I need help proving the first one via Fermat's little theorem.
I need a hint, or a good starter!
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What are you stuck on? – Alexander Gruber Apr 24 '14 at 00:41
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$n(n-1)(n+1)(n^2+1)=n^5-n$. – André Nicolas Apr 24 '14 at 00:43
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Why is FLT required? Homework says so? – MT_ Apr 24 '14 at 00:49
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No, you see in fermats little theorem is about primes. So if the gcd(n, 6) = 1 then i know that n is odd and since its equal to 1 its relatively prime. so I wanted to show that 3 divides n^2-1 and same for 4 (thus 12)... but I dont know how to apply the theorem! thats where I'm stuck – Sarah cartenz Apr 24 '14 at 00:59
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As Michael asked, why do you insist on using Fermat's little theorem? It doesn't make sense here. – user21820 Apr 24 '14 at 01:06
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its just that I saw that they were relatively prime, I can connect FLT somehow. However,you are telling me that its not connected, thus that answered my question, thanks! Any ideas concerning the second one though? – Sarah cartenz Apr 24 '14 at 01:10
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"Relatively prime" is very different from "prime". Andre already gave an answer to the second one, where he used Fermat's little theorem on $n^5-n$. But if you understood why my solution works for the first question, you should see that it works on the second one as well! (By the way if you want anyone to receive a notification you should put an @ followed by the user-name. I also didn't know that until someone told me.) – user21820 Apr 24 '14 at 01:38
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Oh i did not know that!, thanks! – Sarah cartenz Apr 24 '14 at 07:26
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"This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Michael Rozenberg Mar 28 '19 at 13:33
1 Answers
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Just notice that if it works for $n$ it also works for $n+6$ and $n-6$, so it is enough to prove the theorem for $n \in \{0,1,2,3,4,5\}$.
user21820
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A shortcut: note that $n^2\equiv 1 \pmod3$(By FLT or "brute force", as you wish) and that $\text{odd}^2\equiv1 \pmod 4$ – chubakueno Apr 24 '14 at 01:31
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@chubakueno: I know that of course, but I didn't want to give the 'standard answer' but rather show that there are many ways of looking at the same problem. And sometimes it's easier to just reduce to finitely many cases and 'whack' than to come up with an ad-hoc solution. Notice that the same method works on the second question too. Haha.. – user21820 Apr 24 '14 at 01:34
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Yeap, it will work on every problem with a finite number of cases :) (a reasonable number of them actually, for practical purposes). I just posted my comment to "whack" the problem as Sarah cartenz wanted, just to show it was possible(but quite and overkill, though) – chubakueno Apr 24 '14 at 02:24