Show that 5^e +6^e is congruent 0 (mod 11) for all odd numbers e
Prove that 6(4^n) is congruent 6(mod 9) for any n>=0
for $\displaystyle n=0$, we have $\displaystyle 6(4^0) \equiv 6\mod 9$
for for $\displaystyle n=1$, we have $\displaystyle 6(4^1) \equiv 24\mod 9 \equiv 6\mod 9$
for suppose it is true that for $\displaystyle n=k$, $\displaystyle 6(4^k) \equiv 6\mod 9$..
then, $\displaystyle 6(4^{k+1}) = 6(4^k)(4) \equiv (6)(4)\mod 9 \equiv 24\mod 9 \equiv 6\mod 9$.. QED.