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 $105$?

The generating function is $$\sum_{k\ge 0} f(k)x^k=\frac{320}{(x-1)^3}+\frac{3672}{(x-1)^5}+\frac{1540}{(x-1)^4}+\frac{4584}{(x-1)^6}+\frac{1}{(x-1)}+\frac{2880}{(x-1)^7}+\frac{720}{(x-1)^8}+\frac{29}{(x-1)^2}$$ with $$f(k)=\frac12 k^7+\frac15 k^5+\frac23 k^3-\frac{1}{105}k$$. Since all the factors of this polynomial of $1/(x-1)$ are integers, all $f(k)$ are integers.
Another method of proof is that the polynomial has the recurrence $$f(k) = 8f(k-1)-28f(k-2)+56f(k-3)-70f(k-4)+56f(k-5)-28f(k-6)+8f(k-7)-f(k-8)$$ and to start an induction on $k$ by showing that the first 8 terms of $f(k)$ are integers. [Of course the recurrence is valid for any polynomial $f(k)$ of order 7 and the signed binomial coefficients are a well-known property of the recurrence.]
