Prove $k^7/7 + k^5/5 + 2k^3/3 - k/105$ is an integer I tried to prove this using induction.
Let $k=1$; then the equation gives
$$1/7 + 1/5 +2/3 – 1/105 = 105/105 = 1,$$
which is an integer. So it is true for $k=1$. Now let it be true for $n>k$. This gives
$$105|(15n^7 + 21n^5 + 70n^3 – n).$$
For $(n+1)$ we have
$$105|(15(n+1)^7 + 21(n+1)^5 + 70(n+1)^3 – (n+1)),$$
which is the same as
$$105|(15n^7 + 105n^6 + 336n^5 + 630n^4 + 875n^3 + 945n^2 + 629n + 175)$$
since $105|(15n^7 + 21n^5 + 70n^3 – n)$.
If $105|(105n^6 + 315n^5 + 630n^4 + 805n^3 + 945n^2 + 630n + 175),$
which is the same as if $$105|((105n^6 + 315n^5 + 630n^4 +945n^2 + 630n) + 805n^3 + 175),$$
which is the same as if $$105|((\text{multiple of } 105) + 805n^3 + 175).$$
I’m stuck at this part, because neither $805$ nor $175$ is a multiple of $105$, so how to prove that they are multiples of $105$?

I think you made a  mistake, and it should have been
$15n^7+105n^6+336n^5+630n^4+805n^3+735n^2+419n+105$
where you typed $15n^7 + 105n^6 + 336n^5 + 630n^4 + 875n^3 + 945n^2 + 629n + 175$,
so it should have been
$105n^6 + 315n^5 + 630n^4 + 735n^3 + 735n^2 + 420n + 105$
where you typed $105n^6 + 315n^5 + 630n^4 + 805n^3 + 945n^2 + 630n + 175$.
Can you take it from here?
