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:
$$\frac{1}{15} \left(5 (k+1)^3+3 (k+1)^5+7 (k+1)\right)$$
 A: 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.
A: This problem becomes trivial when thought of in terms of division.  Realize that the common denominator of the fractions is 15.  Thus, all we need to show is that $$15 \mid 3n^5+5n^3+7n\;\;\forall n$$  Since $15=3 \cdot 5$, all we have to show is that both 3 and 5 will divide this for all $n$.  To make this easy, we simply show the equation is 0 modulo 3 and modulo 5 for all $n$.
Note that mod 3, we are left with $2n^3+n$ and need only consider $n=0,1,2$.  This checks out easily.  
Similarly, mod 5, we are left with $3n^5+2n$ and must consider $n=0,1,2,3,4$.  Again, this checks out.  
Since $\gcd(3,5)=1$ it follows that this equation will be divisible by their product, 15, for all $n$ as well.
A: 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.
A: Expand the binomials and group the resulting fractions such that you get integers.
In other words, you know that
$$(k + 1)^3 = k^3 + 3k^2 + 3k + 1$$
and
$$(k + 1)^5 = k^5 + 5k^4 + 10k^3 + 10k^2 + 5k + 1$$
That should do the trick.
A: 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.
