I have found this question answered before but I am stuck going from  $$m^{\phi(n)}+n^{\phi(m)}\equiv1 \pmod{n}\\ n^{\phi(m)}+m^{\phi(n)}\equiv1 \pmod{m}$$ to $$m^{\phi(n)}+n^{\phi(m)}\equiv1 \pmod{mn}$$  Why can we combine $\pmod{m}$ and $(\pmod{n}$ into $\pmod{mn}$? I know it's overall because $\gcd(m,n)=1$, but I don't understand why it is so. Is there a theorem for this specifically? I found this, but  seems to not multiply what is in the modulus together.   Here is the beginning of my proof:  $\gcd(m,n)=1\implies m^{\phi(n)}\equiv1 \pmod{n}$ and $n^{\phi(m)}\equiv1 \pmod{m}$ by Euler's theorem.  $ m^{\phi(n)}\equiv1 \pmod{n} \equiv  m^{\phi(n)}+n^{\phi(m)}\equiv1+n^{\phi(m)}\pmod{n}$ and $n^{\phi(m)}\equiv1 \pmod{m}\equiv n^{\phi(m)}+m^{\phi(n)}\equiv1+m^{\phi(n)} \pmod{m}$.  But $m^{\phi(n)} \equiv 0\pmod{m}$ and $n^{\phi(m)}\equiv 0\pmod{n}$.  So we have $m^{\phi(n)}+n^{\phi(m)}\equiv1\pmod{n}$ and $n^{\phi(m)}+m^{\phi(n)}\equiv1 \pmod{m}$...  ... which implies $m^{\phi(n)}+n^{\phi(m)}\equiv1\pmod{mn}$.
I would like to prove the following equality: $$k^{\phi(l)} + l^{\phi(k)} \equiv1\pmod{lk}$$if $\gcd(l,k)=1$. What methods can I use? Thank you for your help.
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I'm confused about it, as some sites say it is only for those with a new passport. If I do need a vignette transfer, what is the procedure and what documents are required? How to list user installed applications (not packages)? Not packages and not all applications. Just the applications user installed by himself from whatever source (Software Center, manually added PPA, etc.). If the above is not possible - a list of all installed applications or at least a GUI which lists the applications so I can take screenshots of it.  I've read a dozen of similar questions and people posting answers usually don't even get close to OP question merit. Please note that my question includes 'user installed'.   Answer: It is not currently possible in Ubuntu Linux.  (choosing tijybba's answer as the closest one though) I'm trying to install programs/files from the getdeb repo but get a "check your internet connection" error.  Programs than I've tried include Qbittorrent and Mozilla-Plugin-VLC.    Through the terminal I get the following errors:  Err http://archive.getdeb.net/ubuntu/ natty-getdeb/apps qbittorrent amd64 2.8.2-1~getdeb2   Could not connect to archive.getdeb.net:80 (209.105.191.78). - connect (113: No route to host) E: Failed to fetch http://archive.getdeb.net/ubuntu/pool/apps/q/qbittorrent/qbittorrent_2.8.2-1~getdeb2_amd64.deb: Could not connect to archive.getdeb.net:80 (209.105.191.78). - connect (113: No route to host)   &amp;  Err http://archive.getdeb.net/ubuntu/ natty-getdeb/apps mozilla-plugin-vlc amd64 1.1.10-1~getdeb1   Could not connect to archive.getdeb.net:80 (209.105.191.78). - connect (113: No route to host) E: Failed to fetch http://archive.getdeb.net/ubuntu/pool/apps/v/vlc/mozilla-plugin-vlc_1.1.10-1~getdeb1_amd64.deb: Could not connect to archive.getdeb.net:80 (209.105.191.78). - connect (113: No route to host)   I'm not having any problems adding/removing other programs/files from my machine.  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Or is there another way to reset my password? 'This post has been deleted' appears above the flag bar Elementary derivation of certian identites related to the Riemannian Zeta function and the Euler-Mascheroni Constant Is the proof of these identities possible, only using elementary differential and integral calculus? If it is, can anyone direct me to the proofs? ( or give a hint for the solution )  1)$$\int_0^\infty { e^{-x^2} \ln x }\,dx = -\tfrac14(\gamma+2 \ln 2) \sqrt{\pi} $$  2)$$\int_0^\infty { e^{-x} \ln^2 x }\,dx = \gamma^2 + \frac{\pi^2}{6} $$  3) $$\gamma = \int_0^1 \frac{1}{1+x} \sum_{n=1}^\infty x^{2^n-1} \, dx$$  and lastly,  4) $$\zeta(s) = \frac{e^{(\log(2\pi)-1-\gamma/2)s}}{2(s-1)\Gamma(1+s/2)} \prod_\rho \left(1 - \frac{s}{\rho} \right) e^{s/\rho}\!$$  I personally think the last is obtained from a simple use of the Weierstrass factorization theorem. I'm unsure as to what substitution is used.  $\gamma$ is the Euler-Mascheroni constant and $\zeta(s)$ is the Riemannian Zeta function.  Thanks in advance.