Problem: Let $a_n=6^n+8^n$. Determine the remainder on dividing $a_{83}$ by 49 . Solution: Check this out. Write $a_n = (7-1)^n+(7+1)^n$. Via a binomial expansion, we see that $(7-1)^{83} = \binom{83}{0}7^{83}+\binom{83}{1}7^{82}- \dots +\binom{83}{82}7-\binom{83}{83} \equiv 83\cdot 7-1 \pmod {7^2}$ and $(7+1)^{83} = \binom{83}{0}7^{83}+\binom{83}{1}7^{82}+ \dots +\binom{83}{82}7+\binom{83}{83} \equiv 1+83 \cdot 7$ so together, we have that $a_{83} \equiv 14\cdot 83 \pmod {7^2}$ so $a_{83} \equiv 35 \pmod {7^2}$.