Question

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.

Answer 1

familiar with fermat's little thm ?

Answer 2

@ganeshie8 I have read about it but I am not familiar with it. Any hints would be helpful

Answer 3

For all integers \(a\) and primes \(p\), we have \[a^p\equiv a\pmod{p}\]

Answer 4

\[n^5\equiv n\pmod{5} \implies n^5 = n+5k\]
\[n^3\equiv n\pmod{3} \implies n^3 = n+3l\]

plug them int he given expression and simplify

Answer 5

\[\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n=\frac{ 1 }{ 15 } (3n^5+5n^3+7n)\]
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?

Answer 6

simply replace n^5 by n+5k
and n^3 by n+3l

Answer 7

\[\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n = \frac{1}{5}(n+5k)+\frac{1}{3}(n+3l)+\frac{7}{15}n \\~\\=k+l+\frac{n}{5}+\frac{n}{3}+\frac{7n}{15}\]

simplify

Answer 8

@ganeshie8 Thanks a bunch

Answer 9

yw