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

Here is a proof not by induction, but which may be instructive.
Using repeated differences on the first few values of $P(n)$ we get
$$
\begin{array}{llll}
0 & 1 & 10 & 59 & 228 & 669 & 1630 & \\
1 & 9 & 49 & 169 & 441 & 961 & \\
8 & 40 & 120 & 272 & 520 & \\
32 & 80 & 152 & 248 & \\
48 & 72 & 96 & \\
24 & 24 & \\
0 & \\
\end{array}
$$
Newton's interpolation formula then gives us
$$
P(n)=
0 \binom{n}{0} + 1 \binom{n}{1} + 8 \binom{n}{2} + 32 \binom{n}{3} + 48 \binom{n}{4} + 24 \binom{n}{5} 
$$
This is clearly an integer for all $n\ge 0$.
In general, if a polynomial of degree $d$ and with rational coefficients takes integer values for $d+1$ consecutive integers, then it takes integers values for all integer arguments because all repeated differences are integers and so are the coefficients in Newton's interpolation formula.
