+14 Prove that \(5^{3^{n}} + 1\) is divisible by \(3^{n + 1}\) for all nonnegative integers \(n\).
I've done part of this problem but can't seem to complete it. If you want to see what I've done already just ask
+9673 For the base case: n = 0.
\(\text{When }n = 0,\\ 5^{3^{0}} + 1 = 6\\ 3^{0 + 1} = 3\\ 3|6\)
Assume \(5^{3^k} + 1\text{ is divisible }3^{k + 1}\).
\(5^{3^{k + 1}} + 1 = 5^{3(3^k)} + 1 = (5^{3^k} + 1)\left(5^{2(3^k)} - 5^{3^k} + 1\right)\)
Now, \(5^{2(3^k)} \equiv 2^{2(3^k)} \equiv 4^{3^k} \equiv 1^{3^k} \equiv 1 \pmod 3\)
\(5^{3^k}\equiv 2^{3^k}\equiv 2^{2K + 1}\equiv 2 \pmod 3\) for some \(K\in \mathbb Z\)
Therefore \(3|\left(5^{2(3^k)} - 5^{3^k} + 1\right)\)
The proposition is proved.
P.S.: I cheated and used some modular arithmetic ^_^
