## 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.

1. ganeshie8

familiar with fermat's little thm ?

2. anonymous

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

3. ganeshie8

For all integers $$a$$ and primes $$p$$, we have $a^p\equiv a\pmod{p}$

4. ganeshie8

$n^5\equiv n\pmod{5} \implies n^5 = n+5k$ $n^3\equiv n\pmod{3} \implies n^3 = n+3l$ plug them int he given expression and simplify

5. anonymous

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

6. ganeshie8

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

7. ganeshie8

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

8. anonymous

@ganeshie8 Thanks a bunch

9. ganeshie8

yw