$\frac{1}{5}n^5+\frac{1}{3}n^3+\frac{7}{15}n$ 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?
 A: 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 $15.$
A: You can use Little Fermat in the form $n^p\equiv n\bmod p$ ($p$ prime, which works for all $n$ as opposed to $n^{p-1}\equiv 1 \bmod p$ for $n$ not divisible by $p$)
For the prime $5$, we have $n^5\equiv n \bmod 5$ so you get $$3n^5+5n^3+7n\equiv 3n+7n\equiv 0\bmod 5$$
For the prime $3$ we have $n^3\equiv n\bmod 3$ so that $$3n^5+5n^3+7n\equiv 5n+7n\equiv 0 \bmod 3$$
Combine these and you have your answer.
A: For a degree $n$ polynomial $f$ to be "integer-valued", all you need
to check is that $f(a)\in\Bbb Z$ for $n+1$ consecutive integers $a$. This is because
$$\sum_{j=0}^{n+1}(-1)^j\binom{n+1}jf(x+j)=0\tag{*}$$
and if you have $n+1$ consecutive integer values for $f$, then upwards
and downwards induction gives $f(a)\in\Bbb Z$ for all $a\in\Bbb Z$.
In your example, $f(0)=0$, $f(\pm1)=1$, $f(\pm2)=\pm10$
and $f(\pm3)=\pm 59$ which is more than enough.
A: Hint: Your term is equal $$\frac{n}{15}(3n^4+5n^2+7)$$
and now Show that that $$15|n(3n^4+5n^2+7)$$ for all $n$
A: 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.
