2023/022) Show that $\frac{n^7}{7} + \frac{n^{13}}{13} + \frac{71n}{91}$ is an integer for any integer n

 We have $\frac{n^7}{7} + \frac{n^{13}}{13} + \frac{71n}{91}$

$=\frac{n^7-n +n }{7} + \frac{n^{13}-n + n}{13} + \frac{71n}{91}$ 

$=\frac{n^7-n} {7} + \frac{n^{13}-n}{13} + \frac{n} {7}+ \frac{n}{13} + \frac{71n}{91}$ 

$=\frac{n(n^6-1)}{7} + \frac{n(n^{12}-n) }{13} + n$

If n is divisble by 7 1st term is integer and if n is not divisibl by 7 $n^6-1$ is divsible by 7 as per Fermats Little Theorem is 1st term is integer similarly the 2nd term   and 3rd term is integer for any integer n and hence the given expression