I am trying to solve an exercise but I don't know how to do it at all :  Given $m,n \in \mathbb N$ with $m&gt;n&gt;0$ and $r = m \bmod n$ Find ($2^m-1) \bmod (2^n-1)$ (as a function of $r$)   I don't know at all where to start or what to do... How can the value of the remainder help me to find the required result ? Thank you.
For all $a, m, n \in \mathbb{Z}^+$,   $$\gcd(a^n - 1, a^m - 1) = a^{\gcd(n, m)} - 1$$
When I type uname -a, it gives the following output.  Linux mars 2.6.9-67.0.15.ELsmp #1 SMP Tue Apr 22 13:50:33 EDT 2008 i686 i686 i386 GNU/Linux   How can I know from this that the given OS is 32 or 64 bit?  This is useful when writing configure scripts, for example: what architecture am I building for? Changes to parameter not showing on Report Server after deployment Prove or disprove: For every integer a, if a is not congruent to 0 (mod 3), the a^2 is congruent to 1 (mod 3) Prove or disprove: For every integer a, if a is not congruent to 0 (mod 3), the a^2 is congruent to 1 (mod 3)  SO this is for abstract algebra and I am really struggling with this. Here are some of the definitions and theorems I think would apply.  -an integer a divides and integer b if there is an integer q such that b=aq -Let a and b be integers, with a>0. Then there exist unique integers q and r such that b= aq+r and 0&lt;=r  The question says hint: use 2 cases but I'm not really sure what that means. I'm guessing it wants me to look at it when a is congruent to 0 and when it's not? so i tried to do that but just ended up stuck. below is as far as i got with that attempt  Proof: Case 1 There should be some information about letting variables exist in certain number sets here  assume a is not congruent to 0 mod 3, then 3 does not divide a-0, therefore a-0 does not equal 3q for some q in the integers  I was trying to get this to wind down to thus 3 divides a^2-1 but even tho it seems true (I couldn't find a counter example) I couldn't figure out how to prove it with what I have.  Next i tried reversing it thinking I could contradict it or something.  I think this would be case 2? assume a is congruent to 0 mod 3, then 3 divides a-0 and a-0=3q for some q in the integers, then 3q=a since 0 is the additive identity, therefore 3 divides a,  and again stuck and my brain hurts a little..  Can anyone help me figure out how to work this. Dirac Delta function inverse Fourier transform We know that the Fourier transform of the Dirac Delta function is defined as $$\int_{-\infty}^{\infty} \delta(t) e^{-i\omega t} dt = 1,$$  and if I were to reconstruct the function back in time domain, the inverse Fourier transform is defined as  $$\delta(t) = \frac{1}{2\pi}\int_{-\infty}^{\infty} e^{i\omega t} d\omega.$$  How do I compute this integral analytically? Suppose you have a physical . It is true that as amplitude increases, the period increases. Can we demonstrate this fact without explicitly finding the period (which is pretty involved and pretty messy) in:    an intuitive fashion,  rigorously? Pivot rows into multiple columns On Windows 7 can one make UAC on an admin account prompt for the password, just like a limited account? I am one of the very few who likes the security of UAC and so I have the UAC set to "Always Notify" on my Admin account. But I would be even happier if the system would prompt me for a password, when notifying me.  I can get what I want by adding a second non-admin user and using that account.  But somehow it seems wasteful to have two accounts set up when I will always use the non-admin account day-to-day.  So I am wondering if there is any setting that would cause an Admin account to behave password-prompting-wise as a standard account? Ubuntu 16.04 has been hanging on me ~1x per day.  This happens when I am in the middle of web browsing or using a desktop application, not when booting.  When it does, the mouse pointer will still move freely, but clicking or keystrokes have no effect on my system until I do a hard reboot.  What is the best way for me to debug this?  Here is some information:  selah@selah-Precision-Tower-5810:~$ uname -a Linux selah-Precision-Tower-5810 4.4.0-59-generic #80-Ubuntu SMP Fri Jan 6 17:47:47 UTC 2017 x86_64 x86_64 x86_64 GNU/Linux   Also, in case it is relevant, I have a "very big" monitor, a Dell 42" at 3840x2160 resolution.  selah@selah-Precision-Tower-5810:~$ lspci | grep VGA 03:00.0 VGA compatible controller: NVIDIA Corporation GM107GL [Quadro K2200] (rev a2)     UPDATE:  Following Artyom's advice I found the following message in my error logs:   Apr 27 09:47:25 selah-Precision-Tower-5810 kernel: nouveau 0000:03:00.0: fifo: SCHED_ERROR 0a [CTXSW_TIMEOUT] Apr 27 09:47:29 selah-Precision-Tower-5810 kernel: nouveau 0000:03:00.0: fifo: SCHED_ERROR 0a [CTXSW_TIMEOUT] Apr 27 09:47:33 selah-Precision-Tower-5810 kernel: nouveau 0000:03:00.0: fifo: SCHED_ERROR 0a [CTXSW_TIMEOUT]   Which has let me to this bug which describes similar behavior: Show that $\sum_{n = 1}^{+\infty} \frac{n}{2^n} = 2$ Success of Bernoulli trials with different probabilities The cardinality of a language over a set of variables $V$ with $\#V= κ$ In predicate logic, to define a language, we consider the following $3$ sets:  (1) $V$, the set of variables;  (2) $F$, the sets of function symbols;  (3) $P$, the set of predicate symbols with their arities.   Now I am concidering about determine the cardinality of a language over a set with arbitary cardinality, $κ$(Note that $κ$ can be either finite, countablely infinite or uncountablely infinite ).  May I please ask:  If the cardinality of $V$, $\#V = κ$, what conditions on $V$ , $F$ and $P$ will give us a language over $V$, $F$ and $P$ with cardinality $κ$?  And may I please ask if it is the fact that if $\kappa$ is finite, then to get a language with cardinality $\kappa$ we need $(V+F+P)^{&lt;\omega}=(\kappa+F+P)^{&lt;\omega}=\kappa$, so we can never have a language of cardinality $\kappa$. I think it is something weird here. Am I wrong with something?  I had googled it but I have get nothing from websites, now I am confused. Could someone please help? Thanks so much!  EDIT: Latter I am told that in my course, the cardinality of a language refers to the set of expression instead of this of the set of strings. So now I need to consider further about when a string makes sense.  EDIT':Now I am considering that if we can find out a general way to determine the cardinality of expression, which may related to combinatorics. See this question:  Any help would be appreciate. Find all functions $f$ such that $f:\mathbb{N}\rightarrow\  \mathbb{N}$ and $f(f(n))+f(n+1)=n+2$   Let us plug in $n=1$  $f(f(1))+f(2)=3$  Since the function is from $\mathbb{N}$ to $\mathbb{N}$, $f(2)$ can only take the values $1,2$. Now we divide the problem into cases.  Case-1: $f(f(1))=2,f(2)=1$  We can assume that $f(1)=c$ for the time being. Then plugging in $n=3$ and using $f(2)=1$ gives $$f(3)=4-c$$ and again since the range of the function is positive integers,then $4-c$ has to be positive and hence $c$ belongs to  {$1,2,3$}. Now, $$f(1)=c$$$$\implies f(f(1))=f(c)$$$$\implies 2=f(c)$$ by the assumption that $f(f(1))=2$ . Now,since $c$ can only take the values $1,2,3$,we start treating cases. If $c=1$,$$f(c)=2$$$$\implies f(1)=2$$ but we know from the deinition of $c$ that $f(1)=c=1$,a contradiction.If $c=2$,then $2=f(c)=f(2)$ but $f(2)=1$ by assumption. Finally,if $c=3$ $$2=f(c)=f(3)$$ but $$f(3)=4-c=4-3=1$$ which is once again a contradiction. Therefore there are no such functions in this case.  Case-2: $f(f(1))=1,f(2)=2$  Again assuming $f(1)=c$ and using $f(n)\le n$ along with plugging $n=c-1$ will give us that $f(1)=1$ and then it is easy to prove that such a function exists by recursion. I can only give a "sort of recursive" way to define the function. Here it goes $$f:\mathbb{N}\rightarrow \mathbb{N}$$$$f(1)=1$$$$f(n)=n+1-f(f(n-1))$$  But this case is harder to deal with.Some help will be appreciated. ffmpeg is not even cutting this and other mp4s to the nearest second Suppose one is trying to find the radius of convergence of  $$ \exp(x)=\sum\limits_{k=0}^{\infty} \frac{x^k}{k!} $$  This "proof" was given in the lecture: $$ \begin{align} r =&amp; \frac{1}{\limsup\limits_{n \rightarrow \infty}\sqrt [n]{|a_n|}} \\ =&amp; \frac{1}{\limsup\limits_{n \rightarrow \infty}\sqrt [n]{\frac{1}{n!}}} \\ =&amp;\lim\limits_{n \rightarrow \infty} \sqrt [n]{n!}\\ r=&amp; \infty \end{align}$$  Other proofs involving the ratio test also seem to inspect the convergence of $\frac{1}{k!}$ instead of $\frac{x^k}{k!}$. Why is one allowed to substitute $\frac{1}{k!}$? I have to move a MongoDB Collection (Size: 1TB+) from Sharded MongoDB Cluster to another[5 shards on both clusters]. During this operation, it is assumed that the accompanying program will be offline, so no need to worry about inconsistent data, but I need to minimize the downtime.  I've tried testing it with mongodump and mongorestore, but looks like that is taking ages to finish.  Please share your experiences for such kind of scenarios.  Note: Cluster has only one collection, so I am open at cluster level sync as well. We have to prove that  "the product of two numbers ending in 6 also ends with 6" mathematically.  I have no clue how to start. I don't want you to prove it for me! but some hints would be very helpful as I'm totally stuck on how to start this prove. Requesting review of technically inaccurate answers What are good out of the box alternative geocoding solutions to Esri's services? Esri's  will retire at the end of the year, what are good GIS services that can replace the North American and US Streets geocoders?  I have totally excluded Esri's replacement, the  as it is a subscription service. How do I sort a list of dictionaries by a value of the dictionary? I have a list of dictionaries and want each item to be sorted by a specific value. Take into consideration the list: [{'name':'Homer', 'age':39}, {'name':'Bart', 'age':10}]  When sorted by name, it should become: [{'name':'Bart', 'age':10}, {'name':'Homer', 'age':39}] I would change the font of section and subsection of my document. I have created it using book class. I would use this font:    According to Adobe pdf propriety, this font is AvantGarde-Demi. How can I use it in Latex. what command I should use to recreate this in latex. I think I should use titlesec package right?