Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$ Show that $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{63}$
According to Fermat's theorem:
$$1^7+7^7+13^7+19^7+23^7\equiv{1+7+13+19+23}\pmod{7}\equiv{63}\pmod{7}\equiv{0}\pmod{7}$$
Now we need to show: $1^7+7^7+13^7+19^7+23^7\equiv{0}\pmod{9}$ , but how??

Just to be a bit different, note that a binomial expansion tells us
$$7^7=(1+6)^7=1+7\cdot6+(\text{stuff})\cdot6^2\equiv43\equiv-2\mod 9$$
since $6^2\equiv0$ mod $9$, and thus, since $13\equiv4$, $19\equiv1$, and $23\equiv-4$ mod $9$, we have
$$1^7+7^7+13^7+19^7+23^7\equiv1^7-2+4^7+1^7+(-4)^7=1-2+4^7+1-4^7=0\mod 9$$
