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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

 Jun 11, 2020
 #1
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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 ^_^

 Jun 11, 2020

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