Prove that $$ N = \frac{5^{125} - 1}{5^{25} - 1} $$ is a composite number.   I just proved it will be a natural number with  quotient= $5^{100} +5^{75}+5^{50}+5^{25}+1$  But I do not know how to factorise it for proving it composite.
Prove that $n=\dfrac{5^{125}-1}{5^{25}-1}$ is a composite number   My attempt,  Let $x=5^{25}$, so that $5^{125}-1=x^5-1=(x-1)(x^4+x^3+x^2+x+1)$  $=(x^4+9x^2+1+6x^3+6x+2x^2-5x^3-10x^2-5x)(x-1)$  $=((x^2+3x+1)^2-5x(x+1)^2)(x-1)$  I'm stuck at this point and don't know how to continue anymore. Hope someone can provide a detailed solution. Thanks a lot.
If $p,q$ are positive quantities and $0 \leq m\leq 1$ then Prove that $$(p+q)^m \leq p^m+q^m$$    Trial: For $m=0$, $(p+q)^0=1 &lt; 2= p^0+q^0$  and for $m=1$, $(p+q)^1=p+q =p^1+q^1$.   So, For $m=0,1$ the inequality is true.How I show that the inequality is also true for $0 &lt; m &lt; 1$.   Please help. Why is my USB drive showing corrupted data when plugged as an internal SATA drive? I have an experiment where I measure the effect of a manipulation on two different groups of participants, and the variable that I manipulate also has two levels. Thus, I can essentially use a 2x2 ANOVA to analyze the results. But it is also possible for me to measure the effect of the different levels of the variable first on Group 1 and then on Group 2, by using paired-samples t-tests.   Is there a rule of thumb to decide what needs to be done in this case? I think one advantage of using an ANOVA could be the option to investigate the interaction between the different groups and the different levels of the variable. Perhaps one could also argue that an omnibus analysis is also a better choice from a stylistic point of view.   But is there a very compelling conceptual reason to prefer a 2x2 ANOVA over two separate pairwise comparisons in my case? And if so, what would it be? The set of integers $n$ expressible as $n=x^2+xy+y^2$ Let $S$ be the set of integers $n$, such there exist integers $x,y$ with $$n=x^2+xy+y^2$$   Is the implication $$a,b\in S\implies ab\in S$$ true? If yes, how can I prove it?  I worked out $$n\in S\iff 4n\in S$$ and $$n\in S\iff 3n\in S$$ I tried two approaches. The first is to express $$(a^2+ab+b^2)(c^2+cd+d^2)$$ in the form $$f^2+fg+g^2$$ with polynomials $f,g$ with integer coefficients. I however could not find suitable $f$ and $g$. The second approach is based on $$x^2+xy+y^2=\frac{(2x+y)^2+3y^2}{4}$$ If we have $n=x^2+xy+y^2$ , we have $u^2+3v^2=4n$ for some integers $u,v$ with equal parity. The main problem of this approach is to consider the equal parity. Any ideas ? How to clone ArrayList and also clone its contents? How can I clone an ArrayList and also clone its items in Java?  For example I have:  ArrayList&lt;Dog&gt; dogs = getDogs(); ArrayList&lt;Dog&gt; clonedList = ....something to do with dogs....   And I would expect that objects in clonedList are not the same as in dogs list. Callback functions in Java How do I fix this "E: Type '*' is not known on line * in source list ..." update error? Every once in a while I see users having issues to update due to errors like this:  E:Type 'ain' is not known on line 1 in source list /etc/apt/sources.list.d/some-ppa.list'   with varying types/line numbers/source list files (often after removing a PPA).  How can such an error be fixed? biblatex: Handling secondary ("quoted in") citations This question and the answer became part of a package:      How can I qoute secondary quotes which I’ve taken form another book.  Example  Jon Doe writes a book an cites Fred Bloggs in it. Now I read the book of Doe an want to cite his cite of Bloggs. So my bibliography entry must look like     Bloggs, Fred: A nice old book, 1850, p. 35. Quoted in Doe, John: A bad new Book, 2011, p. 67.   Doe’s book should not be listed separately in the bibliography.  I know that there’s a bibstring quotedin containing “quoted in” (p. 159 of biblatex’s manual) but I don’t know if theres a (predefined) way to use it …  Update  It seems that there’s no predefined way, so I decided to use userc for crossreferences like this. This works for the bibliography but not for cite commands. At the moment I create the “in text cite” manually.  It would be great to have a macro like  \QIautocite[&lt;prenote a&gt;][&lt;postnote a&gt;]{&lt;author a&gt;}[&lt;prenote b&gt;][&lt;postnote b&gt;]{&lt;author b&gt;}   that expands to the above given example.  That’s my new code, which gives the right results for bibliography and “in text cite” but the latter not automatically …  \RequirePackage{filecontents} \begin{filecontents}{\jobname.bib} @BOOKLET{autora,   title = {A nice old book},   userc={autorb},   author = {Fred Bloggs},   year = {1850}, } @BOOKLET{autorb,   title = {A bad new book},   author = {Jon Doe},   year = {2011}, } \end{filecontents}  \documentclass[ngerman]{scrartcl}  \usepackage[T1]{fontenc} \usepackage[latin1]{inputenc}  \usepackage{babel,csquotes,blindtext}  \usepackage[%     bibstyle=authortitle,     citestyle=verbose-ibid, ]{biblatex} \addbibresource{\jobname.bib}  \NewBibliographyString{bibcitedas} \DefineBibliographyStrings{german}{%   bibcitedas = {{zitiert als}{zit. als}}, }  % Put the 'quoted in text' behind the bibentry. \renewbibmacro*{finentry}{%   \ifbibliography{%     \iffieldundef{userc}{}{%       \newunit\bibstring{quotedin}\addspace\fullcite{\thefield{userc}}     }   }{}%   \finentry% }  \begin{document} \blindtext \begin{quotation}     \blindtext\footnote{\cite[36]{autora}. Zit. nach \cite[89]{autorb}} \end{quotation} \blindtext\footnote{\cite[36]{autora}. Zit. nach \cite[89]{autorb}}  \printbibliography \end{document}   Example 2  In addition to my comment (Hi @Audrey. This works fine :-) but theres a little thing …) on Audrey’s answer:  If I use the quotecite command like this  \quotecite[1]{reese}[5]{weinberg} \quotecite[1]{reese}[5]{weinberg} \pagebreak \quotecite[1]{reese}[5]{weinberg} \quotecite[1]{reese}[5]{weinberg}   the footnotes should look like       1 Reese [full cite], p. 1. Zit. nach Weinberg [full cite], p. 65   2 Ebd.   ———— page break ————   3 Rese [short cite], p 1. Zit nach Weinberg [short cite], p. 65   4 Ebd.   Everything works like expected except footnote 3 which is     3 Rese [short cite], p 1. Zit nach Weinberg [full cite], p. 65     at the moment  Update 2  Since it seems not possible to react to autocite I created a new plain command with  \DeclareMultiCiteCommand{\quoteciteplain}[\cbx@qcwrapperplain]{\cbx@quotecite}{\cbx@qcdelim}  \newrobustcmd{\cbx@qcwrapperplain}[1]{%   \global\toggletrue{cbx@isquote}%   #1%   \global\togglefalse{cbx@isquote}%   \global\togglefalse{cbx@iflastibid}%   \global\togglefalse{cbx@ifquoteepost}%   \global\togglefalse{cbx@ifquoterpost}} How can I make Firefox open magnet-links in Transmission? Fourier series identity I need to prove that $\dfrac{a \sin(bx)}{1 - 2a \cos(bx) + a^2} = \sum_{n=1}^\infty a^n \sin(nbx)$ where $|a| &lt; 1$.  It seems that this can be proved by using Euler's formula identities for $\cos(bx)$ and $\sin(bx)$ and substituting $z = e^{ibx}$. From that, I get $$ \dfrac{a \sin(bx)}{1 - 2a \cos(bx) + a^2} = \dfrac{a (z - 1/z)}{2i(1 - a(z + 1/z) + a^2)}$$ $$= \dfrac{a (z^2 -1)}{2i(z - a(z^2 +1) + za^2)}$$ $$= \dfrac{a (z^2 -1)}{-2ai(z^2 - z(1-a^2)/a + 1)}$$  I don't understand what I am supposed to do next. Note: I will expand this question with more specific points when I have my own internet connection and more time (we're moving in, so I'm at a friend's house).  This question is broad, involved, and to some degree subjective.  (I started out as a physics-only student, but eventually decided to add a mathematics major. I am greatly interested in mathematics; the typical curriculum required for physics students is not deep or thorough enough; mathematics is more general (that means work!); and it only requires a few more classes. Naturally, I enjoy  immensely.)  This question asks mainly of undergraduate-level study, but feel free to discuss graduate-level study if you like.  Please do not rush your answer or try to be comprehensive. I realize the StackOverflow model rewards quick answers, but I would rather wait for a thoughtful, thorough (on a point) answer than get a fast, cluttered one. (As you probably know,  produces clear, useful writing; and a properly-done comprehensive answer would take more than a reasonable amount of time and effort.) If you think an overview is necessary, that is fine.  For a question this large, I think the best thing to do is focus on a specific area in each answer.    Update: To Sklivvz, Cedric, Noldorin and everyone else: I had to run off before I could finish, but I wanted to say I knew I would regret this; I was cranky and not thinking clearly, mainly from not eating enough during the day. I am sorry for my sharp responses and for not waiting for my reaction to pass. I apologize.  Re: Curricula:   Please note that I am not asking about choosing your own curriculum in college or university. I did not explicitly say that, but several people believed that was my meaning. I will ask more specific questions later, but the main idea is how a physics student should study mathematics (on his or her own, but also by choosing courses if available) to be a competent mathematician with a view to studying physics.  I merely mentioned adding a mathematics major to illustrate my conclusion that physics student need a deeper mathematical grounding than they typically receive.   And now I have to run off again. My fiancée and I are looking for a good Caribbean cruise in October and were wondering which islands are best to see and which Cruise line to take?  It seems like a lot of the cruises don't run in this month due to Hurricane season so I'm looking for other good options.  EDIT We'll be travelling in 2012. During self-study, I ran across the question of how many ways six numbers can be chosen from the numbers 1 - 49 without replacement, stipulating that no two of the numbers be consecutive.  I can obtain a simple lower bound by saying that, in the worst-case scenario, when you choose a particular number, there are now three numbers that cannot be chosen next.  For example, if I first pick 17, then I can't choose 16, 17, or 18 for the remaining choices.  This gives me the lower bound  $$\frac{49*46*43*40*37*34}{6!} = 6,773,770 \frac{8}{9}$$  This is about 48% of ${}_{49}C_6 =  13,983,816$.  The real answer must be bigger (and an integer).  I haven't found a way to calculate it, though.  The original problem asked to show that the probability of having non-consecutive integers when you choose six is greater than 50%, so if the problem is complicated to count exactly, better approximations that show the answer is above 50% would also be appreciated.   Of course, I can use a computer to do the counting, but I'm interested in learning what methods I'm missing. $\mathbb{R}P^4$ and $\mathbb{R}P^6$ do not admit fields of tangent $2$-planes See the related question . This is the second part of question 4-C in Milnor and Stasheff's book on characteristic classes. In the solution to the first part, we rely on the fact that having a field of tangent $1$-planes allows us to split the tangent bundle into a direct sum. It doesn't seem like this should be true for a field of tangent $2$-planes (otherwise the question wouldn't just be asking about $P^4$ and $P^6$), so I'm not sure exactly where to start.  edit: actually, since $P^n$ can be given a Riemannian metric, any subbundle should be a Whitney summand; this doesn't answer the question as to why they only ask for $P^4$ and $P^6$ though, so I'm still confused. Show that $2^n > n^2$ whenever n>4. Let $S(n)$ be the proposition "$2^n &gt; n^2$"  Basis step: $S(5)$ is true, since $2^5 = 32$, $ 32 &gt; 25 = 5^ 2$  Inductive step:  Induction hypothesis:   Assume $S(K)$ is true, $2^K &gt; K^2$.  Then  $2^{K+1} = 2.2^K &gt; K^2 + K^2$   I don't know how to proceed after this. How to complete this proof? How can I download an entire website? How can I download all pages from a website?  Any platform is fine. $f(x)=x^3$ is convex nor concave without to derivate Package in Java double r = 11.631; double theta = 21.4;   In the debugger, these are shown as 11.631000000000000 and 21.399999618530273.  How can I avoid this? When Ron broke his wand, why didn't he apply for Hogwarts financial aid?