:            rtl8723be   State:             unavailable   Default:           no   HW Address:        14:2D:27:21:8D:47    Capabilities:    Wireless Properties     WEP Encryption:  yes     WPA Encryption:  yes     WPA2 Encryption: yes    Wireless Access Points      nmcli nm:  RUNNING         STATE           WIFI-HARDWARE   WIFI       WWAN-HARDWARE   WWAN       running         disconnected    disabled        disabled   enabled         disabled       sudo rfkill list:  0: ideapad_wlan: Wireless LAN     Soft blocked: no     Hard blocked: yes 1: ideapad_bluetooth: Bluetooth     Soft blocked: no     Hard blocked: yes 2: hci0: Bluetooth     Soft blocked: no     Hard blocked: no 3: phy0: Wireless LAN     Soft blocked: no     Hard blocked: no Comparing integers: arithmetic expression or conditional expression In Bash, two integers can be compared using conditional expression     arg1 OP arg2      OP is one of -eq, -ne, -lt, -le, -gt, or -ge. These   arithmetic binary operators return true if arg1 is equal to, not equal   to, less than, less than or equal to, greater than, or greater than or   equal to arg2, respectively. Arg1 and arg2 may be positive or negative   integers.   or arithmetic expression:     &lt;= &gt;= &lt; &gt; comparison      == != equality and inequality   Why do we have two different ways for comparing two integers? When to use which?  For example, [[ 3 -lt 2 ]] uses conditional expression, and (( 3 &lt; 2 )) uses arithmetic expression. Both returns 0 when the comparison is true  When comparing two integers, can these two methods always be used interchangeably? If yes, why does Bash have two methods rather than one? error while loading shared libraries: libssl.so.1.1 (and libcrypto.so.1.1) As you can see I'm missing libcrypto and libssl files. I locked out myself from my server after running apt-update and apt-upgrade, I'm running this from web console, because ssh nor any other program won't work because of missing SSL files.  I tried removing and installing apt install openssl so did libssl-dev and/or compiling own openssl from a distro. None of these create the needed .so files, I ran locate, those files are not present anywhere on the server. There's also libssl1.0.0 which cannot be reinstalled and removing it would cause removal of 700MB of packages that would literally kill server and delete all data.