Question

prove k^7/7 + k^5/5 + 2k^3/3 -k/105 is an integer for every positive integer k

Answer 1

Recall from little fermat we have $p\mid(a^{p}-a)$ for all integers $a$ and prime $p$

$\frac{k^7}{7}+\frac{k^5}{5}+\frac{2k^3}{3}-\frac{k}{105}=\frac{15k^7+21k^5+70k^3-k}{105}$

$105=3\cdot5\cdot7$, so it is sufficient to show that $15k^7+21k^5+70k^3-k$ is divisible by $3,5,7$

$15k^7+21k^5+70k^3-k\equiv 0+0+k^3-k\equiv k^3-k\equiv{0}\pmod{3}$ 
$15k^7+21k^5+70k^3-k\equiv 0+k^5+0-k\equiv 0\equiv k^5-k\equiv 0\pmod{5}$ 
$15k^7+21k^5+70k^3-k\equiv k^7+0+0-k\equiv 0\equiv k^7-k\equiv 0\pmod{7}$ 

Therefore $15k^7+21k^5+70k^3-k$ is divisible by $3,5,7$ and consequently the expression main question is an integer for all value sof $k$