This is an obscure bug that only can be seen by moderators. I found a flag on a few comments of a user that were just spam. When I looked at the user, it was obvious he was only created to spam, and I destroyed the user's account, as is usually done for such people. The comments are now deleted, however, the flags still remain, with the only option to dismiss them, which I believe will give the user who correctly flagged an invalid flag. This doesn't seem right.  Proposal: If a user is destroyed, any flags associated with the account should be dismissed and marked as valid.
This appears to be related to the behavior originally suggested in .  It appears to be the same issue raised in , but this post has specific steps to reproduce the issue.  Here's what happened on one site in the network earlier today:   A question was posted much earlier Someone came along, and as an unregistered, new user with no other activity, posted a spam answer to that question at 04:36:54 UTC That same user posted a comment on their own answer with the same spam link at 04:37:31 UTC The answer was deleted as spam via community votes at 07:32:07 UTC A high-rep user saw the comment and raised a custom flag on it as being spammy at 07:32:54 UTC The unregistered user account was destroyed by a moderator at 08:13:10 UTC The comment now shows as "deleted by Anonymous" with a deletion time of 08:13:10 UTC The flag on the comment remains, but the only options are "Edit" and "Decline"; there is no "Delete" action   That there is no "Delete" action possible on the flag is reasonable, since the comment is now deleted. However, the flag is/was legitimate, so to decline the flag would be wrong. There's also nothing to leave while editing out the spammy part of the comment, as the comment consists solely of the same spam URL.  There is also no undelete option on the comment, so it's not possible to undelete the comment and then mark the flag as helpful by re-deleting the comment.  Effectively, this means that while the comment itself is effectively gone, the flag on it is in limbo and there is no good choice for what to do with the flag.  When deleting comments normally, my understanding is that any flags associated with those comments are automatically marked as helpful. However, it appears that the deletion of comments when destroying the user who posted them does not do that.  Therefore, I suggest that whenever comments are deleted as part of the owning user being destroyed, any pending flags against those comments should be marked as helpful at the same time.  It would also be a nice touch if those comments showed as deleted by Community♦, instead of Anonymous, but that's less of a concern in my case.
Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ is reducible over $\mathbb{Q}$? We know that $f(x)=x^4+1$ is a .  My question is:     Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ has a linear factor over $\mathbb{Q}$?   Edit: Thanks for @Jyrki Lahtonen's answer, I want to do some modifications:     Is it true that if $f(x)$ has a linear factor over $\mathbb{F}_p$ for every prime $p$, then $f(x)$ is reducible over $\mathbb{Q}$?   Thanks in advance! How should the "and/or" text of a card be interpreted? How should the "and/or" text of a card be interpreted? For example on a card like  the text reads:     Nahiri's Wrath deals damage equal to the total converted mana cost of the discarded cards to each of up to X target creatures and/or planeswalkers.   For the remainder of this question, let's fix X at 2 to keep things simple.  Should the "and/or" be interpreted such that the effect is applied to a combination of targets the player chooses. That is to say for Nahiri's Wrath the player could perform one of the following options:   Nahiri's Wrath deals damage to each of two target creatures. Nahiri's Wrath deals damage to each of two target planeswalkers. Nahiri's Wrath deals damage to each of one target creature and one target planeswalker.   Or should the card's effect be expanded and applied to both sides of the "and/or". With this interpretation the player would have to the following two options when playing Nahiri's Wrath:   Nahiri's Wrath deals damage to each of X target creatures and Nahiri's Wrath deals damage to each of X target planeswalkers Nahiri's Wrath deals damage to each of X target creatures or Nahiri's Wrath deals damage to each of X target planeswalkers   The confusion stems from  (which was formerly formerly worded using the same phrasing) where some commenters have said things like:     So this lets you do a total of 12 damage? 3 each to 2 target creatures AND 3 each to 2 target players, one of which could be a planeswalker? That's how the "and/or" works, right? You can choose. Otherwise it would just be "or"... How to prove $\vdash\neg P\to (P\to Q)$? I am trying to prove,  $$\vdash\neg P\to (P\to Q)$$  where $P$ and $Q$ are arbitrary but otherwise fixed formulas. The only axioms and rule of inference that I can use are,  $\color{crimson}{\text{Axiom 1.}}\  P\to (Q\to P)$ $\color{crimson}{\text{Axiom 2.}}\  (S\to (P\to Q))\to((S\to P)\to (S\to Q))$ $\color{crimson}{\text{Axiom 3.}}\  (\neg Q\to\neg P)\to(P\to Q)$ $\color{crimson}{\text{Rule of Inference.}}$ .  I know that in Angelo Margaris's book First Order Mathematical Logic there is a proof (page 53) but as I have mentioned , the dot notation seems very confusing and translating them to the nested parentheses notation seems very difficult to me. So, I am trying on my own to prove the result. Can anyone help? Could not load 'vboxdrv' after upgrade to Ubuntu 16.04 (and I want to keep secure boot) I upgrade from Ubuntu 15.10 to 16.04 and since then VirtualBox 5.0.18 isn't starting my VMs anymore. It complains that 'vboxdrv' isn't loaded. So I try to load it and get the following error:  $ sudo modprobe vboxdrv modprobe: ERROR: could not insert 'vboxdrv': Required key not available   I believe it is related to secure boot which I use and which I want to continue using. Actually with Ubuntu 15.10 secure boot and VirtualBox were working just fine.  Also I tried $ sudo apt-get --reinstall install virtualbox-dkms which built the kernel module successfully but didn't solve this issue.  Any idea how to get vboxdrv loaded while keeping secure boot enabled?  Update 2: Also I tried executing sudo mokutil --disable-validation. When executing this command, during the next boot I get prompted to disable secure boot, add a key or hash from disk. Since I don't want to disable secure boot, it seems that this doesn't solve my issue either. Also I want to keep UEFI activated for a parallel Windows installation.  Note: If you don't mind disabling secure boot, see  instead. Where do I place my own .sty or .cls files, to make them available to all my .tex files? I intend to use fsck for checking my Linux partition of the main disk, because its file system is suspicious of being not unmounted cleanly by shutting down the system. I have searched in my environment variables: none of them except of PATH should control the fsck program. But the program apparently needs to be used during a boot: :~$ fsck -V fsck from util-linux 2.20.1 Checking all file systems. [/sbin/fsck.ext4 (1) -- /] fsck.ext4 /dev/sda6  e2fsck 1.42 (29-Nov-2011) /dev/sda6 is mounted. WARNING!!!  The filesystem is mounted.   If you continue you WILL cause SEVERE filesystem damage. Do you really want to continue? no check aborted.  (There is no /forcefsck file in my root according to  . My etc/fstab file contains    # /etc/fstab: static file system information. # # Use 'blkid' to print the universally unique identifier for a # device; this may be used with UUID= as a more robust way to name devices # that works even if disks are added and removed. See fstab(5). # #                 proc            /proc           proc    nodev,noexec,nosuid 0       0 # / was on /dev/sda6 during installation UUID=1ac55d8d-c112-4bc7-9e79-921d196f9f79 /               ext4    errors=remount-ro 0       1 # swap was on /dev/sda7 during installation UUID=54f7e314-50e2-419b-a45d-47c3058ecc00 none            swap    sw              0       0 /dev/fd0        /media/floppy0  auto    rw,user,noauto,exec,utf8 0       0   ). According to an answer to  I have tried tune2fs hoping that I'll get the present value of the max_mounts_count parameter to be able to set it to 1 for checking the partition during the next boot. But after sudo rm /var/lib/update-notifier/fsck-at-reboot and a restart I have only got a response  :~$ tune2fs -l /dev/sda6 tune2fs 1.42 (29-Nov-2011) tune2fs: Permission denied while trying to open /dev/sda6 Couldn't find valid filesystem superblock. Can I use sudo tune2fs -l /dev/sda6 safely? Thanks. Can I install Android on my non-Android device? C# - how can I get owner's name for a Mutex I have a shared mutex between 2 applications. I want to create an administration console that shows the current owner's name of the mutex.  How can I get the application name that currently owns the Mutex? Adding a link with a space in front causes the link to not work When you paste a link into the "Insert Hyperlink" box, it adds it automatically as a citation link:  [Example 1][1]    [1]: http://example.com   But when you paste a link starting with a space , it tries to URL encode it, to change the space to %20, and then sticks a http:// on the front to make it a valid link:  [Example 2][2]    [2]: http://%20http://example.com   This, however, means the link doesn't work.    Please could leading whitespace be stripped from the URL? How to output string completions to stdout? "Users" folder shared with everyone Today I found something strange on my newly installed Windows 7 OS - my "users" folder C:\Users was shared and "Everyone" had full access... I disabled that and now it seems fine, however I'm a bit scared actually, I'd like to know what caused that and if my system is compromised. How to restart X Window Server from command line? How can I restart X Window Server from the command line?  I'd really like to be able to restart my GUI without having to do a full system reboot. I want to use symbols from wasysym and marvosym packages:  \usepackage{wasysym} \usepackage{marvosym}   but it gives an errors as both packages have the command of \CheckedBox. I do not want to use this command, but since both packages have it, it makes confusion for latex. How can I ignore one command to have both packages? Problem: Prove that $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.  My work: So I think I have to do a proof by induction and I just wanted some help editing my proof.  My attempt:  Let $P(n)=1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ for $n \in \mathbb{N}$.  Then $$P(1)=1^2=\frac{1(1+1)(2+1)}{6}$$ $$1=\frac{6}{6}.$$ So $P(1)$ is true.  Next suppose that $P(k)=1^2+2^2+\cdots+k^2=\frac{k(k+1)(2k+1)}{6}$ for $k \in \mathbb{N}$. Then adding $(k+1)^2$ to both sides of $P(k)$ we obtain the following: $$1^2+2^2+\cdots+k^2+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$$ $$=\frac{2k^3+3k^2+k+6(k^2+2k+1)}{6}$$ $$=\frac{2k^3+9k^2+13k+6}{6}$$ $$=\frac{(k^2+3k+2)(2k+3)}{6}$$ $$=\frac{(k+1)(k+2)(2k+3)}{6}$$ $$=\frac{(k+1)((k+1)+1)(2(k+1)+1)}{6}$$ $$=P(k+1).$$ Thus $P(k)$ is true for $k \in \mathbb{N}$. Hence by mathematical induction, $1^2+2^2+\cdots+n^2=\frac{n(n+1)(2n+1)}{6}$ is true for $n \in \mathbb{N}$. What does // mean in a path? I was working in my terminal, when I misspelled while cd to root (/). Accidentally, I typed // and I got in terminal:  root@weblocalhost://#   When i type pwd, i get response // but when I type ls -s i got same list as in / directory.  When you type more than 2, i.e. /// you get:  root@weblocalhost:/#   Always 1 / except for 2. Is there any difference and what is // for? How to typeset some text including math content in sans serif How come such different methods result in the same number, $e$? I guess the proof of the identity $$ \sum_{n = 0}^{\infty} \frac{1}{n!} \equiv \lim_{x \to \infty} \left(1 + \frac{1}{x}\right)^x $$ explains the connection between such different calculations. How is it done? Motivation: I once had one surplus card on my hand, and knew the robbers would come soon. So I gave it to someone for free, to avoid losing more cards when they hit. I was surprised by how badly this was received. Two of the three other players (the hosts of the game) acted upset, like I had tried to cheat. I had to take the card back, get robbed the next turn and 'giving cards away for free' was banned immediately. (Oh, how I hate mid-game additions to the rules. But that's beside the point.)  So now the question: Is this really illegal in Settlers of Catan? Is it generally considered underhanded in any way? If so, why?   My reasoning was like this:        Consider players A and B.   A wants to give B a sheep.   A trades a sheep and one rock to B, in exchange for wheat.   A trades the received wheat for the (formerly his) rock.   A and B now both have the same resources they had in the beginning, only the sheep has moved from A to B. Prove that $n$ is a sum of two squares? Problem   Let $n = p_1.p_2.p_3 \cdots p_k.m^2$, where $p_1, p_2, p_3 \cdots p_k$ are distinct primes. Prove that n is sum of two squares if and only if $p_i$ is either 2 or $p_i \equiv 1 \pmod{4}$   For $p_i = 2$ , this is trivial case since $2m^2 = m^2 + m^2$. For $p_i \equiv 1 \pmod{4}$, I tried to use the fact that the product of a number of the form $4k + 1$ is also in this form. So I come up with: $$n = (4k + 1) \cdot m^2 = 4km^2 + m^2$$ Apparently, $m^2$ is a square, but I could not figure out how to prove $4km^2$ is a square, since k is in unknown form. Any idea?    Thanks, TAR 6 GB worth of files into separate 1.9 GB TAR files I have 28,000 files of size 6.7 GB in the directory aaa/input. I need create TAR files in the directory aaa/output, such that each TAR file cannot exceed 1.9 GB, so I have to tar the 28,000 files into 4 TAR files (6.7/1.9) as Log_(current date)_01.tar Log_(current date)_02.tar ... etc.  such that each tarball is less than 1.9 GB. Is there a way to do this using a Unix script?