<s>[INST] $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for every integer n. Q: Prove that $\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ is an integer for every integer n.
In fact, I already have a method, but it seems too long. So I would first (1)is there any problem with my logic? (2) Is there any better/faster method to do these kinds of questions?
My solution:
$$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$$
$$=\frac{1}{15}(3n^5+5n^3+7n)$$
Then my next time is to use M.I to prove that $(3n^5+5n^3+7n)$ is divisible by $15$ for positive integers. And for negative integer, I switch it into -$(3n^5+5n^3+7n)$ and it also satisfies the divisibility of $15$.
But the above seems too clumsy. This course is "Intro to number theory". Is there any beter method to do it?
 [/INST] Modulo $3$, your numerator is:
$$3n^5+5n^3+7n \equiv -n(n^2-1) $$
$$\equiv -n(n-1)(n+1)\pmod{3}.$$
The last expression is a product of three consecutive integers and hence divisible by $3$.
Modulo $5$, you have
$$3n^5+5n^3+7n \equiv -2n^5+2n $$  $$\equiv -2n(n^4-1) \pmod{5}.$$
If $n$ is divisible by $5$, then so is the last expression.  If $n$ is not divisible by $5$, then $n^4 \equiv 1 \pmod{5}$ and still the last expression is divisible by $5$.  
Since the numerator is divisible by both $3$ and $5$, it's divisible by $15.$
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