Find all pairs of integers $(n,k)$ that satisfy $(n+1)^k - 1 = n!$ Find all pairs of integers $(n,k)$  that satisfy $(n+1)^k - 1 = n!$  It is easy to see that (1,1) and (2,1) are solutions by inspection,  but how do we prove that these are the only solutions? After sometime, I also saw that $(4,2)$ was also a solution, as it wasn't so obvious.  I was thinking that maybe $n! + 1$ cannot be a perfect power of a number? Though I also can't prove it. I also tried binomial theorem but still cant see anything.
To find all $n$ such that $(n-1)!+1$ is a perfect power of $n$ How to find all positive integers $n$ such that $(n-1)!+1$ can be written as $n^k$ for some positive integer $k$?
Determine whether $712! + 1$ is a prime number or not Let $n = 712! + 1$ If $n$ was a prime number then, by Wilson's theorem:  $ (712!)! \equiv -1 \pmod{712}$  The double factorial makes it seriously more difficult...  But We can require:  $$712!! + 1 \equiv 0 \pmod{712}$$  $$712 = 2\cdot 356$$   Hence: $712!! = 2^{356} 356!$  Lets compute: $356! \pmod{712}$. Obviously:  $$356! = (356 \cdot 2) \cdot (355!/2)  = 712 \cdot 355!/2$$  Then: $356! \pmod{712} \equiv 0$.   Now, $2^{356} \pmod{712}$  $2^{356} = 2^{6 + 350}$  Nevermind, what should I do? Understanding dummy (manual or automated) variable creation in GLM If a factor variable (e.g. gender with levels M and F) is used in the glm formula, dummy variable(s) are created, and can be found in the glm model summary along with their associated coefficients (e.g. genderM)  If, instead of relying on R to split up the factor in this way, the factor is encoded in a series of numeric 0/1 variables (e.g. genderM (1 for M, 0 for F), genderF (1 for F, 0 for M) and these variables are then used as numeric variables in the glm formula, would the coefficient result be any different?  Basically the question is: does R use a different coefficient calculation when working with factor variables versus numeric variables?  Follow-up question (possibly answered by the above): besides just the efficiency of letting R create dummy variables, is there any problem with re-coding factors as a series of numeric 0,1 variables and using those in the model instead? Finding $ \int^1_0 \frac{\ln(1+x)}{x}dx$ There is supposed to be a clean solution to the integral below, maybe involving some symmetry $$ \int^1_0 \frac{\ln(1+x)}{x}dx$$ I have tried integration by parts as followed: $\ln(x+1)=u$ ,$\frac{1}{x+1} dx = du$ and also $\frac{1}{x}dx=dv$, $\ln(x)=v$. Then, the integral becomes $$\int^1_0 \frac{\ln(1+x)}{x}dx=- \int^1_0 \frac{\ln x}{1+x}dx$$ which does not make this easier. I have also tried using the identity $\int_a^bf(x)~dx=\int_a^bf(a+b-x)~dx$. So let $I = \int^1_0 \frac{\ln(1+x)}{x}dx$ and also $I = \int^1_0 \frac{\ln(2-x)}{1-x}  $. Then $$2I= \int^1_0 \ln(1+x)+\ln(2-x)$$ which is not any easier, either. Any ideas? :) How do I add the kernel PPA? Publicity badges might be taking internal links into account? For some complicated diagrams, I need to use something like Adobe Illustrator or Inkscape, and would like to include nicely typeset labels. (I'm aware of post-processing solutions, such as pinlabel.) One way to do this is to use a closely-matching font, and then convert fonts to outlines. Better, however, would be to use Computer Modern directly in Illustrator.  Where can I find a Computer Modern font compatible with Adobe Illustrator? I've built a current sense differential amplifier that measures the voltage across a 1 Ohm resistor, to give me a reading for the current through it.  I've connected it as follows, all resistors are matched 10k to give a gain of 1:    (Image source: )  I am powering it from a single rail 12 volt source. The problem is even though this is a rail to rail opamp (Output Low swing from 5-20mV) I'm getting a constant error of 0.62V and it seems the opamp is not able to go below this value.   I have tried to resolve this by connected Vout to ground using a 10k resistor, but that makes no difference whatsoever.  I tried it with dual supply configuration (+8V, -8V) and that seems to resolve the issue. However I want to run this off a single supply rail, so this is not a viable option.  How can I resolve this? I'm trying to measure the voltage drop across a 1 ohm resistor to give me a reading for current through the circuit (0-1V represents 0-1 A) so I can set up a current limit. A better way (or practice) to doing this is also welcomed. The initial (but not ideal) fix I can think of, is to offset the readings by a few volts?  Edit:  Changed R1-R4 from 10k to 100k and I'm down to about 30-50mV! Return row only if value doesn't exist I am not mathematician by any means so this question might be rather stupid. I came across this Wikipedia article on  and found this bewildering formula,   $$1 + 2 + 3 + \dots  = - \frac1{12}$$  The article also says that "Ramanujan summation of a divergent series is not a sum in the traditional sense". I am wondering then what this summation actually implies?  Is there any interpretation of this confounding result in the physical world that someone not so math-savvy can understand? How to show via inductive proof (and without the use of the binomial theorem) that $ {n\choose k} \in \mathbb{N} $? How do I show via inductive proof (and without the use of the binomial theorem) that $$ {n\choose k} \in {\mathbb{N}} $$ $$\text{ for  all }\: n,k \in \mathbb{N}_0\: \text{ with }\: n≥k .$$ For my induction proof, I would show that$$  {n\choose k-1}+{n\choose k} = {n+1 \choose k}   $$ is true. However, how do I start the induction correctly with the base cases and the hypothesis? As far as I know, I cannot simply assume $$ {n\choose k-1}+{n\choose k}, $$  so how do I come to this derivation? What permissions should my website files/folders have on a Linux webserver? I have a problem called "The Prison Problem" that I need to explain to my 9-year-old cousin. I would think that he has just started learning about divisors and perfect squares, and as such, I have a proposed solution. Any input from you guys would be welcome, as to what is the best way to go about this.  Problem:  There was a jail with 100 cells in it, all in a long row. The warden was feeling very jolly one night and told his assistant that he wanted to give all the prisoners a wonderful surprise. While they were sleeping, he wanted the assistant to unlock all the cells. This should be done, he told the assistant, by putting the key in each lock, turning it once. The assistant did this, then came back to report that the job was done. Meanwhile, however, the warden has second thoughts. "Maybe I shouldn't let all the prisoners free," he said. "Go back and leave the first cell open, but lock the second one (by putting the key in and turning it once). Then leave the third open, but lock the fourth, and continue in this way for the entire row." The assistant wasn't very surprised at this request as the warden often changed his mind. After finishing this task, the assistant returned, and again the warden had other thoughts. "Here's what I really want you to do," he said. "Go back down the row. Leave the first two cells as they are, and put your key in the third cell and turn it once. Then leave the fourth and fifth cells and turn the key in the sixth. Continue down the row this way." The assistant again did as instructed. Fortunately, the prisoners were still asleep. As a matter of fact, the assistant was getting pretty sleepy, but there was no chance for rest yet. The warden changed his mind again, and the assistant had to go back again and turn the lock in the fourth cell and in every fourth cell down the row. This continued all through the night, next turning the lock in every fifth cell, and then in every sixth, and on and on, until on the last trip, the assistant just had to turn the key in the hundredth cell. When the prisoners finally woke up, which ones could walk out of their cells?  Proposed solution:  Ten cells are left open after this process. Every cell that is a perfect square will remain open (1, 4, 9, 16, 25, 36, 49, 64, 81, and 100). If a number is not a perfect square, then it has an even number of divisors, therefore it will be "toggled" an even number of times and end up where it started (closed). Perfect squares have an odd number of divisors, so they will end up the opposite of where they started (open). How to reset Canon 70D WiFi? Number of isosceles triangle. Dropdownlist control with s for asp.net (webforms)? Can anyone recommend a dropdownlist control for asp.net (3.5) that can render option groups? Thanks Why I can't connect to VPS after changing OS from CentOS to Ubuntu? How do I create an Excel (.XLS and .XLSX) file in C# without installing Microsoft Office? I'm trying modeling things like this using oracle:  An account can have 1~5 members  So I'll have a account table and a member table with a FK of account. But is that possible to check if the account have less than 5 members?  I've tried constraint and trigger but oracle doesn't allow sub queries in constraint or condition of trigger like  CHECK( SELECT ...) or IF count(*) &gt;= 5  which make things getting hard. I have complicated math expression:    Vertical spacing between text and fraction line is too small. I tried to define new \frac command (with amsmath package) with adjustable vertical spaces. As you can see, the result is not satisfying:    \documentclass[a4paper,fleqn,leqno]{article} \usepackage{polyglossia} \usepackage{color,pifont,fontspec} \usepackage[top=2.54cm,bottom=2.54cm,left=2.54cm,right=2.54cm]{geometry} \usepackage{tikz,tikz-3dplot,graphicx} \usepackage{mathtools,breqn,amsmath} \usepackage[fontsize=12]{scrextend} \usepackage{array,multirow} \usepackage{pgfplots} \DeclareMathSizes{12}{14}{14}{10} \newcommand{\myfrac}[3][0pt]{\genfrac{}{}{}{}{\raisebox{#1}{$#2$}}{\raisebox{-#1}{$#3$}}} \begin{document} \begin{dmath*}[breakdepth={0},style={\mathindent=0pt},spread={10pt}] {n=\myfrac[3pt]{\myfrac[3pt]{0,0625N+0,0215}{2}-0,043}{\myfrac[3pt]{\myfrac[3pt]{\scriptstyle 0,0625N+0,0215}{\scriptstyle 2}-0,043}{\myfrac[3pt]{\scriptstyle N}{\scriptstyle 2}-1}+0,0215}} \end{dmath*} \end{document}   How can I enlarge verticall spacing in whole fraction? How to derive this binomial identity?