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anonymous
 one year ago
Use congruence theory to prove that
\[\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n\]
is an integer for every integer n.
anonymous
 one year ago
Use congruence theory to prove that \[\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n\] is an integer for every integer n.

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ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0familiar with fermat's little thm ?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 I have read about it but I am not familiar with it. Any hints would be helpful

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0For all integers \(a\) and primes \(p\), we have \[a^p\equiv a\pmod{p}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[n^5\equiv n\pmod{5} \implies n^5 = n+5k\] \[n^3\equiv n\pmod{3} \implies n^3 = n+3l\] plug them int he given expression and simplify

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n=\frac{ 1 }{ 15 } (3n^5+5n^3+7n)\] Plugging into the equation to get \[n^5=3(n+5k)+7n=10n+15k\] \[n^3=5(n+3l)+7n=12n+15l\] @ganeshie8 Am I doing this correct?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0simply replace n^5 by n+5k and n^3 by n+3l

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\[\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n = \frac{1}{5}(n+5k)+\frac{1}{3}(n+3l)+\frac{7}{15}n \\~\\=k+l+\frac{n}{5}+\frac{n}{3}+\frac{7n}{15}\] simplify

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@ganeshie8 Thanks a bunch
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