<s>[INST] Prove: $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for $n \ge 0$ I am attempting to prove the following problem: 

Prove that $\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n {30}$ is an integer for all integers $n = 0,1,2,...$

I attempted to solve it by induction, but when proving for $n= x+1$ the algebra gets very messy very fast. I was wondering if this is the only way or if there is a quicker way to prove this. I guess I am a little unsure as to how to prove something is an integer. 
I also noticed that letting $f(x) = \frac{x^5}5 + \frac{x^4}2 + \frac{x^3}3 - \frac x{30}$ and deriving $f(x)$ yields a fairly clean result, but I don't know if this helps me at all. Any help would be great.
 [/INST] First write over a common denominator:
$$\frac{n^5}5 + \frac{n^4}2 + \frac{n^3}3 - \frac n{30} = \frac{1}{30}(6n^5+15n^4+10n^3-n)$$
Thus we want to show that 30 always divides expression in parenthesis. Since $\gcd(5,6)=1$, it suffices to prove divisibility by 5 and 6 individually. My first attempt was to factor this polynomial (using a computer, as I'm lazy):
$$6n^5+15n^4+10n^3-n=n(n+1)(2n+1)(3n(n+1)-1).$$
Observe that $6\sum_{k=1}^n k^2=n(n+1)(2n+1)$. Hence the expression is divisible by 6. You can check that 5 does not divide the remaining factor $3n(n+1)-1$, so we need a different approach.
Instead, we reduce the original expression mod 5, that is,
$$6n^5+15n^4+10n^3-n \equiv n^5-n \;\text{mod}\;5.$$
The computer tells me that the expression on the right appears to be divisible by 5. Can you prove this by induction (or even easier Fermat's little theorem)?
</s>