Determine all positive integer solutions for $\frac 1x + \frac 1y + \frac 1z + \frac 1{xy} + \frac1{yz} + \frac 1{xz} + \frac 1{xyz} = 1$
Math contest integer triplet problem: $x + y + z + xy + yz + xz = xyz − 1$
Submodule of free module over a p.i.d. is free even when the module is not finitely generated? Let $A$ and $B$ be two real $n\times n$ matrices such that $AB-BA=A$ Prove that $A$ is not invertible and that $A$ is nilpotent.  My attempt is the following. It holds that $AB=(B+I)A$ If $A$ were invertible, $B=A^{-1}(B+I)A$. Taking trace on both sides yields $tr(B)=tr(B)+n$, hence $n=0$, which is a contradiction. I can't prove that $A$ is nilpotent though. I'm reading the proof of the multiplicative property of $$s_k(n)=\sum_{d|(n,k)}f(d)g\bigg( \frac kd\bigg)$$ The book wrote that in order to understand the proof, we need to know if $a,b,k,m$ are integers such that $(a,k)=(b,m)=1$, then $$(ab,km)=(a,m)(b,k).$$  I have written proof, but I don't know is it consistent, and this seems used more tools than it necessarily has to.   My proof: Let $$S_1=\{d: d|(a,m), d&gt;0\}\\S_2=\{d: d|(b,k), d&gt;0\}\\S=\{d:d|(ab,km), d&gt;0\}.$$ I will prove the statement by showing that the function $\phi:S_1\times S_2\to S$ defined as $$\phi(d_1,d_2)=d_1d_2$$ is bijective, and complete the proof by connecting the maximal elements from these partial ordering sets.  Obviously, for $d_1\in S_1, d_2\in S_2$, this tells us $d_1 $ is both divisor of $a$ and $m$, and $d_2$ is both divisor of $b$ and $k$, hence $d_1d_2$ is both divisor of $ab$ and $km$, so we can tell $\phi(S_1\times S_2)\subseteq S$.  To show one-to-one, if we have $c_1,d_1\in S_1$ and $c_2,d_2\in S_2$ such that $$\phi(c_1,c_2)=\phi(d_1,d_2)$$ then $$c_1c_2=d_1d_2$$  In particular, $c_1|d_1d_2$, we need to show $c_1|d_1$. Indeed, if $(c_1,d_2)&gt;1$, this implies $a,k$ have common factor larger than $1$, contradicts the assumption. Therefore $c_1|d_1$. A similar argument implies $d_1|c_1$, therefore $c_1=d_1$. Again a similar argument implies $c_2=d_2$, which shows one-to-one.   To show onto, let $d\in S$, since $d|ab$, we can split $d$ into $d_1d_2$ such that $d_1|a$ and $d_2|b$. Since $(a,k)=1$, we see $(d_1,k)=1$, by property of $S$ we see also $d_1|m$, therefore $d_1\in S_1$, a similar argument implies $d_2\in S_2$. So every number in $S$ has a preimage in $S_1\times S_2$  This completes the proof. Is there any mistake in the proof?     Edit: A proof about the statement: If $d|ab$ then there are $d_1,d_2$ such that $d=d_1d_2$ and $d_1|a, d_2|b$.  Let $ab=p_1^{e_1}\cdots p_r^{e_r}$ for distinct prime factors $p_1,\cdots,p_r$, then we can write $$a=p_1^{x_1}\cdots p_r^{x_r}\\b=p_1^{y_1}\cdots p_r^{y_r}$$ where some of $x_i, y_i$ can be zero, and $x_i+y_i=e_i$ for all $i=1,\cdots,r$.  Since $d|ab$, we can write $$d=p_1^{f_1}\cdots p_r^{f_r}$$ with $f_i\leq e_i$ for all $i=1,\cdots,r$. Then since $$f_i\leq x_i+y_i$$ Split into two cases:   Case 1: $f_i&gt; x_i$, then we can choose $g_i=x_i$ and $h_i=f_i-x_i$.  Case 2: $f_i\leq x_i$, then we can choose $g_i=f_i$ and $h_i=0$.  Then the choice $$d_1=\prod_{i=1}^r p_i^{g_i}, \quad d_2=\prod_{i=1}^r p_i^{h_i}$$ will prove the statement. What are the differences between "su", "sudo -s", "sudo -i", "sudo su"? I'm trying to downgrade my iPhone from the current iOS to an older one, as per .  But every time I try, I always receive this error:     The iPhone could not be restored. This device isn't eligible for the requested build.     I have tried many times with and without DFU mode. Any ideas regarding this? cp - cannot create regular file : permission denied I want copy to /lib/udev but gives error  cp -f /tmp/ok_pcscd_hotplug.sh /lib/udev  cp:cannot create regular file /lib/udev/ok_pcscd_hotplug : Permission denied   I am root , but I can't understand . what can I do ? tabular input by columns (i.e. transpose a table) I would like to be able to transpose a tabular(x) table like in the following example. It is not necessary to be completely automated, but I don't want to change the data order, which is a painful and stupid. By "not completely automated" I mean that e.g. the headers can be done seperately etc.  Let's see the MWE:  \documentclass{article}  \usepackage{booktabs} \def\Midrule{\midrule[\heavyrulewidth]}  \begin{document}  \centering  \begin{tabular}{*6c}\toprule \bfseries Name &amp; Alice &amp; Bob &amp; Chuck &amp; Dave &amp; Eve\\\Midrule \bfseries Sex &amp; Female &amp; Male &amp; Male &amp; Male &amp; Female\\\midrule \bfseries Age &amp; 18 &amp; 19 &amp; 20 &amp; 21 &amp; 22\\\bottomrule \end{tabular}  \bigskip  Which I want to look like  \bigskip  \begin{tabular}{*3c}\toprule \bfseries Name &amp; \bfseries Sex &amp; \bfseries Age\\\Midrule Alice &amp; Female &amp; 18\\\midrule Bob &amp; Male &amp; 19\\\midrule Chuck &amp; Male &amp; 20\\\midrule Dave &amp; Male &amp; 21\\\midrule Eve &amp; Female &amp; 22\\\bottomrule \end{tabular}   \end{document}     Note: I know that , but I would like to get it without expl3 (mostly because people would might use my code quite often have an old LaTeX installed and I cannot be convincing them to change this). Is there a difference between Bayesian and Classical sufficiency? The title pretty much says it all. I wonder whether there is any difference in the way Bayesians understand sufficiency vs. the way orthodox statistics understands sufficiency, or are they equivalent?  If there is a difference, what is it? Prove that the equation $x^4+(2k^2)x^2+2k^4$ is not a perfect square How does the current know how much to flow, before having seen the resistor? With the following circuits as examples :    and    How will the current I know how much to flow? Would any other wave travel first in the circuit and then come back and say so much current should flow? How can one prove that the series $\displaystyle \sum\limits_{n=1}^\infty \frac{1}{(n!)^2}$ converges to an irrational number? There's no need to use Taylor expansion, integrals or any advanced/professional techniques. It can be proved using only basic techniques. My first attempt was to assume that the series does converge to a rational number $ \frac{p}{q}$ and then it follows by definition that for every positive epsilon there exists an integer k such that $ {\sum_{n=1}^{k} \frac{1}{(n!)^2} -\frac{p}{q}}&lt;=\epsilon$ I tried to get a contradiction but I failed. How can I get a pure white background in jewelry photography, without losing sparkle and shine? I've read through several threads similar to this, but I'm looking for a comparison of two options here.   I currently take photos on a raised glass platform against a white foam board background and run about 300w of lights (3 100x LEDs @ 5000k, Home Depot Specials in softboxes and umbrellas). As you can imagine, it takes me about 15 minutes of post processing to achieve a perfect white background, and sometimes I have a particular photo that I am unable to get there.  I want to upgrade to either a photography table with a translucent background that I can place a speedlight under, or a cyclorama infinity background such as the MyStudio MS20. The latter option may require an upgrade to my lighting, which I am planning to do anyways.  The idea of course is to get a blown out 255 background with no post processing, other than color correction on the jewelry or product as needed. This would cut my time after the shoot to a ridiculously small amount of time.   Which option would make my life easier?   For reference, I currently use a D5300 with a Tamron 90mm f2.8 macro lens.  This is the best shot I got after quite a lot of effort, before and after processing:      Note that the silver is not shiny like it should be, and the sparkle on the gems is lost. is a light side power that many Jedi, including  are capable of. In Episode I, Qui-Gon uses it twice (once to outrun droids and once on Jar Jar's tongue), Obi-Wan uses it once and Anakin uses high speed reflexes for the entire duration of the pod race with no formal training.  According to  the power has a high drain and requires a longer recovery period than other powers. And then of course there are hilarious drawbacks such as .  But why wasn't this power used in the following scenarios (and many, many others):   The Redshirt Jedi who accompanied Windu to arrest Palpatine, before they were summarily dispatched to the Nether. Qui-Gonn, to avoid a lightsaber through his gut. Yoda, to avoid flying senate seats. By Windu, who is supposedly a notable practitioner. How do you detect a network disconnection when downloading a file in java? I was trying to experiment with docker, Ubuntu 14.04, 64-bit. I managed to run an image, and I gave it the path on one of my removable drives, /media/Drive1.  Now, whenever I reboot, I see there is a /media/Drive1 with only a docker directory inside it, and the actual drive which should be Drive1 ends up mounted on /media/Drive11, and this happens every reboot - even when I unmount these drives, and do a sudo rm -rfv /media/Drive1 before rebooting!  So something probably creates /media/Drive1 before it is fully mounted, and I guess that something is docker, because when I boot, I have:  $ ps axf | grep docker  2085 ?        Ssl    0:00 /usr/bin/docker -d   So, how can I disable the docker autostart, so it doesn't interfere with the mounting of external USB drives at boot? Where are some good places to find royalty free stock images? I'm always on the hunt for free stock images to use while I design, what are some good resources for images that are under a a royalty free or creative commons licence? Is there a rule in British English about how to pronounce "either"? There are two common pronunciations of "either": British /ˈaɪðər/ and American /ˈiːðər/. If Americans are more or less consistent in this regard, then the Brits seem to be freely using both. In fact, from what I can tell, "either this or that" is more often in the first form, whereas "me either" is in the second. But I may be wrong. Is there any kind of an informal rule in the modern British English with regard to this? singular/ plural in the attributive A doll house or a dolls house? Which is correct for the toy? ... There's that thing called Same Sex Couples Act; how about calling it Same Sex Couple Act?  The question is - when should I put attributives in plural? As far as I remember, Bill Bryson claimed in his Dictionary of Troublesome Words that - and that's what I vaguely recall from my classes - attributives should be in singular. Except for some exceptions. If so, what are they? I'm looking for some rule of thumb here other than "when in doubt, JFGI", of course.  NOTE: I leave out forms like a doll's house/dolls' house as they're pretty clear to me, as well as cases like The 'Big Boys and Their Motors' Club in which attributives are separate entities. This might be subjective, I don't know. However, it's a serious problem for me. My electrical bill is outrageous. It's three times as large as my next-door neighbor, and four or five times the size of my neighbor's across the street. It's worse in the summer (I live in central TX), but I've had the air conditioner inspected, and while the house is right at the maximum capacity for our unit, the unit should still be able to take care of the load. That also doesn't explain why our winter bills are larger than the comparables.  What I would like to do is check the amount of current drawn by each running appliance and calculate where all my money is going. I just don't know how to go about doing this. I'm using Ubuntu 12.04 &amp; everything about it is great except that it just won't resume after suspend. I've tried the suggestions listed in this thread  but none worked. I'm using Lenovo Y510P. I'm new to Ubuntu, or unix for that matter, so please let me know if there's anymore  information that I should be posting. Thanks.  update  I tried the suggestion by Dima on this post . The machine wakes up from hibernation but immediately the screen gives a screen like this     Wondering if this is an issue with the graphics card or something.