<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] Use induction for $n\in \mathbb Z^+$. 
Base case:
$$P(1)=\frac{1}{15}(3\cdot 1^5+5\cdot 1^3+7\cdot 1)=1.$$
Inductive hypothesis:
$$P(n)=\frac{1}{15}(3n^5+5n^3+7n) \in \mathbb Z^+.$$
Inductive step:
$$P(n+1)=\frac{1}{15}(3(n+1)^5+5(n+1)^3+7(n+1))=\\
=\underbrace{\frac{1}{15}(3n^5+5n^3+7n)}_{P(n)\in \mathbb Z^+}+\\
\frac{1}{15}(15(n^4+2n^3+2n^2+n)+15(n^2+n)+15) \in \mathbb Z^+.$$
Since the powers of $n$ in $P(n)$ are odd, it will be true for negative integers as well.
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