For any $n$ prove that $n^5-n$ is divisible by $30$.
Problem on divisibility, number theory
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elementary-number-theory
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2What do we win if we solve that problem? – 2017-02-28
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1Please post your attempts. We are here to help you, but not to make your homework. – 2017-02-28
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0Try to factorize it and to play a little bit with remainders – 2017-02-28
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0@Math_QED, Sorry, I tried induction to solve the problem and was able to eliminate 5 from the denominator, but was still stuck with 2*3, if u break it up as n*(n-1)*(n+1)*(n^2+1), it's obvious that it divisible by 2*3, but what about 5? So I wondering how one would go about solving it. By Induction- n^5 -n is divisible for k and for k+1 after expansion, we get k^5+5*k^4+10*k^3+10*k^2 + 5k +1-(k+1) – 2017-02-28
1 Answers
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By Fermat, $5|n^5-n$, Manifestly $n^5-n=n(n-1)(n+1)(n^2+1)$ is clearly divisible by $6=2\cdot 3$ as it has as factors three consecutive integers.
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0Does the downvoter care to comment where he/she thinks there is an error? – 2017-02-28
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0I see no errors, so this is probably one of those downvotes by a user who doesn't like answers to "no effort" questions. – 2017-02-28