If $r_1, ... r_{p-1}$ and $s_1, ... s_{p-1}$ are two reduced residue systems modulo an odd prime p, prove that $r_1s_1, r_2s_2, ..., r_{p-1}s_{p-1}$ is not a reduced residue system modulo p.  don't know where to start, any help appreciated!
If $a_1, a_2, ..., a_n$ is a reduced residue system $\mod p$ and $a_1', a_2', ..., a_n'$ is another reduced residue system $\mod p$ then $a_1 a_1', a_2 a_2', ..., a_n a_n'$  is not a reduced modulo system $\mod p$.  Not able to show a counter to it. Help Needed.
Stopping xcode from indexing How do I Uninstall Windows 10 or Windows 10 preview? Proving 'Law of Excluded Middle' in Fitch system What is the name of this castle part? Moments of the truncated normal distribution (univariate) away from the mean How do you sort du -sh /dir/* by size? I read one site that said use | sort -n but that's obviously not right. Here's an example that is wrong.  [~]# du -sh /var/* | sort -n 0       /var/mail 1.2M    /var/www 1.8M    /var/tmp 1.9G    /var/named 2.9M    /var/run 4.1G    /var/log 8.0K    /var/account 8.0K    /var/crash 8.0K    /var/cvs 8.0K    /var/games 8.0K    /var/local 8.0K    /var/nis 8.0K    /var/opt 8.0K    /var/preserve 8.0K    /var/racoon 12K     /var/aquota.user 12K     /var/portsentry 16K     /var/ftp 16K     /var/quota.user 20K     /var/yp 24K     /var/db 28K     /var/empty 32K     /var/lock 84K     /var/profiles 224M    /var/netenberg 235M    /var/cpanel 245M    /var/cache 620M    /var/lib 748K    /var/spool When to use IMG vs. CSS background-image? In what situations is it more appropriate to use an HTML IMG tag to display an image, as opposed to a CSS background-image, and vice-versa?  Factors may include accessibility, browser support, dynamic content, or any kind of technical limits or usability principles. NewForm.aspx disappeared SEO of  in  header Long lay over in Seoul. Do I need a visa as a Filipino citizen? I'm a Filipino immigrant from USA about to visit home in December.  I have a 14-hour lay over in Seoul,  South Korea.  Do I need a visa to leave the airport to do a little sightseeing? Output the common lines (similarities) of two text files (the opposite of diff)? Finding an angle within an 80-80-20 isosceles triangle The following is a geometry puzzle from a math school book. Even though it has been a long time since I finished school, I remember this puzzle quite well, and I don't have a nice solution to it.  So here is the puzzle:    The triangle $ABC$ is known to be isosceles, that is, $AC=BC$. The labelled angles are known to be $\alpha=\gamma=20°$, $\beta=30°$. The task is to find the angle labelled "?".  The only solution that I know of is to use the sine formula and cosine formula several times. From this one can obtain a numerical solution. Moreover this number can be algebraically shown to be correct (all sines and cosines are contained in the real subfield of the 36th cyclotomic field). So in this sense I solved the problem, but the solution is kind of a brute force attack (for example, some of the polynomials that show up in the computation have coefficients > 1000000). Since the puzzle originates from a book that deals only with elemetary geometry (and not even trigonometry if I remember correctly) there has to be a more elegant solution. How to convert an MKV to AVI with minimal loss To convert an MKV to AVI, I do two things. The first thing I do is this:  ffmpeg -i filename.mkv -vcodec copy -acodec copy output.avi   or this:  ffmpeg -i filename.mkv -sameq -acodec copy output.avi   Either of these will convert the MKV to an AVI, but the problem is that the video does not play smoothly for some reason. That's fine though, because if I do one more thing it gets fixed:  ffmpeg -i output.avi -vcodec mpeg4 -b 4000k -acodec mp2 -ab 320k converted.avi   After I do this then the file plays without problem. I had success doing it this way for one file, but then I tried it on another file, and there is a slight, but noticeable loss in video quality. This is the output I get when doing the second step:  FFmpeg version 0.6.1, Copyright (c) 2000-2010 the FFmpeg developers   built on Dec 29 2010 18:02:10 with gcc 4.2.1 (Apple Inc. build 5664)   configuration:    libavutil     50.15. 1 / 50.15. 1   libavcodec    52.72. 2 / 52.72. 2   libavformat   52.64. 2 / 52.64. 2   libavdevice   52. 2. 0 / 52. 2. 0   libswscale     0.11. 0 /  0.11. 0  Seems stream 0 codec frame rate differs from container frame rate: 359.00 (359/1) -&gt; 29.92 (359/12) Input #0, avi, from 'output.avi':   Metadata:     ISFT            : Lavf52.64.2   Duration: 00:04:17.21, start: 0.000000, bitrate: 3074 kb/s     Stream #0.0: Video: mpeg4, yuv420p, 704x480 [PAR 229:189 DAR 5038:2835], 29.92 fps, 29.92 tbr, 29.92 tbn, 359 tbc     Stream #0.1: Audio: vorbis, 48000 Hz, stereo, s16 Output #0, avi, to 'converted.avi':   Metadata:     ISFT            : Lavf52.64.2     Stream #0.0: Video: mpeg4, yuv420p, 704x480 [PAR 229:189 DAR 5038:2835], q=2-31, 4000 kb/s, 29.92 tbn, 29.92 tbc     Stream #0.1: Audio: mp2, 48000 Hz, stereo, s16, 320 kb/s Stream mapping:   Stream #0.0 -&gt; #0.0   Stream #0.1 -&gt; #0.1   I just used arbitrarily large settings on the second step and it worked nicely before but not in this case. What settings should I use? How do I prove/disprove that $F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1}$, where $F_n$ is the $n$th Fibonacci number? I was looking at the Fibonacci sequence:  $$1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144,\ldots, F_n : F_n + F_{n - 1} = F_{n + 1} \land F_0 = 0$$  Here I noticed something; $\qquad F_n^2 + F_{n\pm1}^2 = F_{2n\pm1} : n\in \mathbb{N}$  For example; $$\ \ F_3^2 + F_{3 + 1}^2 = F_3^2 + F_4^2 = 2^2 + 3^2 = F_{2\times3 + 1} = F_{6 + 1} = F_7 = 13$$ $$F_3^2 + F_{3 - 1} = F_3^2 + F_2^2 = 2^2 + 1^2 = F_{2\times3 - 1} = F_{6 - 1} = F_5 = 5$$ However, no matter how true this can seem, I cannot seem to prove/disprove it.    My Attempt: $$\because (a + b)^2 = (a + b)(a + b) = a^2 + ab + ba + b^2 = a^2 + b^2 + 2ab : (a,b)\in \mathbb{Z}$$ $$\implies F_n^2 + F_{n\pm 1}^2 = (F_n + F_{n\pm 1})^2 - 2F_nF_{n\pm 1}$$ $$\implies F_n^2 + F_{n - 1}^2 = F_{n + 1}^2 - 2F_nF_{n - 1} : F_n + F_{n - 1} = F_{n + 1}$$ $$\implies F_n^2 + F_{n + 1}^2 = F_{n + 2}^2 - 2F_nF_{n + 1} : F_n + F_{n + 1} = F_{n + 2} \iff F_n + F_{n - 1} = F_{n + 1}$$ $$\implies F_{2n\pm 1} + 2F_nF_{n\pm 1} = F_{n + 2\lor1}^2 : F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1} \tag{+2 if +, +1 if -}$$ $$\implies F_{2n\pm 1} + 2\sqrt{F_{2n\pm 1} - F_{n\pm 1}^2}\sqrt{F_{2n\pm 1} - F_n^2}= F_{n + 2\lor1}$$ $$\implies F_{2n\pm 1} + 2\sqrt{(F_{2n\pm 1} - F_{n\pm 1}^2)(F_{2n\pm 1} - F_n^2)} = F_{n + 2\lor1}^2$$ $$\implies F_{2n\pm 1} + 2\sqrt{F_{2n\pm 1}(F_{2n\pm 1} - F_n^2 - F_{n\pm 1}^2) - (F_nF_{n\pm 1})^2} = F_{n + 2\lor1}^2$$.  $$\because F_n^2 + F_{n\pm 1}^2 = F_{2n\pm 1}$$ $$\implies F_{2n\pm 1} + 2\sqrt{-(F_nF_{n\pm 1})^2} = F_{n + 2\lor1} \neq F_{2n\pm 1} + 2\sqrt{\big(-(F_nF_{n\pm 1})\big)^2}$$ $$\therefore F_{2n\pm 1} + 2F_nF_{n\pm 1}i = F_{n + 2\lor1} : i = \sqrt{-1}$$  I have looked over every step and cannot see where I am wrong. The only part that must be wrong is that:  $$F_{2n\pm 1} + 2F_nF_{n\pm 1}i = F_{n + 2\lor1} \iff F_n^2 + F_{n\pm1}^2 = F_{2n\pm 1}$$  I can seperate $F_{2n\pm 1}$ as $F_n^2 + F_{n\pm 1}^2$ (if true) so I turn the equation into:  $$F_n^2 + F_{n\pm 1}^2 + 2F_nF_{n\pm 1}i = F_{n + 2\lor1}$$ $$\therefore (F_n + F_{n\pm 1})\bigg(\frac{F_n}{F_{n\pm 1}} + \frac{F_{n\pm 1}}{F_n} + 2i\bigg) = F_{n + 2\lor1}$$ Now I know that the greater the value of $n$, one of the fractions in the above equation (depending on which sign is used in $F_{n\pm 1}$) will get closer to the golden ratio $φ = \frac{1 + \sqrt{5}}{2}$ but $i = \sqrt{-1}$ is not a "real" number, although $F_n$ is.  I then searched for other formulae for finding $F_n$ and came across the one below (being Binet's Formula):  $$F_n = \bigg\{\frac{φ^n - ψ^n}{φ - ψ} : ψ = φ - \frac{\sqrt{20}}{2} = \frac{1 - \sqrt{5}}{2} \ \land \ φ - ψ = \sqrt{5}\bigg\}$$  But I do not know what how to turn that into $F_{n\pm 1}$. Could you please show me where I went wrong (because I must have gone wrong somewhere), and also if you wish, could you please show me how to turn the above fraction into $F_{n\pm 1}$ (if possible).  I only used the $(a + b)^2$ expansion rule because my skill level in mathematics is not particularly high compared to most of the users on this site (especially when logging into the MSE and finding fancy symbols like $\nabla$ and $\partial$ and $\sum$ which all look Greek to me, however the only "fancy" thing I know is that $i = i^5 = \cdots = \sqrt{-1}$) so if you used any other techniques in trying to prove this formula, could you please show me step by step?  Much appreciated. Thank you in advance.    Edit: I believe this question is a duplicate of this other question $\longrightarrow$   and I can go here to serve as a little hint $\longrightarrow$ Why are effect size estimates (such as $R^2$) difficult in mixed-effects models? 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When connected to the Internet, the system checks if it's in a walled garden, which is simply trying to connect to Google. If it fails, in Android 5, there'll be an exclamation mark on the WiFi icon or signal icon.   In some situations, this is annoying. Say, in China, Google is blocked, the exclamation mark will always be there, which is meaningless.  How can I disable this check, and make the system assume, connected WiFi/mobile data is connected Internet?