Using congruences, show $\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n$ is integer for every $n$ Using congruences, show that the following is always an integer for every integer
value of $n$:
$$\frac{1}{5}n^5 + \frac{1}{3}n^3 + \frac{7}{15}n.$$

Lets show that $P(n)=3n^5+5n^3+7n$ is divisible by 15 for every $n$.  To do this, we will show that it is divisible by $3$ and $5$ for every $n$.
Recall that for a prime $p$, $x^p\equiv x \pmod{p}$. (Fermat's Little Theorem)  Then, looking modulo 5 we see that
$$P(n)\equiv 3n^5+7n\equiv 3n+7n=10n\equiv 0.$$
Now looking modulo 3 we see that $$P(n)\equiv 5n^3+7n\equiv 5n+7n=12n\equiv 0.$$ Thus $P(n)$ is divisible by 15 for every $n$ as desired.
