Proving that $f(n)$ is an integer using mathematical induction I want to prove that

$$\frac{n^3}{3}+\frac{n^5}{5}+\frac{7 n}{15}$$

is an integer for every integer $n \geq 1$.

I define P(n) to be: 

$$\frac{n^3}{3}+\frac{n^5}{5}+\frac{7 n}{15}$$ is an integer.

For my basis step, P(1) is true because

$$\frac{1^3}{3}+\frac{1^5}{5}+\frac{7}{15}=1$$
which is an integer.

The inductive step is what's tripping me up...

Let k be an arbitrary positive integer.  Assume that P(k) is true, that is,

$$\frac{k^3}{3}+\frac{k^5}{5}+\frac{7 k}{15}$$
is an integer.

So based on that assumption, I need to now show that P(k+1) is true, i.e., that

$$\frac{(k+1)^3}{3} +\frac{(k+1)^5}{5} +\frac{7 (k+1)}{15}$$
is an integer.

At this point, I am stuck as to where to go next...

I have tried rewriting the assumption:

$$\frac{k^3}{3}+\frac{k^5}{5}+\frac{7 k}{15}=15 m$$
for some integer m.  Then I solve for m:

$$\frac{1}{15} \left(\frac{k^3}{3}+\frac{k^5}{5}+\frac{7 k}{15}\right)=m$$

But this looks like a dead-end, seems there's nothing I can do with this to the "to prove" equation.

I have also tried re-writing the "to show" equation as this, but I get a dead end there and am not sure where to go next:

$$\frac{1}{15} \left(5 (k+1)^3+3 (k+1)^5+7 (k+1)\right)$$
Why do you think that $P(k) = 15m$ for some integer $m$ if it does not hold for, say $k=1$? If you assume that $P(k)$ is integer then the strategy is to show that 
$$
P(k+1) - P(k) \in\mathbb Z
$$
and let us do it:
$$
P(k+1) - P(k) = \frac{1}{5}((n+1)^5-n^5)+\frac13((n+1)^3 - n^3)+\frac7{15} = 
$$
$$
= \frac15(5n^4+10n^3+10n^2+5n+1) +\frac13(3n^2+3n+1)+\frac7{15}
$$
$$
= n^4+2n^3+2n^2+n +\frac15+n^2+n+\frac13+\frac{7}{15}
$$
$$
= n^4+2n^3+3n^2+2n+1
$$
$$
=(n^2+n+1)^2\in \mathbb Z
$$
and you're done.
You are needed to use the fact that integer is another integer.
Now put $k$ and consider the result true then go for $k-1$. From step in which you put $k$ you get an equation that give an $\frac{k^3}{3}+\frac{k^5}{5}+\frac{7k}{15}=p$ (an integer). Using it and fact stated in line $1$ solves the problem.