I want to prove that
$$ n^3/3 + n^5/5 + 7n/15 $$
is an integer for every integer n >= 1.
I define P(n) to be: $$ n^3/3 + n^5/5 + 7n/15 $$ is an integer.
For my basis step, P(1) is true because $$ 1^3/3 + 1^5/5 + 7(1)/15 = 1 $$ 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, $$ k^3/3 + k^5/5 + 7k/15 $$ is an integer.
So based on that assumption, I need to now show that P(k+1) is true, i.e., that $$ (k+1)^3/3 + (k+1)^5/5 + 7(k+1)/15 $$ is an integer.
At this point, I am stuck as to where to go next...
I have tried rewriting the assumption:
$$ k^3/3 + k^5/5 + 7k/15 = 15m $$ for some integer m. Then I solve for m: $$ (k^3/3 + k^5/5 + 7k/15)/15 = m $$ 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:
$$ (5(k+1)^3 + 3(k+1)^5 + 7(k+1)) / 15 $$