Prove that $\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$ is an integer. Prove that $$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}$$ is an integer using mathematical induction.
I tried using mathematical induction but using binomial formula also it becomes little bit complicated.
Please show me your proof.
Sorry if this question was already asked. Actually i did not found it. In that case only sharing the link will be enough.

We have:
$$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}
=\frac{15k^7+21k^5+70k^3-k}{3\cdot 5\cdot 7}
$$
To prove this is an integer we need that:
$$15k^7+21k^5+70k^3-k\equiv 0 \pmod{3\cdot 5\cdot 7}$$
According to the Chinese Remainder Theorem, this is the case iff
$$\begin{cases}15k^7+21k^5+70k^3-k\equiv 0 \pmod{3} \\
15k^7+21k^5+70k^3-k\equiv 0 \pmod{5}\\
15k^7+21k^5+70k^3-k\equiv 0 \pmod{7}\end{cases} \iff
\begin{cases}k^3-k\equiv 0 \pmod{3} \\
k^5-k\equiv 0 \pmod{5}\\
k^7-k\equiv 0 \pmod{7}\end{cases}$$
Fermat's Little Theorem says that $k^p\equiv k \pmod{p}$ for any prime $p$ and integer $k$.
Therefore the original expression is an integer.
