How to remove a password I know from a PDF, with no degradation, on Linux? Someone sent me a password-protected PDF. I know the password, but I want to make the PDF password-less, for convenience.  However - I want to make sure I don't lose any of the information in there. For example, I don't want images to be resampled, or vectorized content rasterized, or text runs/paragraphs broken up, or ancillary/non-critical non-printing data to be lost, I want the document properties meta-data to stay the same (well, perhaps I'm a little flexible about that).  Also, since I'm not a PDF format wiz, I'm not sure I can tell when a PDF has been degraded. So, for example, if I print to PDF from a PDF viewer, I'm not sure if the result is satisfactory - until perhaps someday later I'm trying to import it in a document editor or select some of the text and then find out something has changed.  My question is: Is there a way for me to do this which is qualitatively better than printing to a PDF? Perhaps I can simply remove the password somehow, in-place? Or at least ensure maximum-conservatism in the processing?  Notes:   I'm using Devuan GNU/Linux 3.0 (which is about the same as Debian Buster). There's a near-duplicate question,  - but the answers there are Windows and Mac utilities. I've tried using pdftk ... input_pw, but that doesn't work.
Removing the password from a PDF file I have a couple of ebooks as PDFs with passwords, however my ebook reader (sony prs600) doesn't seem to support PDFs with passwords.  What is the easiest of removing the password from a PDF (I know the password, which presumably helps a lot).  It's a bit annoying buying a book and then only being able to read it in front of a computer.
In  when Ambassador Sarek became critically ill, why didn't they send him home or send for a Vulcan medical expert? Since they were hosting many prominent political figures from many races, it seems strange that they did not have much medical support for a race so common as Vulcans (who are part of the Federation). Prove that $\mathbb{Z}$ is not isomorphic to additive group of any vector space over any field.  Proof.  Surpose that: $\phi : (A, +) \rightarrow  \mathbb{Z} $ is an isomorphism. Then there is some $r∈A $ such that $ϕ(r)=1_\mathbb{Z}$. We have $$ 1_\mathbb{Z}=ϕ(r)=ϕ(2(r/2))=2ϕ(r/2)$$  or equivalently that $ϕ(r/2)=\frac{1}{2}$. But this is not in $ \mathbb{Z}$, so there can be no such morphism.  Is my proof above correct? Thanks Let $K$ be a field and $f \in K[x]$ a polynomial of degree $n&gt;0$. Assume that $E$ is an extension of $K$ such that $f$ can be expressed as a product of linear factors over $E$:  $$f(x)=\gamma (x-\alpha_1)(x-\alpha_2)\dots(x-\alpha_n).$$  I want to show that $\lvert K(\alpha_1,\dots,\alpha_2) : K \rvert$ is a divisor of $n!$. I also want to show that if $\lvert K(\alpha_1,\dots,\alpha_n): K \geq (n-1)!$ then $f$ is irreducible over $K$ or $f$ already has a root in $K$.  The problem has a hint. It says: "Use induction on $n=\deg f$. If $f=gh$ in $K[x]$ with $\deg g =r$ and $\deg h=n-r$, then $r!(n-r)!$ is a divisor of $n!$.  Any idea for solving this problem would be appreciated. How to use JACK and Pulseaudio/ALSA at the same time on the same audio device? Whenever I launch hydrogen it breaks my audio in all other applications which use Pulse, for example, I can no longer play videos in Chrome after having launched Hydrogen once, until I restart my PC or do sudo alsa force-reload. I have found people with similar issues, but no actual guide to fix this, often there are links to  (which has moved to  ) I am using Ubuntu Studio 14.04 64bit.  ALTERNATIVELY if this is truly impossible to achieve, can someone tell me how to use Jack on one soundcard and Pulse on another? This way I could plug in a USB sound card only when I need JACK.  Please don't make me use windows. If background applications can't launch automatically how does Cardcase launch on a location change? How to add/subtract months from given date in Excel How to spawn a Giant with equipment? I really want to spawn Giant with armor and swords. Because Giants are strong, and great challenge if spawned with enchanted diamond armor and a diamond sword. Not only in ortho: top, front, back.  Please don't link this answer   I want a background image like this guy has -> What is $48\div2(9+3)$? There is a huge debate on the internet on the value of $48\div2(9+3)$. I believe the answer $2$ as I believe it is part of the bracket operation in BEDMAS.  yields the same answer. I also believe that if $48\div2\times(9+3)$ was asked it would be $288$ which Mathway agrees with as well. However,  says it is $288$ either way. A friend of mine (who is better at math) told me that there is no such thing as 'implicit multiplication', only shorthand so that is in fact done after the division (going left to right, not necessarily because division occurs before multiplication. But he didn't explicitly give a reason) What is the answer and why? CAST and IsNumeric Why would the following query return "Error converting data type varchar to bigint"?  Doesn't IsNumeric make the CAST safe?  I've tried every numeric datatype in the cast and get the same "Error converting..." error.  I don't believe the size of the resulting number is a problem because overflow is a different error.  The interesting thing is, in management studio, the results actually show up in the results pane for a split second before the error comes back.  SELECT CAST(myVarcharColumn AS bigint)   FROM myTable   WHERE IsNumeric(myVarcharColumn) = 1 AND myVarcharColumn IS NOT NULL   GROUP BY myVarcharColumn   Any thoughts? How does SO's (or other SE sites) versioning work of questions and answers? Do they use some standard piece of code or is it something homebrew?  With versioning I meant if for example a question is edited you'll see all the changes (a diff). Right-click on Touchpad not working after update to 18.04 LTS proof that on a compact manifold a vector field is complete I would like to know if my proof of the following statement is correct     If $M$ is a compact manifold, then every vector field $X$ over $M$ is complete.   Proof:  I take $p\in M$ and $(\gamma_p,I_p)$ a maximal integral curve through $p$. $I_p$ is open and non empty so if $I_p$ is closed, it is equal to $\mathbb{R}$.  So take $s\in \text{Adh}(I_p)$. There exists $t_i\rightarrow s, t_i\in I_p$. Define $q$ as $\text{lim}(\gamma_p(t_i))$ and take a maximal integral curve passing by $q$ as $(\gamma_q,I_q)$. Then since the vector fields are $C^\infty$, $\gamma_q(I_q)\cap\gamma_p(I_p)$ is nonempty (intuitively, this is clear but don't really know how to show it).  On the intersection, $\gamma_p$ and $\gamma_q$ coincide. So we can extend the maximal curve $\gamma_p$ to include $\gamma_q$. This implies that $q\in\gamma_p(I_p)$ and is actually equal to $\gamma_p(s)$  I think the main idea is there even though it probably lacks a bit of rigour. Deleted comments are persisting in the global inbox I realise that there is other questions out there about the  (First open one I found is ), but this is marked .  I often get the below issue where it will notify me about comments on my  (in this case what appears to be the user re-writing after deleting their comment) that no longer exist.  Also, occasionally, I see comment snippets with text that doesn't match what is on the page when I view it.  Let me know if I should be trying to edit that one to remove the tag indicating that it has been fixed, and add the above as another answer. After installing Ubuntu Studio 12.04, I have found that it uses a low latency kernel.   I searched for an explanation and how to change back it to a real-time or generic one. However, it looks like this part of Linux hasn't been covered to explain the details.   Q: Why choose a low latency kernel over a generic or real-time one?  PS: I have already read the answers from  question and  post. When does $\,n\mid q^3 \Rightarrow n\mid q\,$? I think this is true because of prime factorisations, i.e.  If $3$ a factor of the prime factorisation of $q^3$, then $3$ is a factor of the prime factorisation of $q$.  Therefore If $3$ divides $q^3$, then $3$ divides $q$.  Or for a general case, if $n$ divides $q^3$; then $n$ divides $q$. Is this correct? Word for people who live in the same city What are the people who live in the same city are called? Any words for that? I want to use it in the following context:      I and my ____ are happy. How do programmatically create fields for content types, and ADD them to the the content type form Say I have this field, "map_description". I know I would this function to define the field:   $field = array(   'field_name' =&gt; 'map_description',   'cardinality' =&gt; 1,   'type' =&gt; 'text', ); field_create_field($field);   And I have this code which I'm not sure what it does but I'm told that I'll need it:   $instance = array(     'field_name' =&gt; 'map_description',     'label' =&gt; 'The map description.',     'bundle' =&gt; 'my_content_type',     'entity_type' =&gt; 'node',     'widget' =&gt; array(     'type' =&gt; 'text_textfield',  );  field_create_instance($instance)   Those two code bits are both in my install hook and run when I install the module. But while the fields are in fact created, I have to manually assign them to the content type via "manage fields", is there a way to have the fields automatically assigned to the content type? Changing affiliation on publication I am moving from one job to another (postdoc to tenure track) this summer and I have papers in several stages of being written, these go from handwritten notes with all the necessary arguments to under revision at a journal. All the papers will have my new contact information put on them but where is the line for changing affiliation? It seems obvious to me that the one that is still in the handwritten notes stage will get the new affiliation and that the ones under revision won't. So somewhere in between is a line.   Where is it?  There are some similar questions so far (see  and  but they do not seem to be quite the same situation since I am not changing fields and the appointments are not concurrent. Do Stack Exchange sites allow editing of comments and posts from other users?