NT prac 1

$\begin{aligned} a_n &= (7-1)^n+(7+1)^n \\ &=\sum_{r=0}^n \binom{n}{r} 7^r \left( (-1)^{n-r}+1 \right) \\ &\equiv \sum_{r=0}^1 \binom{n}{r} 7^r \left( (-1)^{n-r}+1 \right) \pmod{49} \\ &\equiv (-1)^n+1+7n\left((-1)^{n+1}+1\right) \pmod{49}\end{aligned}$

So $a_{83}\equiv 7\cdot 83 \cdot 2 \equiv \boxed{35} \pmod{49}$

Similar solution with @Chris Lewis 's

$\begin{aligned} \blue{6^{83}} + \red{8^{83}} & = \blue{(7-1)^{83}} + \red{(7+1)^{83}} \\ & = \blue{7^{83} - 83 \cdot 7^{82} + \frac {83 \cdot 82}2 \cdot 7^{81} + \cdots - 1} + \red{7^{83} + 83 \cdot 7^{82} + \frac {83 \cdot 82}2 \cdot 7^{81} + \cdots + 1} \\ & = 2\left(7^{83} + \frac {83 \cdot 82}2 \cdot 7^{81} + \cdots + \frac {83\cdot 82 \cdot 81}6\cdot 7^3 + 83 \cdot 7 \right) \quad \quad \small \blue{\text{Twice the odd terms}} \end{aligned}$

Since $\dbinom {83}k 7^k \equiv 0 \text{ (mod 83)}$ for $k \ge 2$ , we have $6^{83} + 8^{83} \equiv 2 \cdot 83 \cdot 7 \equiv \boxed{35} \text{ (mod 83)}$ .